University Press of Florida
Concepts in Calculus II
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Title: Concepts in Calculus II
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Language: en-US
Creator: Bona, Miklos
Shabanov, Sergei
The University of Florida Department of Mathematics
Publisher: University Press of Florida
Place of Publication: Gainesville, FL
Publication Date: 12/3/2011
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Subjects / Keywords: Calculus II
MAC 222
MAC 231
Equations (Mathematics)
Mathematics
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Abstract: From the University of Florida Department of Mathematics, this is the second volume in a three volume presentation of calculus from a concepts perspective. The emphasis is on learning the concepts behind the theories, not the rote completion of problems. WebAssign online ancillaries are available. Most questions from this textbook are available in WebAssign. The online questions are identical to the textbook questions except for minor wording changes necessary for Web use. Whenever possible, variables, numbers, or words have been randomized so that each student receives a unique version of the question. This list is updated nightly. Downloads are freely available (no cost). CLICK THE SHOPPING CART BUTTON TO PURCHASE A LOW COST PRINTED TEXTBOOK ONLINE OR YOU MAY ALSO PHONE THE UNIVERSITY PRESS OF FLORIDA TO ORDER A TEXTBOOK ON THEIR TOLL FREE NUMBER: 800-226-3822.
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ConceptsinCalculusIIBetaVersion UNIVERSITYPRESSOFFLORIDAFloridaA&MUniversity,Tallahassee FloridaAtlanticUniversity,BocaRaton FloridaGulfCoastUniversity,Ft.Myers FloridaInternationalUniversity,Miami FloridaStateUniversity,Tallahassee NewCollegeofFlorida,Sarasota UniversityofCentralFlorida,Orlando UniversityofFlorida,Gainesville UniversityofNorthFlorida,Jacksonville UniversityofSouthFlorida,Tampa UniversityofWestFlorida,Pensacola OrangeGroveTexts Plus

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ConceptsinCalculusIIBetaVersion Mikl osB onaandSergeiShabanov UniversityofFloridaDepartmentof MathematicsU niversity P ressof F lorida Gainesville € Tallahassee € Tampa € BocaRaton Pensacola € Orlando € Miami € Jacksonville € Ft.Myers € Sarasota

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Copyright2012bytheUniversityofFloridaBoardofTrusteesonbehalfoftheUniversityof FloridaDepartmentofMathematics Thisworkislicensedunderamodi“edCreativeCommonsAttribution-Noncommercial-No DerivativeWorks3.0UnportedLicense.Toviewacopyofthislicense,visithttp:// creativecommons.org/licenses/by-nc-nd/3.0/.Youarefreetoelectronicallycopy,distribute,and transmitthisworkifyouattributeauthorship. However,allprintingrightsarereservedbythe UniversityPressofFlorida(http://www.upf.com).PleasecontactUPFforinformationabout howtoobtaincopiesoftheworkforprintdistribution .Youmustattributetheworkinthe mannerspeci“edbytheauthororlicensor(butnotinanywaythatsuggeststhattheyendorse youoryouruseofthework).Foranyreuseordistribution,youmustmakecleartoothersthe licensetermsofthiswork.Anyoftheaboveconditionscanbewaivedifyougetpermissionfrom theUniversityPressofFlorida.Nothinginthislicenseimpairsorrestrictstheauthorsmoral rights. ISBN978-1-61610-156-5 OrangeGroveTexts Plus isanimprintoftheUniversityPressofFlorida,whichisthescholarly publishingagencyfortheStateUniversitySystemofFlorida,comprisingFloridaA&M University,FloridaAtlanticUniversity,FloridaGulfCoastUniversity,FloridaInternational University,FloridaStateUniversity,NewCollegeofFlorida,UniversityofCentralFlorida, UniversityofFlorida,UniversityofNorthFlorida,UniversityofSouthFlorida,andUniversityof WestFlorida. UniversityPressofFlorida 15Northwest15thStreet Gainesville,FL32611-2079 http://www.upf.com

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Contents Chapter6.ApplicationsofIntegration136.TheAreaBetweenCurves137.Volumes738.CylindricalShells1639.WorkandHydrostaticForce2140.AverageValueofaFunction25Chapter7.MethodsofIntegration2941.IntegrationbyParts2942.TrigonometricIntegrals3243.TrigonometricSubstitution3644.IntegratingRationalFunctions4045.StrategyofIntegration4546.IntegrationUsingTablesandSoftwarePackages4847.ApproximateIntegration5248.ImproperIntegrals59Chapter8.SequencesandSeries6949.In“niteSequences6950.SpecialSequences7651.Series8252.SeriesofNonnegativeTerms8753.ComparisonTests9354.AlternatingSeries9755.RatioandRootTests10256.Rearrangements10957.Power Series11558.RepresentationofFunctionsas Power Series12059.TaylorSeries126Chapter9.FurtherApplicationsofIntegration13560.ArcLength13561.SurfaceArea139 Chapterandsectionnumberingcontinuesfromthepreviousvolumeintheseries, ConceptsinCalculusI .

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viCONTENTS 62.ApplicationstoPhysicsandEngineering14563.ApplicationstoEconomicsandtheLifeSciences15164.Probability156Chapter10.PlanarCurves16565.ParametricCurves16566.CalculuswithParametricCurves17267.PolarCoordinates17868.ParametricCurves:TheArcLengthandSurfaceArea18569.AreasandArcLengthsinPolarCoordinates19170.ConicSections196

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CHAPTER6 ApplicationsofIntegration 36.TheAreaBetweenCurves36.1.TheBasicProblem.Inthepreviouschapter,welearnedthatif thefunction f satis“es f ( x ) 0forallrealnumbers x intheinterval [ a,b ],thentheareaofthedomainwhosebordersarethegraphof f thehorizontalaxis,andtheverticallines x = a and x = b isequalto b af ( x ) dx .Ifthereisnodangerofconfusionastowhat a and b are, thenthisfactissometimesinformallyexpressedbythesentencethe integralof f isequaltotheareaofthedomainthatis under thegraph of f .Ž Whatcanwesayabouttheareaofthedomain betweentwocurves ? Thereareseveralwaystoaskthisquestion.Theeasiestversion,discussedbythefollowingtheorem,diersfromtheprevioussituation onlyinthatthehorizontallineisreplacedbyanotherfunction g Theorem 6.1 Let f and g betwofunctionssuchthat,forallreal numbers x [ a,b ] ,theinequality f ( x ) g ( x ) holds.Thenthedomain whosebordersarethegraphof f ,thegraphof g ,andtheverticallines x = a and x = b hasarea A = b a( f ( x ) Š g ( x )) dx. SeeFigure6.1foranillustrationofthecontentofTheorem6.1. Thereaderisinvitedtoexplainwhythistheoremisadirectconsequenceofthefactthatwerecalledinthe“rstparagraphofthissection. Thereaderisalsoinvitedtoexplainwhythetheoremholdsevenif f and g takenegativevalues. Example 6.1 Computethearea A ( D ) ofthedomain D whose bordersarethegraphofthefunction f ( x )= x3+1 ,thefunction g ( x )= x2+2 ,andtheverticallines x =2 and x =3 .SeeFigure6.2foran illustrationofthisspeci“cexample. Solution: InordertoseethatTheorem6.1isapplicable,wemust “rstshowthat,forall x [2 3],theinequality f ( x ) g ( x )holds.1

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26.APPLICATIONSOFINTEGRATION Figure6.1. Areaenclosedby f ( x )and g ( x )between x = a and x = b Figure6.2. Areaenclosedbythegraphsof f ( x )= x3+ 1and g ( x )= x2+2between x =2and x =3. Thisisnotdicult,sinceweonlyneedtoshowthatif x [2 3],then f ( x ) g ( x ),thatis, x3+1 x2+2 x3Š x2 1 x2( x Š 1) 1 andthisisclearlytruesince x 2,so x2 4,and x Š 1 1,forcing x2( x Š 1) 4.

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36.THEAREABETWEENCURVES3 Therefore,Theorem6.1applies,andwehave A ( D )= 3 2( x3+1) Š ( x2+2) dx = 3 2( x3Š x2Š 1) dx = x4 4 Š x3 3 Š x 3 2=8 11 12 36.2.IntersectingCurves.Sometimes,thepoints a and b aredeterminedbythecurvesthemselves,andnotgiveninadvance.Inthat case,wehavetocomputethembeforewecanapplyTheorem6.1. Example 6.2 Findthearea A ( D ) ofthedomain D whoseborders arethegraphsofthefunctions f ( x )= x2+3 x +5 and g ( x )=2 x2+ 7 x +8 .SeeFigure6.3foranillustration. Solution: Letus“ndthepointsinwhichthegraphsof f and g intersect.Inthesepoints,wehave x2+3 x +5=2 x2+7 x +8 0= x2+4 x +3 0=( x +3)( x +1) Thatis,thetwocurvesintersectintwopoints,andthesepointshave horizontalcoordinates a = Š 3and b = Š 1.Furthermore,if x Figure6.3. Areaenclosedbythegraphsof f ( x )and g ( x )between x = Š 3and x = Š 1.

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46.APPLICATIONSOFINTEGRATION [ Š 3 Š 1],thatis,if x isbetweenthosetwointersectionpoints,then f ( x ) Š g ( x )=( x2+3 x +5) Š (2 x2+7 x +8) = Š ( x2+4 x +3) = Š ( x +3)( x +1) 0 since x +3 0and x +1 0.Therefore,if x [ Š 3 Š 1],then f ( x ) g ( x ),andTheorem6.1applies.Sowehave A ( D )= Š 1 Š 3( f ( x ) Š g ( x )) dx = Š ( x2+4 x +3) dx = Š x3 3 Š 2 x2Š 3 x Š 1 Š 3=1 1 3 Thesituationbecomesslightlymorecomplicatedif f g doesnot holdthroughouttheentireinterval[ a,b ].Forinstance,itcouldhappen that f ( x ) g ( x )atthebeginningoftheinterval[ a,b ],andthen,froma givenpointon, g ( x ) f ( x ).Inthatcase,wesplit[ a,b ]upintosmaller intervalssothatoneachofthesesmallerintervals,either f ( x ) g ( x ) or g ( x ) f ( x )holds.ThenwecanapplyTheorem6.1toeachof theseintervals.Asonsomeoftheseintervals f ( x ) g ( x )holds,while onsomeothers g ( x ) Š f ( x )holds,theapplicationofTheorem6.1will sometimesinvolvethecomputationof ( f ( x ) Š g ( x )) dx andsometimes ( g ( x ) Š f ( x )) dx.Thefollowingtheoremformalizesthisidea. Theorem 6.2 Let f and g betwofunctions.Thentheareaofthe domainwhosebordersarethegraphof f ,thegraphof g ,thevertical line x = a andtheverticalline x = b isequalto b a| f ( x ) Š g ( x ) | dx. NotethatTheorem6.1isaspecialcaseofTheorem6.2,namely, thespecialcasewhen f ( x ) Š g ( x )= | f ( x ) Š g ( x ) | forall x [ a,b ]. Example 6.3 Let f ( x )= x3+3 x2+2 x andlet g ( x )= x3+ x2. Computethearea A ( D ) ofthedomainwhosebordersarethegraphsof f and g andtheverticallines x = Š 2 and x =1 .SeeFigure6.4for anillustration.

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36.THEAREABETWEENCURVES5 Figure6.4. Graphsof f ( x )and g ( x )on[ Š 2 1]. Solution: InordertouseTheorem6.2,weneedtocompute | f ( x ) Š g ( x ) | .Wehave f ( x ) Š g ( x )=( x3+3 x2+2 x ) Š ( x3+ x2) 2 x2+2 x =0 2 x ( x +1)=0 Thatis,thereareonlytwopointswherethesetwocurvesintersect, namely,at x = Š 1and x =0.If x Š 1orif x 0,then f ( x ) Š g ( x )= 2 x ( x +1) > 0,so | f ( x ) Š g ( x ) | = f ( x ) Š g ( x )=2 x2+2 x .If Š 1
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66.APPLICATIONSOFINTEGRATION Figure6.5. Graphof | f ( x ) Š g ( x ) | on[ Š 2 1]. 36.3.CurvesFailingtheVerticalLineTest.Sometimeswewanttocomputetheareabetweentwocurvesthatdonotpasstheverticalline test;thatis,theycontaintwoormorepointsonthesameverticalline. Suchcurvesare not graphsoffunctionsofthevariable x .Iftheypass the horizontallinetest ,thatis,iftheydonotcontaintwopointson thesamehorizontalline,thentheycanbeviewedasfunctionsof y Wecanthenchangetherolesof x and y inTheorems6.1and6.2and proceedasintheearlierexamplesofthissection. Example 6.4 Computethearea A ( D ) ofthedomainbetweenthe verticalline x =4 andthecurvegivenbytheequation y2= x .See Figure6.6foranillustration. Solution: Neithercurvesatis“estheverticallinetest,butbothsatisfy thehorizontallinetest.Therefore,weset f ( y )=4and g ( y )= y2.Itis Figure6.6. Graphof y2= x and x =4.

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37.VOLUMES7 clearthatthetwocurvesintersectatthepointsgivenby y = Š 2and y =2.Betweenthesetwocurves,thevalueof f ( y )islarger.Therefore, Theorem6.1applies(withtherolesof x and y reversed).Sowehave A ( D )= 2 Š 2( f ( x ) Š g ( x )) dx = 2 Š 24 Š y2dy = 4 y Š y3 3 2 Š 2=16 Š 16 3 =10 2 3 Notethatthegeometricmeaningofreversingtherolesof x and y issimplyre”ectingallcurvesthroughthe x = y line.Thatre”ection doesnotchangetheareaofanydomain,soonecanexpectanalogous methodsofcomputingareasbeforeandafterthatre”ection.36.4.Exercises.(1)Findtheareaofthedomainwhosebordersaretheverticalline x =0,theverticalline x =2,andthegraphsofthefunctions f ( x )= x2+3and g ( x )=sin x (2)Findtheareaofthedomainwhosebordersaretheverticalline x =1,theverticalline x =3,andthegraphsofthefunctions f ( x )= x3and g ( x )= eŠ x. (3)Findtheareaofthedomainbetweenthegraphsofthefunctions f ( x )= x2+2and g ( x )=4 x Š 1. (4)Findtheareaofthedomainbetweenthegraphsofthefunctions f ( x )= x3Š 3 x2and g ( x )= x2. (5)Findtheareabetweenthecurvesgivenbytheequations x = 5 y and x = y2+6. (6)Computetheareabetweenthe three curves f ( x )= x g ( x )= Š x ,and h ( x )=4.37.Volumes37.1.ExtendingtheDe“nitionofVolumes.Ifasolid S canbebuiltup usingunitcubes,thenwecansimplysaythatthevolume V ( S )of S is thenumberofunitcubesusedtobuild S .However,ifthebordersof

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86.APPLICATIONSOFINTEGRATION S arenotplanes,thenthismethodwillhavetobemodi“ed.Aballor aconeisanexampleofthis. Sowewouldliketode“nethenotionofvolumesothatitisapplicabletoalargeclassofsolids,notjusttothosesolidsthatarebordered byplanes.Thisde“nitionshouldagreewithourintuition.Itshould alsobeinaccordancewiththefactthatwecan approximate allsolids withacollectionofverysmallcubes;therefore, V ( S )mustbecloseto thenumberofunitcubesusedintheapproximation. Withthesegoalsinmind,werecallthatwealreadyde“nedthe area ofadomainintheplanewhosebordersarethegraphsofcontinuous functions.Buildingonthatde“nition,wesaythatthe volume ofa prism isitsbaseareatimesitsheight.Moreformally,let S beasolid whosebaseandcoverareidenticalcopiesoftheplate P ,locatedat distance h fromeachother,ontwoparallelplanesthatareatdistance h fromeachother.Thenwede“nethevolumeof S tobe V ( S )= A ( P ) h, where A ( P )istheareaof P .SeeFigure6.7foranillustration.In particular,thevolumeofacylinderwhosebaseisacircleofradius r andwhoseheightis h is r2h Nowlet S beanysolidlocatedbetweentheplanesgivenbythe equations x = a and x = b .Inordertode“neandcomputethevolume V ( S )of S ,wecut S into n partsbytheplanes x = xifor i =0 1 ,...,n where a = x0
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37.VOLUMES9 plateof Piareontheplanes x = xi Š 1and x = xi,respectively,andthe baseandcoverplatesof Piarecongruenttotheintersection Tiof S and theplane x = x iforsomepoint x i [ xi Š 1,xi].Wecanassumethat x iisthemidpointof[ xi Š 1,xi],butthatwillturnouttobeinsigni“cant. Notethat Pihasvolume A ( Ti) x If n islarge,thentheunionoftheprisms Piapproximates S well, so V ( S )shouldbede“nedinawaythatassuresthat V ( S )iscloseto (6.1)ni =1V ( Pi)=ni =1A ( Ti) x. As n goestoin“nity,theRiemannsumontheright-handsideof (6.1)hasalimit.Wede“ne thatlimit tobethevolume V ( S )of S ,so V ( S )=limn ni =1A ( Ti) x. Ontheotherhand,bythede“nitionofthede“niteintegral,wehave limn ni =1A ( Ti) x = b aA ( t ) dt, where A ( t )istheareaoftheintersectionof S andtheplane x = t Therefore,thisintegralisequaltothevolume V ( S )ofthesolid S Thatprovesthefollowingtheorem,whichwillbeourmaintoolinthis section. Theorem 6.3 Let S beasolidlocatedbetweentheplanes x = a and x = b ,andlet A ( t ) betheareaoftheintersectionof S andthe plane x = t .Thenthevolume V ( S ) of S satis“estheequation V ( S )= b aA ( t ) dt. Example 6.5 Computethevolumeoftheball B whosecenteris attheoriginandwhoseradiusis r Solution: Foranygiven t [ Š r,r ],theintersectionoftheplane x = t andtheball B isacircle.Let Ctdenotethiscircle.Bythe Pythagoreantheorem, Cthasradius r2Š t2,andtherefore,thearea of Ctis A ( t )=( r2Š t2) .SeeFigure6.8foranillustration.

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106.APPLICATIONSOFINTEGRATION Figure6.8. Thevolumeoftheball B approximatedbycylinders. NowwecanuseTheorem6.3tocomputethevolumeof B .Weget V ( B )= r Š r r2 Š t2 dt = r2t Š 1 3 t3r Š r=2 r3 Š 2 3 r3 = 4 3 r3. Thereisnothingmagicalaboutthe x axisasfarasTheorem6.3is concerned.Theargumentthatyieldedthattheoremcanberepeated forthe y axisinsteadofthe x axis,yieldingthefollowingtheorem. Theorem 6.4 Let S beasolidlocatedbetweentheplanes y = a and y = b ,andlet B ( t ) betheareaoftheintersectionof S andthe plane y = t .Thenthevolume V ( S ) of S satis“estheequation V ( S )= b aB ( t ) dt. Example 6.6 Let S betherightcircularconewhosesymmetry axisisthe y axis,whoseapexisat y = h ,andwhosebaseisacircle intheplane y =0 withitscenterattheoriginandwithradius r .Find thevolumeof S Solution: Thecone S isbetweentheplanes y =0and y = h ,and B ( t )ofTheorem6.4iseasiertocomputethan A ( t )ofTheorem6.3,so weusetheformer.

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37.VOLUMES11 Theintersectionoftheplane y = t and S isacircle.Theradius rtofthiscircle,bysimilartriangles,satis“es rt r = h Š t h showingthat rt= r ( h Š t ) /h .Therefore, B ( t )= r2( h Š t )2/h2,and Theorem6.4implies V ( S )= h 0( r2( h Š t )2/h2) dt = r2 h2h 0( h2Š 2 ht + t2) dt = r2 h2 h2t Š ht2+ t3 3 h 0= r2 h2 h3 3 = 1 3 hr2. SeeFigure6.9foranillustration. 37.2.AnnularRings.Intheexamplesthatwehavesolvedsofar,the computationof A ( t )or B ( t ),thatis,thecomputationoftheareaof theintersectionbetweenasolidandahorizontalorverticalplane,was notdicult.Thatcomputationcouldbedonedirectly. Therearesituationsinwhichthedomainswhoseareasweneedto computearenot convex ;thatis,visuallyspeaking,thereisaholein Figure6.9. Rightcircularcone.

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126.APPLICATIONSOFINTEGRATION them.Thishappensparticularlyoftenwhen S isobtainedbyrotating adomain D aroundaline. Example 6.7 Let D bethedomainbetweenthetwocurves y = x2= f ( x ) and y =2 x = g ( x ) andlet S bethesolidobtainedby rotating D abouttheline x = Š 3 .Findthevolumeof S Solution: Thetwocurvesintersectatthepoints(0 0)and(2 4).The intersectionofthehorizontalplane y = t with S hastheformofan annularring ,whichissometimesinformallycalleda washer .Thisis simplyasmallercirclecutoutothemiddleofalargercircle,sothat thetwocirclesareconcentric.Ifthelargercirclehasradius r1andthe smallercirclehasradius r2,thentheannularringhasarea ( r2 1Š r2 2). Thisgeneralrecipeenablesustocompute B ( t )intheexampleat hand.Thepointsin D satisfy x [0 2]and y [0 4].As0 x 2, theinequality x2 2 x holds.So,for t [0 4],thepoint Pi=( t/ 2 ,t ) onthegraphof g ( x )= y =2 x isclosertothe y axisthanthepoint Po=( t,t )onthegraphof y = x2= f ( x ).(Ittakesalargervalue of x togetthesamevalue t = y by f thanittakestogetthesame valueby g .)Sotheoutercircleoftheannuluswillbegivenbythe rotatedimageofthecurveof f (theparabola),andtheinnercircleof theannuluswillbegivenbytherotatedimageofthecurveof g (the straightline). Inparticular,for“xed t ,theradiiareobtainedasthedistanceof Po(resp. Pi)fromtheaxisofrotation,thatis,theline x = Š 3.For theinnerradius,thisyields r2= Š 3 Š t 2 = t +6 2 whilefortheouterradius,thisyields r1= Š 3 Š t = t +3 Therefore,theareaoftheannularringthatistheintersectionof S andtheplane y = t isgivenby B ( t )= ( r2 1Š r2 2)= t +3 2Š t +6 2 2= 6 t0 5Š t2 4 Š 2 t .

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37.VOLUMES13 NowwecanapplyTheorem6.4tocompute V ( S ).Weobtain V ( S )= 4 0B ( t ) dt = 4 06 t0 5Š t2 4 Š 2 tdt = 4 t1 5Š t3 12 Š t2 dt 77 01 37.3.SpecialCasesofTheorems6.3and6.4.Notethatthesolidswe discussedsofarinthissectioncouldbeobtainedbyrotatingthegraph ofafunctionaroundanaxis.Suchvolumesarecalled volumesofrevolution .Indeed,theballofExample6.5canbeobtainedbyrotating thegraphofthefunction f ( x )= r2Š x2(asemicircle)aboutthe x axis.TheconeofExample6.6canbeobtainedbyrotatingthegraph ofthefunction f ( x )= Š xh r+ r = y (astraightline)aboutthe y axis. Forsuchsolids,theareas A ( t )and B ( t )appearinginTheorems6.3 and6.4areeasytocompute,sincetheintersectionsappearinginthose theoremswillbe circles or annularrings .If S isasolidobtainedby rotatingacurveaboutthe x axis,thentheintersectionoftheplane x = t and S isacircleofradius f ( x ),andhence A ( t )= f ( x )2 .If S is asolidobtainedbyrotatingthecurveofthefunction g ( y )= x about the y axis,thentheintersectionof S andtheplane y = t isacircle ofradius g ( t ),andso B ( t )= g ( t )2 .Thisyieldsthefollowingspecial versionsofTheorems6.3and6.4. Theorem 6.5 Let S beasolidbetweentheplanes x = a and x = b obtainedbyrotatingthegraphofthefunction f ( x )= y aboutthe x axis.Thenwehave V ( S )= b af ( t )2dt. Theorem 6.6 Let S beasolidbetweentheplanes y = a and y = b obtainedbyrotatingthegraphofthefunction g ( y )= x aboutthe y axis. Thenwehave V ( S )= b ag ( t )2dt. Theexercisesattheendofthissectionwillprovidefurtherexamples fortheusesofthesetheorems.

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146.APPLICATIONSOFINTEGRATION Ifthedomaintoberotateddoesnotincludethe entire areabetween thecurveandandcoordinateaxis(forinstance,becauseitisadomain between twocurves),thenwegetannularrings,whichwediscussedin thelastsection.37.4.ASolidNotObtainedbyRevolution.Whilevolumesofrevolution areaveryfrequentapplicationofTheorems6.3and6.4,theyarenot theonlyapplicationsofthosetheorems. Example 6.8 Let S beapyramidwhosebaseisasquareofside length a andwhoseheightis h .Computethevolumeof S Solution: The“rststepistoplace S inacoordinatesystemsothat Theorem6.3canbeapplied.Letusplacetheaxisof S onthe x axisof thecoordinatesystem,sothatthecenterofthebaseof S isattheorigin andthecuspof S isat x = h .Thisdoesnotcompletelydeterminethe positionof S ,because S couldstillrotatearoundthe x axis.However, suchrotationswillnotchangethevalueof A ( t )forany t [0 ,h ],and sotheyareinsigni“cantforthecomputationof V ( S ). Nownotethat,forany t [0 ,h ],theintersectionof S andtheplane x = t isasquareofsidelength a ( h Š t ) /h (bysimilartriangles).See Figure6.10. Pyramid.

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37.VOLUMES15 Figure6.10foranillustration.So A ( t )= a2( h Š t )2/h2,andTheorem 6.3implies V ( S )= h 0a2( h Š t )2 h2dt = a2 h2 h2t Š ht2+ t3 3 h 0= a2h 3 37.5.TheBigPicture.Notethatthetheoremspresentedinthissectiononintegralscon“rmourintuitionthatifacertainfunctionmeasuresaquantity,then,undertheappropriateconditions,theintegral ofthatfunctionmeasuresaquantitythatissomehowinaspacethat isonedimensionhigher.Forinstance,wesawearlierthatif f ( x )measuredtheheightofacurveatagivenhorizontalcoordinate x ,then, undertheappropriateconditions, f ( x ) dx measuredthe area under thecurve.Sotakingintegralsmeantmovingfromonedimensionto two.Inthissection,thefunctions A ( t )and B ( t )measuredareas ofdomainsinagivenplane,while A ( t ) dt and B ( t ) dt measured volumes .Sotakingintegralsmeantmovingfromtwodimensionsto three.37.6.Exercises.(1)Computethevolumeofthesolidbetweentheplanes x = Š 1 and x =1obtainedbyrotatingthegraphofthefunction f ( x )= x2= y aboutthe y axis. (2)Computethevolumeofthesolidbetweentheplanes x = Š 1 and x =1obtainedbyrotatingthegraphofthefunction f ( x )= x2= y aboutthe x axis. (3)Computethevolumeofthesolidbetweentheplanes x =0 and x = obtainedbyrotatingthegraphofthefunction f ( x )=sin x = y aboutthe x axis. (4)Computethevolumeofthesolidobtainedbyrotatingthe domainbetweenthecurves y = x and y = x abouttheline y = Š 2. (5)Computethevolumeofthesolidobtainedbyrotatingthe domainbetweenthecurves y = x4and y = x abouttheline x = Š 1.

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166.APPLICATIONSOFINTEGRATION (6)Computethevolumeofa regulartetrahedron ofsidelength z Aregulartetrahedronisasolidthathasfourfaces,eachof whichisaregulartriangle. (7)Computethevolumeofthesolidbetweentheplanes x =0and x =1obtainedbyrotatingthecurve f ( x )= ex= y aboutthe y axis. (8)Computethevolumeofthesolidbetweentheplanes x =0 and x =1obtainedbyrotatingthecurveofthefunction f ( x )= x (1 Š x )aboutthe x axis.38.CylindricalShells38.1.AnAlternativeMethodtoComputeVolumes.Inprinciple,Theorems6.3and6.4aresimplemethodstocomputethevolumesofsolids. Inpractice,however,theareas A ( t )and B ( t )thatappearinthesetheoremsmaybediculttoexplicitlyevaluate.Onesituationinwhich theseareasareoftendiculttocomputeiswhenthesolidinquestion isobtainedbyrotatingadomainaroundsomeline;thatis,itisasolid ofrevolution. Asanexample,letustrytocomputethevolumeofthesolid S obtainedbyrotatingaboutthe y axisthedomainborderedbythelines x =0, x =3,and y =0andthegraphofthefunction f ( x )=3 x3Š x4. IfwetrytosolvethisproblemusingTheorems6.3or6.4,weruninto diculties,because A ( t )and B ( t )willnotbeeasytocompute.For instance,ifwewantedtouseTheorem6.4,then,inordertocompute B ( t ),wewouldneedtodescribetheintersectionof S andtheplane y = t .Forthis,wewouldhaveto“ndthe x coordinatesofthepoints ofthatintersection;thatis,wewouldneedto“ndallrealnumbers x [0 3]forwhich y =3 x3Š x4= t .Thisisafourth-degreeequation for x ,whichisverydicultandcumbersometosolve.Ifwewanted touseTheorem6.3,then,inordertocompute A ( t ),wewouldneed todescribetheintersectionoftheplane x = t and S ,whichisnot straightforwardtodo. Insituationslikethis,thatis,whentheapplicationofTheorems 6.3and6.4leadstotechnicaldiculties,itoftenhelpstouseanother methodcalledthemethodof cylindricalshells .Acylindricalshell C issimplyacylinder C1ofwhichasmallercylinder C2isremoved,so that C1and C2havethesamesymmetryaxis.SeeFigure6.11foran illustration.If C2isjustalittlebitsmallerthan C1,then C lookslike ashell,explainingthename cylindricalshell .

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38.CYLINDRICALSHELLS17 Figure6.11. Singlecylindricalshell. If C1and C2bothhaveheight h and Cihasradius ri,thenthe volumeof C canbecomputedas V ( C )= V ( C1) Š V ( C2) = hr2 1 Š hr2 2 = h ( r2 1Š r2 2) Notethatthelastformof V ( C )canberearrangedas (6.2) V ( C )= h ( r1Š r2) 2 ( r1+ r2) 2 Thiswayofwriting V ( C )mightseemcontrivedat“rstsight.However,ithasthefollowingmotivation.Notethat r1Š r2isthewidthŽ of C ,while h isitsheight.Finally,ifwe”atten C outintheplane, itwillbecomeabrickwithsidelengths h r1Š r2,and,2 r1+ r2 2,since thelengthofthemissingsideisequaltothecircumferenceofacircle whoseradiusistheaverageoftheradiiof C1and C2. Inotherwords,(6.2)saysthatthevolumeofacylindricalshellis equaltotheproductofitsheight,width,andlengthŽ(ifthelatteris interpretedproperly). Wearenowinapositiontousecylindricalshellstocomputevolumes.Let S beasolidthatisobtainedbyrotatingthedomain D ,

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186.APPLICATIONSOFINTEGRATION whichliesbelowthecurveof f ( x )= y from x = a to x = b ,aboutthe y axis.Inordertoestimate V ( S ),wecut[ a,b ]into n intervalsofequal lengthusingpoints a = x0
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38.CYLINDRICALSHELLS19 Figure6.12. (a)Thecurveof y =3 x3Š x4and(b)the solidobtainedbyitsrotation. Solution: ByTheorem6.7,wehave V ( S )=2 3 0x (3 x3Š x4) dx =2 3 0(3 x4Š x5) dx =2 3 x5 5 Š x6 6 3 0=2 729 5 Š 729 6 =152 677 Theaxisaroundwhichwerotateadomaindoesnothavetobe acoordinateaxisinorderforthemethodofcylindricalshellstobe applicable.Wecanapplythemethodaslongaswecandecomposethe solidinquestionintocylindricalshellswhoseheightandradiuswecan compute. Example 6.10 Let S bethesolidobtainedbyrotatingthedomain whosebordersarethehorizontalaxis,theverticallines x =0 and x =2 ,andthegraphofthefunction f ( x )=2 x Š x2aboutthevertical line x =3 .Compute V ( S ) Solution: Wecandecompose S intocylindricalshellswhosecenter isontheverticalline x =3.Theshellcontainingthepoint x of thehorizontalaxiswillhaveheight f ( x )=2 x Š x2andradius3 Š x .

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206.APPLICATIONSOFINTEGRATION Figure6.13. (a)Thecurveof y = x2/ 2and(b)the solidobtainedbyitsrotation. Therefore,wehave V ( S )= 2 02 (2 x Š x2)(3 Š x ) dx =2 2 0 x3Š 5 x2+6 x dx =2 x4 4 Š 5 x3 3 +3 x22 0=2 8 3 16 755 38.2.Exercises.In(1)…(3),usethemethodpresentedinthissectiontocomputethe volumeofthesolidobtainedbyrotatingthedomainbetweenthegiven curvesaboutthe y axis. (1) f ( x )= x3= y x =1, y =0. (2) f ( x )=1 x2, x =2, x =3, y =0. (3) f ( x )= x g ( x )= Š x x =2. In(4)…(7),usethebestavailablemethodtocomputethevolumeofthe solidobtainedbyrotatingthedomainbetweenthegivencurvesabout thegivenaxis. (4) f ( x )= x = y g ( x )= Š x = y x =2,about x = Š 1.

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39.WORKANDHYDROSTATICFORCE21 (5) f ( x )=6 Š x g ( x )= x = y y =6,aboutthe y axis. (6) f ( x )=6 Š x g ( x )= x = y y =6,abouttheline y =6. (7) f ( x )=6 Š x g ( x )= x = y y =6,abouttheline y =7.39.WorkandHydrostaticForce39.1.WorkMovingaPoint-likeObject.Inphysics,theword work has amorespeci“cmeaningthanineverydaylife.Workinphysicsmeans thata force isextendedtomoveanobjectacertaindistance. Theforce F movinganobjectiscomputedbytheformula F = m d2s dt2, whichiscalled Newtonssecondlaw .Here m isthemassoftheobject, while a =d2s dt2isits acceleration .SoNewtonssecondlawsaysthatthe massofanobjectisindirectproportiontotheforceneededtomove itatconstantacceleration. Ifaconstantforce F isexertedwhileanobjectmovesdistance d thentheworkdonebythatconstantforceiscomputedbytheformula W = Fd. Notethatinthemetricsystem,distanceismeasuredinmeters(m), timeismeasuredinseconds(s),andthereforeaccelerationismeasured inm / s2.Massismeasuredinkilograms(kg),soforceismeasuredin kg m s2,whicharecalled newtons (N),and,“nally,workismeasuredin N m,whicharealsocalled joules (J).Onejouleistheworkthatis donewhenaforceequalto1newtonmovesanobjectadistanceof1 meter. Example 6.11 Howmuchworkisneededtoliftachildof20kgto aheightof0.5meters?Usethefactthatgravitationcausesdownward accelerationof g =9 8m / s2. Solution: Inordertoliftthechild,oneneedstoovercomethedownwardaccelerationcausedbygravity.Thismeansthatanupwardforce of m g =20kg 9 8 m s2=198N hastobeexertedacrossadistanceof d =0 5meters.Thisyields W = Fd =198N 0 5m=99J Sotheworkneededis99J. Iftheforceexertedisnotconstantacrosstheentiredistance,but thedistancecanbesplitupintoafewpartssothattheforceisconstant oneachpart,thenwecancomputetheworkdonebytheforceoneach

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226.APPLICATIONSOFINTEGRATION partjustasinthepreviousexample,andthenwecanaddtheobtained amountstogetthetotalamountofworkdoneacrosstheentiredistance. Iftheforceexertedchangesaccordingtoacontinuousfunction f ( d ), thenwecanapproximatetheworkdoneusingtheideaoftheprevious paragraphandthenuseintegrationtocomputethetotalworkdoneby theforceasfollows. Let a and b berealnumbersandletusassumethatanobjectis movingfrom a to b ,andtheforcemovingtheobjectatagivenpoint x isequalto f ( x ),where f isacontinuousfunction.Inordertocompute theworkdoneacrosstheentiredistance,letussplittheinterval[ a,b ] into n equalintervals,usingpoints a = x0
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39.WORKANDHYDROSTATICFORCE23 Solution: ByTheorem6.8,wehave W = 0 1 070 xdx =[35 x2]0 1 0=12 25J So12.25Jofworkisneededtostretchthespring10cmoveritsnatural length. Wepointoutthatthelawofphysicsthatsaysthattheforceneeded toextendaspringby x unitsoveritsnaturallengthisequalto kx is called Hookeslaw ,and k iscalledthe springconstant .39.2.HydrostaticForce.Letussaythatwewanttopumpwaterout ofatankthathastheshapeofthesouthernhalfofaballofradius 1(m).Howmuchworkisneededtodothat?Thisquestionismore complexthanthepreviousonesincedeeperlayersofthehemisphere aresmaller,andwaterinthoselayershastotravelfartherinorderto reachthetopoftank. Therefore,wewillcutthetankupintosmalllayersandestimate theamountofworkneededtopumpouteachlayerofwater. Let x =0denotethebottomofthetankandlet x =1denote thecenterofthetopcircleofthetank.Cutthetankinto i horizontal layersbyplanesthatareatheights 0= x0
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246.APPLICATIONSOFINTEGRATION Figure6.14. Thetankinacoordinatesystem,andits layeratheight z As n grows,theexpressiondisplayedin(6.5)approximatesthe neededworkbetterandbetter.Wede“nethetotalworkneeded(to pumpallthewateroutofthetank)tobethelimitofthesumshown in(6.5)as n goestoin“nity.AsthatsumisaRiemannsum,itslimit, as n goestoin“nity,isthede“niteintegral W = g 1 0(2 x Š x2)(1 Š x ) dx. Astheintegrandisapolynomial,thisintegralisveryeasytocompute. Wegetthat W = g 1 0x3Š 3 x2+2 xdx = g x4 4 Š x3+ x21 0= g/ 4 =7696 675J Soittakesalmost7700Jofworktopumpoutallthewaterfromthe tank.39.3.Exercises.(1)Howmuchworkisdonewhenabookofmass2kgislifted1.5 metersfromitsoriginallocation?

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40.AVERAGEVALUEOFAFUNCTION25 (2)Ifittakes10Jofworktoliftanobject2meters,howmuch workdoesittaketoliftthatobjectanadditional3meters? (3)Ifittakes20Jofworktostretchaspring20cmoveritsnatural length,howmuchworkdoesittaketostretchthatspringan additional 5cm? (4)Howmuchworkisneededtopumpoutallthewaterfroma tankoftheshapeofaninvertedconeofheight10whosetop circlehasradius2?(Lengthismeasuredinmeters.) (5)Howmuchworkisneededtopumpoutallthewaterfroma tankoftheshapeofacylinderofheight20whosebasecircle hasradius30? (6)Howmuchworkisneededtopumpoutallthewaterfroma tankoftheshapeofaninvertedpyramidofheight15whose topplateisasquareofsidelength10?40.AverageValueofaFunction Theconceptof average isasimpleoneaslongaswetaketheaverage ofa “nitenumber ofvalues,suchastheaveragepriceofahouseina givenneighborhoodortheaveragedailyhightemperatureinagiven cityinagivenmonth.If a1,a2,...,anarerealnumbers,then (6.6) A =( a1+ a2+ + an) /n istheiraverage.However,whatcanwesayabouttheaveragevalue ofafunctionoveragiveninterval[ a,b ]?Wewillclearlyneedanew de“nitionforthatsincetherearein“nitelymanyrealnumbersin[ a,b ], sosummingallofthemandthendividingtheirsumbythenumberof summandsisnotanoption. Hereisanintuitivewayofextendingthede“nitionofaverageto thevaluestakenbyafunctionoveraninterval.Itfollowsfrom(6.6) that A istheonlyrealnumberwiththepropertythatifwereplace each of a1,a2,...,anby A ,thenthesum( a1+ a2+ + an)doesnot change.Thisobservationsuggeststhefollowingde“nition. Definition 6.1 Let f beafunctionsuchthat b af ( x ) dx exists. Thentheaveragevalueof f ontheinterval [ a,b ] istherealnumber c = b af ( x ) dx b Š a Indeed, c istheonlyrealnumberwiththepropertythatifwereplace f bytheconstantfunction f ( x )= c ,thentheintegral b af ( x ) dx does notchange.

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266.APPLICATIONSOFINTEGRATION Amoresystematicapproachisthefollowing.Aswesawwhenwe “rstlearnedaboutintegrals, b af ( x ) dx canbeapproximatedinthe followingway.Split[ a,b ]into n equalintervalsandchooseapoint xiinthe i thsuchinterval.Takearectangleofheight f ( xi)overthe i th interval.Theaveragevalueofthe n valuesof f takenatthepoints xiis,ofcourse, An= f ( x1)+ f ( x2)+ + f ( xn) n Ontheotherhand,thetotalareaofthe n rectangleswehavejust de“nedis Rn= b Š a n ( f ( x1)+ f ( x2)+ + f ( xn)) Comparingthelasttwodisplayedequations,weseethat (6.7) An= Rn b Š a If n goestoin“nity,thenthe n rectangleswillapproximatethedomain underthegraphof f ,andsotheright-handsideof(6.7)willconverge tob af ( x ) dx b Š a,whiletheleft-handsidewillapproximatetheaveragevalue of f on[ a,b ]. Example 6.13 Whatistheaveragevalue A of sin x ontheinterval [0 ] ? Solution: Wehave A = 0sin xdx = [ Š cos x ] 0 = 1 Š ( Š 1) = 2 SeeFigure6.15foranillustration. Itisworthpointingoutthatacontinuousfunction f willactually takeitsaveragevalueoneachinterval.Thisisthecontentofthe followingtheorem. Theorem 6.9 Let f beacontinuousfunctionon [ a,b ] andlet c betheaveragevalueof f on [ a,b ] .Thenthereexistsarealnumber x [ a,b ] suchthat f ( x )= c .

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40.AVERAGEVALUEOFAFUNCTION27 Figure6.15. Theaveragevalueofsin x on[0 ]. Proof. Itsucestoshowthatif m istheminimumof f on[ a,b ] and M isthemaximumof f on[ a,b ],then m c M ,andourclaim followsfromtheintermediatevaluetheorem. Weknowthat m ( b Š a ) b af ( x ) dx M ( b Š a ) forobviousgeometricreasons.Nowdivideallthreetermsby b Š a to get m c M Example 6.14 Thereisarealnumber x [0 ] suchthat sin x = 2 / Solution: ThisfollowsfromthepreviousexampleandTheorem6.9. 40.1.Exercises.(1)Findtheaveragevalueof xnon[0 1]. (2)Findtheaveragevalueoftan x on[0 ,/ 4]. (3)Findtheaveragevalueofln x on[1 ,e ]. (4)Findtheaveragevalueof exon[0 1]. (5)Whatislarger,theaveragevalueofsin x ortheaveragevalue ofcos x ,ifbothaveragesaretakenontheinterval[0 ,/ 2]? (6)Whatislarger,theaveragevalueofsin x on[0 ]ortheaveragevalueofsin x on[ Š 14 17 ]?Canyou“ndananswer thatdoesnotinvolvecomputation?

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CHAPTER7 MethodsofIntegration 41.IntegrationbyParts41.1.MethodofIntegrationbyParts.Let u and v betwodierentiable functionsofthevariable x .Weusedthesimpleproductrule (7.1)( uv )= uv + uvtocomputethe derivative oftheproductofthesetwofunctions.Is thereasimilarruleforcomputingthe integral oftheproductoftwo functions?Ingeneral,theanswerisno.Thereisnorulethatprovides theintegraloftheproductoftwofunctionsthatwouldworkinevery case.However,therearemanycasesinwhicharelativelysimpleway ofreversingŽtheproductruleofdierentiationwillgiveustheanswer wearetryingtoobtain. Indeed,integratingbothsidesoftheproductrule(7.1)ofdierentiation(7.1)withrespectto x ,wegettheidentity u ( x ) v ( x )= ( u( x ) v ( x )) dx + ( u ( x ) v( x )) dx or,afterrearrangement, (7.2) ( u( x ) v ( x )) dx = u ( x ) v ( x ) Š ( u ( x ) v( x )) dx. Formula(7.2)isveryusefulifwewanttocomputetheintegralof theproductoftwofunctions,oneofwhichcanplaytheroleof uand theotheroneofwhichcanplaytheroleof v .Ifwecancompute u and ( uv),thenformula(7.2)enablesustocompute ( uv )aswell.If wecannotcarryoutoneorbothofthesecomputations,thenformula (7.2)willnothelp. Example 7.1 Compute xexdx Solution: Weset u( x )= exand v ( x )= x .Thenformula(7.2)iseasy toapply,since v ( x )= x and v( x )=1.Therefore,(7.2)impliesthat xexdx = ex x Š ex 1 dx = ex x Š ex+ C = ex( x Š 1)+ C. 29

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307.METHODSOFINTEGRATION Thereaderisencouragedtoverifythattheobtainedsolutionis correctbycomputingthederivativeof ex( x Š 1)andcheckingthatit isindeedequalto ex x Atthispoint,thereadermaybeaskinghowweknewthatwe neededtoset u( x )= exand v ( x )= x ,andnottheotherwayaround. Theansweristhattheotherdistributionofroles,thatis, u( x )= x and v ( x )= exwouldnothavehelped.Indeed,ifwehadchosen uand v inthatway,wewouldhaveneededtocompute ( uv) dx = ( exx2) / 2 dx .Thatwouldhavebeenmorecomplexthantheoriginal problem.Weshouldalwayschoose uand v sothat ( uv) dx iseasy tocompute.Thatusuallymeansselecting v sothatitbecomessimpler whendierentiated,andtoselect usothat udoesnotgetmuch morecomplexwhenintegrated(oratleastoneofthesetwodesirable outcomesoccur). Example 7.2 Compute x cos xdx Solution: Weset u( x )=cos x and v ( x )= x ,whichmeansthat u ( x )=sin x and v( x )=1.Soformula(7.2)implies x cos xdx = x sin x Š sin xdx = x sin x +cos x + C. Thetechniqueofintegrationwehavejustexplainediscalled integrationbyparts .41.2.AdvancedExamples.Sometimestheintegranddoesnotseemto beaproduct,butitcanbetransformedintoone.Thefollowingisa classicexample. Example 7.3 Compute ln xdx Solution: Thecrucialobservationisthatwritingln x =1 ln x helps. Let u( x )=1and v ( x )=ln x .Then u ( x )= x and v( x )=1 /x ,so, crucially, u ( x ) v( x )=1.Therefore,formula(7.2)yields ln xdx = 1 ln xdx = x ln x Š 1 dx = x ln x Š x + C. Sometimesintegrationbypartsleadstoanequationorasystemof equationsthatneedstobesolvedinordertogetthesolutiontoour problem. Example 7.4 Compute excos x .

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41.INTEGRATIONBYPARTS31 Solution: Weset u( x )= exand v ( x )=cos x .Then u ( x )= exand v( x )= Š sin x ,andformula(7.2)yields (7.3) excos xdx = excos x + exsin xdx. Sowecouldsolveourproblemifwecouldcomputetheintegral exsin xdx .Wecandothatbyapplyingthetechniqueofintegrationbyparts again,setting u( x )= exand v ( x )=sin x .Weobtain (7.4) exsin xdx = exsin x Š excos xdx. Finally,notethat(7.3)and(7.4)isasystemofequationswithunknowns excos xdx and excos xdx .Wecansolvethissystem,for instance,byaddingthesetwoequationsandnotingthat exsin x cancels.Wegettheequation excos xdx = ex(cos x +sin x ) Š excos xdx or excos xdx = ex 2 (cos x +sin x )+ C. Notethatsubstitutingtheobtainedexpressionfor excos xdx into (7.4),wegetaformulafor exsin xdx ,namely, exsin xdx = ex 2 (sin x Š cos x )+ C.41.3.De“niteIntegrals.Ifweevaluatebothsidesofformula(7.2)from a to b andweapplythefundamentaltheoremofcalculus,wegetthe identity (7.5) b a( uv ) dx =[ uv ]b aŠ b a( uv) dx. Example 7.5 Evaluate 2 1ln xdx Solution: AswesawinExample7.3,wecanset u ( x )= x and v ( x )= ln x .Then u( x )=1, v( x )=1 /x ,andformula(7.5)yields 2 1ln x =[ x ln x ]2 1Š 2 11 dx =2ln2 Š 1

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327.METHODSOFINTEGRATION 41.4.Exercises.(1)Compute x sin xdx (2)Compute x2exdx (3)Compute x ln xdx (4)Compute x2ln x (5)Evaluate 2 1x cos2 x (6)Evaluate 1 0x2sin x .42.TrigonometricIntegrals42.1.Powersofsinandcos.Inthissection,weconsiderfunctionsofthe form f ( x )=sinmx cosnx anddiscusstechniquesfortheirintegration. Itseemsnaturalto“rstconsiderthecaseswhen m or n is0,thatis, when f isjustapowerofsinorcos.Eventhesespecialcaseswillbreak upintofurthersubcases.Theeasiestsubcaseiswhentheexponents are even numbers.Inthatcase,wecanusethetrigonometricidentities (7.6)cos2 x =2cos2x Š 1 and (7.7)sin2 x =2sin x cos x toeliminatehighpowersoftrigonometricfunctionsintheintegrand. Example 7.6 Compute cos4xdx Solution: Using(7.6),wegetthatcos2x =1+cos2 x 2,andso cos4x = 1+cos2 x 2 2= 1 4 + cos2 x 2 + cos22 x 4 Applying(7.6)again,with2 x replacing x ,wegetthatcos22 x =1+cos4 x 2, sothepreviousdisplayedequationturnsinto cos4x = 3 8 + cos2 x 2 + 1+cos4 x 8 Havingeliminatedthepowersofcos,theintegrationiseasytocarry outasfollows: cos4xdx = 3 8 + cos2 x 2 + cos4 x 8 dx = 3 x 8 + sin2 x 4 + sin4 x 32 + C.

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42.TRIGONOMETRICINTEGRALS33 Thecomputationismorecomplexiftheintegrandisan odd power ofsinorcos.Inthatcase,weseparateonefactorandconverttherest intothe other trigonometricfunction,usingtherulecos2x +sin2x =1. Example 7.7 Compute sin3xdx Solution: Wehave sin3x =sin x sin2x =sin x (1 Š cos2x )=sin x Š sin x cos2x. Theadvantageofthisformisthatitmakesintegrationbysubstitutioneasy.Indeed,set u =cos x ,then du/dx = Š sin x ,andso Š sin x cos2xdx = u2du = u3 3 + C = cos3x 3 + C. Comparingthetwodisplayedequationsofthissolutionandnotingthat sin xdx = Š cos x ,weget sin3x = Š cos x + cos3x 3 + C. Themethodsshownabovecanbeusedtocomputetheintegralof products ofpowersofsin x andcos x .Inotherwords,themethodallows ustocompute cosmx sinnxdx asshownbelow. Example 7.8 Compute cos2x sin3xdx Solution: JustasinExample7.7,weseparateonesin x factor.This accomplishestwothings.Itallowsustoconverttheremainingeven numberofsin x factorstocos x factors,anditallowsustointegrateby substitution. Indeed, cos2x sin3xdx = cos2x sin2x sin xdx = cos2x (1 Š cos2x )sin x = cos2x sin x Š cos4x sin x = Š u2du + u4du = Š u3 3 + u5 5 + C = Š cos3x 3 + cos5x 5 + C, whereweusedthesubstitution u =cos x

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347.METHODSOFINTEGRATION Wecanalwaysproceedthiswayifatleastoneof m and n inthe integrandcosmx sinnx isodd.Indeed,inthatcase,afterseparating onefactorfromthatoddpower,anevenpowerremains,andthatcan beconvertedtothe other trigonometricfunctionusingtheidentity sin2x +cos2x =1.Ifboth m and n areeven,thenwecanusethat identityright away. Example 7.9 Compute cos4x sin2xdx Solution: Wehave cos4x sin2xdx = cos4x (1 Š cos2x ) dx = cos4Š cos6xdx. Nownotethatwecomputed cos4dx inExample7.6.Youareaskedto compute cos6x inExercise42.4.2.Thedierenceofthesetworesults thenprovidesthesolutionofthepresentexample. 42.2.Powersoftanandsec.Whenintegratingaproductoftheform tanmx secnx ,wewillusetheidentitysec2x =tan2x +1andthedifferentiationrules(tan x )=sec2x and(sec x )=sec x tan x Therearetwoeasycases,namely,when m isodd(and n isatleast 1)andwhen n 2iseven. Inthe“rstcase,thatis,when m isoddand n 1,weseparateone factoroftan x sec x andexpresstheremainingfactorsintermsofsec x bytheidentity Š 1+sec2x =tan2x .Thenwesubstitute u =sec x whichleadstodu dx=tan x sec x Example 7.10 Compute tan3x sec xdx Solution: Followingthestrategyexplainedabove,wehave tan3x sec xdx = tan2x tan x sec xdx = ( Š 1+sec2x )tan x sec xdx = ( Š 1+ u2) du = Š u + u3 3 + C = Š sec x + sec3x 3 + C.

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42.TRIGONOMETRICINTEGRALS35 Inthesecondcase,thatis,when n 2iseven,weseparateone factorofsec2x ,expresstheremainingfactorsintermsoftan x using theidentitysec2x =1+tan2x ,andsubstitute u =tan x ,whichleads todu dx=sec2x Example 7.11 Compute sec4x dx. Solution: Wehave sec4xdx = sec2x sec2xdx =(1+tan2x )sec2xdx =(1+ u2) du = u + u3 3 + C =tan x + tan3x 3 + C. Ifwearenotinthesetwoeasycases,thenthereisnorecipethat willalwayswork.Wethenneedtohaveaseparatestrategyforeach problem.WewillshowexamplesofthatinExercises42.4.6and42.4.7.42.3.SomeOtherTrigonometricIntegrals.Ifourgoalistocomputeintegralsoftheform cos mx sin nxdx cos mx cos nxdx ,and sin mx sin nxdx ,thenwecanoftenmakeuseofthefollowingidentities: (7.8)sin a cos b = 1 2 sin( a Š b )+ 1 2 sin( a + b ) (7.9)cos a cos b = 1 2 cos( a Š b )+ 1 2 cos( a + b ) (7.10)sin a sin b = 1 2 cos( a Š b ) Š 1 2 cos( a + b ) Example 7.12 Compute cos3 x cos5 xdx Solution: Using(7.9)with a =3 x and b =5 x andnotingthat cos Š 2 x =cos2 x ,wegetthatcos3 x cos5 x =1 2cos2 x +1 2cos8 x ,and so cos3 x cos5 xdx = 1 2 cos2 x + 1 2 cos8 x dx = 1 4 sin2 x + 1 16 sin8 x + C.

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367.METHODSOFINTEGRATION 42.4.Exercises.(1)Compute sin3xdx (2)Compute cos6xdx (3)Compute sin3x cos2xdx (4)Compute tan2x sec4xdx (5)Compute tan3x sec5xdx (6)Compute tan5x byseparatingonefactoroftan2x inthe integrandandexpressingitintermsofsec2x (7)Compute sec3xdx usingintegrationbyparts,with u( x )= sec2x and v ( x )=sec x .43.TrigonometricSubstitution43.1.ReversingtheTechniqueofSubstitutions.Letusassumethatwe wanttocomputetheareaofacirclebyviewingone-fourthofthatcircle asthedomainunderacurve.Let r betheradiusofthecircle,andlet usplacethecenterofthecircleattheorigin.Thenthenortheastern quarterofthecircle,showninFigure7.1,isjustthedomainunderthe graphofthefunction f ( x )= r2Š x2,where x rangesfrom0to r .In otherwords,weneedtocomputetheintegral (7.11) r 0 r2Š x2dx. InChapter5,wepresentedthetechniqueofintegrationbysubstitution.Thistechniqueworkedinsituationswhenthebestwaytocomputeanintegralwastode“neasimplefunctionof x ,suchas y ( x )= x2, andthencontinuetheintegrationintermsofthatnewvariable y Inordertocomputetheintegralin(7.11),weusethe reverse of thestrategymentionedinthepreviousparagraph.Wede“neanother variable y sothat x isasimplefunction f of y .Itisimportantto Figure7.1. Thenortheasternquadrantoftheunitcircle.

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43.TRIGONOMETRICSUBSTITUTION37 de“ne f and y sothat f isone-to-one,sincethatassuresthat f ( y )= x isequivalentto fŠ 1( x )= y Incomputingtheintegralin(7.11),wecanset x = r sin y .Then dx/dy = r cos y ,andthelimitsofintegrationare y =0and y = / 2. Thisyields r 0 r2Š x2dx = / 2 0 r2Š r2sin2yr cos ydy = r2/ 2 0 1 Š sin2y = r2/ 2 0cos2ydy = r2 y +sin2 y 4 / 2 0= r2 4 Notethatwecouldwritecos y for 1 Š sin2y ,since0 y / 2, andinthatinterval,cos y isnonnegative. Wepointoutthatbyconvertingtheindeterminateintegral r2y +sin2 y 4 backtoafunctionof x ,wegetthat r2Š x2dx = r2 4 sinŠ 1 x r + r2 2 x 1 Š x2. Theresultthatwearegoingtocomputeinthenextexamplewillbe usefulinthenextsection,whenwewilllearnatechniquetointegrate rationalfunctions. Example 7.13 Computetheintegral 1 (1+ x2)2dx Solution: Weusethesubstitution x =tan y .Then y =tanŠ 1( x ),and so dy/dx =1 / (1+ x2),andhence dy = dx/ (1+ x2).Thisyields 1 (1+ x2)2dx = 1 1+ x2dy = 1 1+tan2y dy = cos2ydy = y 2 + sin2 y 4

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387.METHODSOFINTEGRATION = y 2 + sin y cos y 2 = 1 2 tanŠ 1( x )+ 1 2 x x2+1 Thelaststepisjusti“edsince (7.12) x x2+1 = tan y 1+tan2y =tan y cos2y =sin y cos y. Figure7.2illustratesthistrigonometricargument. Example 7.14 Computetheintegral 1 x2Š 1. Solution: Thedenominatorremindsusofthetrigonometricidentity tan2y =sec2y Š 1,andso,if y [0 ,/ 2),thentan y = sec2y Š 1. Therefore,weusethesubstitution x =sec y .Then dx/dy =tan y sec y Hence,wehave 1 x2Š 1 = 1 sec2y Š 1 dx = 1 tan y dx = tan y sec y tan y dy = sec ydy =ln | sec y +tan y | =ln | x + x2Š 1 | Figure7.3illustratesthistrigonometricargument. Figure7.2. Someexpressionsfrom(7.12).

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43.TRIGONOMETRICSUBSTITUTION39 Figure7.3. Someexpressionsoccuringinthesolution ofExample7.14.43.2.SummaryoftheMostFrequentlyUsedTrigonometricSubstitutions.Thethreeexamplesthatwehaveseensofarinthissectionshowthe threemostfrequentlyusedreversesubstitutions.Thatis, (i)Tocompute r2Š x2dx ,usethereversesubstitution x = r sin y (ii)Tocomputeintegralsinvolving( r2+ x2)underarootsignor inthedenominatorofafraction,usethereversesubstitution x = r tan y (iii)Tocompute x2Š r2dx ,usethereversesubstitution x = r sec y Finally,awordofcaution.Theavailabilityofthemethodofreverse substitutiondoesnotmeanthatthismethodis always thebestone tocomputeanintegralthatcontainsasquarerootsign.Oneofthe followingexercisescanbesolvedbyanothermethodfaster(and,no, wearenotrevealingwhichone).43.3.Exercises.(1)Usethemethodpresentedinthissectiontocomputethearea ofanellipsedeterminedbytheequation x2 a2+ y2 b2=1 (2)Compute 1 Š 4 x2dx (3)Compute 1+ x2dx (4)Compute x x Š 5dx (5)Compute 1 x2 x2Š 4dx (6)Compute x2Š 2 xdx .

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407.METHODSOFINTEGRATION 44.IntegratingRationalFunctions44.1.Introduction.Recallthata rationalfunction istheratiooftwo polynomials,suchas R ( x )= P ( x ) Q ( x ) = 3 x +5 2 x2+4 x +9 Integratingrationalfunctionsisrelativelysimple,becausemostofthese functionscanbeobtainedassumsofevensimplerfunctions.Ifthe degreeof P ( x )isatleastaslargeasthedegreeof Q ( x ),thenwecan divide P ( x )by Q ( x ),gettingapolynomialasaquotient,andpossibly aremainder.Thatis,ifthedegreeof P isatleastaslargeasthe degreeof Q ,thenthereexistpolynomials P1( x )and P2( x )suchthat thedegreeof P2( x )is less thanthedegreeof Q ( x )and R ( x )= P ( x ) Q ( x ) = P1Q(x )+ P2( x ) Q ( x ) = P1( x )+ P2( x ) Q ( x ) As P1( x )isapolynomial,itiseasytointegrate.Therefore,thedicultyofintegrating R ( x )liesinintegratingR2( x ) Q ( x ),whichisarational functionwhose denominator isofhigherdegreethanitsnumerator. Forthisreason,intherestofthissection,wefocusonintegrating rationalfunctionswiththatproperty,thatis,whenthedegreeofthe denominatorishigherthanthedegreeofthenumerator. Example 7.15 Let R ( x )=x3+2 x +1 x2Š x +1.Thendividing P ( x ) by Q ( x ) usinglongdivision,weget P ( x )=( x +1)( x2Š x +1)+2 x, so R ( x )= P ( x ) Q ( x ) = x +1+ 2 x x2Š x +1 andintegrating Q ( x ) boilsdowntointegrating2 x x2Š x +1.44.2.BreakingUptheDenominator.Inordertodecidehowtobreak uparationalfunction R ( x )intothesumofsimplerterms,weanalyze thedenominator Q ( x )of R ( x ).Atheoremincomplexanalysis,sometimescalledthefundamentaltheoremofalgebra,impliesthatif q ( x ) isapolynomialwhosecoecientsarerealnumbers,then q ( x )canbe writtenasaproductofpolynomialsthatareofdegree1or2. Thisdecomposition,orfactorization,of Q ( x )willdeterminethe wayinwhichwebreakupourrationalfunctionintothesumofsimpler terms.Thereareseveralcasestodistinguish,basedonthefactorization of Q ( x ).

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44.INTEGRATINGRATIONALFUNCTIONS41 44.2.1.DistinctLinearFactors.Theeasiestcaseiswhen Q ( x )factors intotheproductofpolynomialsofdegree1,andeachoftheseterms occursonlyonce. Example 7.16 Compute 1 x2+3 x +2dx Solution: Notethat x2+3 x +2=( x +1)( x +2).Usingthatobservation, wearelookingforrealnumbers A and B suchthat (7.13) 1 x2+3 x +2 = A x +1 + B x +2 asfunctions,thatis,suchthat(7.13)holdsfor all realnumbers x Multiplyingbothsidesby x2+3 x +2,weget (7.14)1= A ( x +2)+ B ( x +1) If(7.14)holdsfor all realnumbers x ,itmustholdfor x = Š 1and x = Š 2aswell.However,if x = Š 1,then(7.14)reducesto1= A andif x = Š 2,then(7.14)reducesto1= Š B .Soweconcludethat A =1and B = Š 1arethenumberswewantedto“nd.Itisnoweasy tocomputetherequestedintegralasfollows: 1 x2+3 x +2 dx = 1 x +1 dx Š 1 x +2 dx =ln( x +1) Š ln( x +2)+ C. Theabovemethodcanalwaysbeappliedif Q ( x )factorsintoa productoflinearpolynomials,eachofwhichoccursonlyonce.In particular,if Q ( x )decomposesas a ( x Š a1)( x Š a2) ( x Š ak),then wecandecompose R ( x )intoasumoftheform A1 x Š a1+ A2 x Š a2+ + Ak x Š ak. Afterdeterminingthenumbers Ai,wecanintegrateeachoftheabove k summands.44.2.2.RepeatedLinearFactors.Thenextcaseiswhen Q ( x )factors intolinearterms,butsomeofthesetermsoccurmorethanonce. Example 7.17 Compute 2 x +7 ( x +1)2( x Š 1)dx Solution: Justasinthepreviouscase,wedecomposetheintegrand intoasumofsimplerfractions.Wearelookingforrealnumbers A B and C suchthat 2 x +7 ( x +1)2( x Š 1) = A x +1 + B ( x +1)2+ C x Š 1 .

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427.METHODSOFINTEGRATION Multiplyingbothsidesbythedenominatoroftheleft-handside,we get 2 x +7= A ( x +1)( x Š 1)+ B ( x Š 1)+ C ( x +1)2. Substituting x =1inthelastdisplayedequationyields9=4 C ,so C =2 25.Substituting x = Š 1yields5= Š 2 B ,so B = Š 2 5.Finally, thecoecientof x2ontheleft-handsideis0,whileontheright-hand side,itis A + C .So A + C =0,yielding A = Š 2 25. Nowweareinapositiontocomputetherequestedintegral. 2 x +7 ( x +1)2( x Š 1) = Š 2 25 x +1 dx + Š 2 5 ( x +1)2dx + 2 25 x Š 1 dx = Š 2 25 x +1 dx + Š 2 5 ( x +1)2dx + 2 25 x Š 1 dx = Š 2 25ln( x +1)+ 2 5 x +1 +2 25ln( x Š 1) Ingeneral,ifaterm( x + a )koccursin Q ( x ),thenthepartialfraction decompositionof R ( x )willcontainonetermwithdenominator( x + a )iforeach i { 1 2 ,...,k } .Forinstance,if Q ( x )=( x +2)3( x +5)2( x Š 10), then R ( x )willhaveapartialfractiondecompositionoftheform A1 x +2 + A2 ( x +2)2+ A3 ( x +2)3+ A4 x +5 + A5 ( x +5)2+ A6 x Š 10 .44.2.3.DistinctQuadraticFactors.Thethirdcaseiswhenthefactorizationof Q ( x )containssomequadraticfactorsthatareirreducible (i.e.,theyarenottheproductoftwolinearpolynomialswithrealcoecients),butnoneoftheseirreduciblequadraticfactorsoccursmore thanonce.Inthatcase,afterobtainingthepartialfractiondecompositionof R ( x ),wemayhavetoresorttotheformulas 1 x2+1 =tanŠ 1x + C and 1 x2+ a2= 1 a tanŠ 1 x a + C. Example 7.18 Computetheintegral 4 x +2 x3+ x2+ x +1dx Solution: Itiseasytonoticethatsetting x = Š 1turnsthedenominatorto0;hence,thedenominatorisdivisibleby x +1.Dividingthe denominatorby x +1,weget x2+1,sothedenominatorfactorsas ( x +1)( x2+1).Thefactor x2+1isirreducible(itisnotdivisibleby

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44.INTEGRATINGRATIONALFUNCTIONS43 x Š b foranyrealnumber b ,sincenorealnumber b satis“estheequation b2+1=0). Therefore,wearelookingforrealnumbers A B ,and C suchthat (7.15) 4 x +2 x3+ x2+ x +1 = A x +1 + B x2+1 + Cx x2+1 Thereaderisinvitedtoverifythatthethirdsummandoftherighthandsideisnecessary;thatis,ifthesummandCx x2+1isremoved,then nopairofrealnumbers( A,B )willsatisfy(7.15). Inorderto“ndthecorrectvaluesof A B ,and C ,multiplyboth sidesof(7.15)by( x +1) ( x2+1)andrearrange,toget 4 x +2=( A + C ) x2+( B + C ) x + A + B. Thecoecientof x2is0ontheleft-handside,soithastobe0onthe right-handside.Therefore, A + C =0.Similarly,thecoecientof x is 4ontheleft-handside,soithastobe4ontheright-handside,forcing B + C =4.Similarly,theconstanttermsofthetwosideshavetobe equal,and,consequently, A + B =2.Solvingthissystemofequations, weget A = Š 1, B =3,and C =1.Therefore, 4 x +2 x3+ x2+ x +1 dx = Š 1 x +1 dx +3 1 x2+1 dx + x x2+1 dx =ln( x +1)+3tanŠ 1x + 1 2 ln( x2+1) Ingeneral,if x2+ ax + b isaquadraticfactorin Q ( x ),thenthepartial fractiondecompositionwillcontainasummandoftheformE x2+ ax + band asummandoftheformFx x2+ ax + b.Again,thelatterisnecessary,since arationalfractionoftheformE + Fx x2+ ax + bwillnotequaloneoftheformE x2+ ax + bforanychoiceof E if F =0.44.2.4.RepeatedQuadraticFactors.Finally,itcanhappenthatthe factorization Q ( x )containsirreduciblequadraticfactors,someofwhich occurmorethanonce. Example 7.19 Computetheintegral x3+2 x2+3 x +7 x4+2 x2+1dx Solution: Itiseasytoseethatthedenominatorfactorsas( x2+1)2. Hence,wearelookingforrealnumbers A B C ,and D suchthat x3+2 x2+3 x +7 x4+2 x2+1 = A x2+1 + Bx x2+1 + C ( x2+1)2+ Dx ( x2+1)2. Multiplyingbothsidesby x4+2 x2+1andrearranging,weget x3+2 x2+3 x +7= Bx3+ Ax2+( B + D ) x +( A + C ) .

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447.METHODSOFINTEGRATION Foreach k ,thecoecientsof xkmustbethesameonbothsides. Hence, A =2and B =1,so C =5and D =2. Nowwecancomputetherequestedintegralusingthepreceding partialfractiondecompositionasfollows: x3+2 x2+3 x +7 x4+2 x2+1 dx = 2 x2+1 + x x2+1 + 5 ( x2+1)2+ 2 x ( x2+1)2 dx =2 tanŠ 1x + 1 2 ln( x2+1)+ 5 x 2( x2+1) + 5 2 tanŠ 1x Š 1 x2+1 = 1 2 ln( x2+1)+4 5tanŠ 1x + 1 2 5 x Š 2 x2+1 Hereweusedtheformulafor 1 ( x2+1)2thatwecomputedinthelast section,inExample7.13. Bynow,thereadermustknowwhatthegeneralversionofthe techniqueoftheprecedingexampleis.Ifthefactorizationof Q ( x ) contains( x2+ ax + b )k,then,foreachinteger i suchthat1 i k thepartialfractiondecompositionof R ( x )willcontainasummandof theformEi ( x2+ ax + b )iandasummandoftheformFix ( x2+ ax + b )i.44.3.RationalizingSubstitutions.Therearesituationswhenafunction thatis not arationalfunctioncanbeturnedintoonebyanappropriate substitution,andthenitcanbeintegratedbythemethodspresented inthissection.Themostfrequentscenarioinwhichthishappensis whentheintegrandcontainsroots,butifthoserootsarereplacedby anothervariable,wegetarationalfunctioninthatothervariable. Example 7.20 Compute x x +1dx Solution: Weusethesubstitution x = y .Then dy/dx =1 2 x=1 2 y. Thisleadsto x x +1 dx = y y +1 2 ydy = 2 y2 y +1 dy = 2( y Š 1)+ 2 y +1 dy

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45.STRATEGYOFINTEGRATION45 = y2Š 2 y +2ln( y +1) = x Š 2 x +2ln( x +1) Notethatthecomputationwouldhavebeenverysimilarifthe integrandcontainedsomeotherrootof x .Indeed,iftheintegrand containedr x insteadof x ,thenwewouldhavesubstituted y =r x andthatwouldhaveturnedtheintegrandintoarationalfunctionof y .Indeed, y = x1 /rimplies dy dx = x(1 /r ) Š 1 r dy dx = y1 Š r r andtherefore dx = ryr Š 1dy. Inotherwords, dx isaequalto dy timesa polynomialfunction of y so,indeed,theintegrandwillbearationalfunctionof y .44.4.Exercises.(1)Compute 5 x2+5 x +4dx (2)Compute 3 x3+ x2Š x Š 1dx (3)Compute 1 x4+5 x2+4dx (4)Compute 1 x4+4 x2+4dx (5)Compute x2+4 x3Š xdx (6)Compute x +3 x3+1dx (7)Compute x +1 x Š 3dx (8)Compute 3 x 3 x +2dx .45.StrategyofIntegration Wepresentedvariousintegrationtechniquesinthischapterand insomeprecedingchapters.Themostgeneraloneswereintegration bypartsandintegrationbysubstitution.Themostfrequentlystudied specialcaseswererelatedtotrigonometricfunctionsandtheirinverses. Reversesubstitutioncameupinsomespecialcases.Wealsodiscussed theintegrationofrationalfunctions,usingthetechniqueofpartial fractions. Inshort,wehavelearnedadecentnumberofmethods.Forthis veryreason,itissometimesnotobviouswhichmethodweshoulduse

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467.METHODSOFINTEGRATION whentryingtointegrateafunction.Whilethereisnogeneralrule,in thissectionwewillprovideafewguidelines. Solet f beafunctionthatisnotequaltooneofthefunctions whoseintegralweeitherknowohandorhaveadeterministicmethod tocompute.Thatis, f isnotapolynomial, f isnotarationalfunction, f isnotthefunction f ( x )= axor f ( x )=logax forsomepositivereal number a ,and f isnotoneofthebasictrigonometricfunctionslike sin x ortan x .Letusalsoassumethatsimplealgebrawillnothelp, thatis,that f cannotbetransformedintooneoftheseelementary functionsbysimplealgebraictransformations.Thenhowdowedecide whichmethodtouse?45.1.Substitution.Themethodthatneedstheleastamountofwork, whenitisavailable,isasimplesubstitution,soitisreasonableto trytousethatmethod“rst.Thereisaparticularlygoodchancefor thisapproachtoworkwhen f isthecompositionoftwofunctions,one ofwhichhasaconstantderivative,orwhen f isoftheform f ( x )= h( g ( x )) g( x ),sincethen f ( x )=dh dx( x ),andso f ( x ) dx = h ( x ).Inthe languageofsubstitutions,thismeansthatsubstituting y = g ( x )will work,since fdx = h( g ( x )) g( x ) dx = h( y ) dy dx dx = h( y ) dy. Inotherwords,theintegralofthecompositefunction f isturnedinto somethingsimpler,theintegralofthefunction h Example 7.21 Let f ( x )=sin2 x .Then,usingthesubstitution y =2 x ,wegetthat sin2 xdx =1 2 sin ydy = Š1 2cos y = Š1 2cos2 x Example 7.22 Let f ( x )=3 x x2+1.Thenweset y = x2+1 ,so dy/dx =2 x .Thisleadsto 3 x x2+1 dx = 3 x y dx = 3 2 1 y dy = 3 2 ln y = 3 2 ln( x2+1) Thereadershouldcomputetheintegral sinnx cos xdx atthis point.

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45.STRATEGYOFINTEGRATION47 45.2.IntegrationbyParts.If f istheproductoftwofunctions,but substitutiondoesnotseemtohelp,thenintegrationbypartsisthe logicalnextstep.Thistechniqueisparticularlyusefulwhenoneof thetwofunctionswhoseproductis f ismadesigni“cantlysimplerby dierentiation. Example 7.23 Computetheintegral xeŠ xdx Solution: Consideringtheintegrand,wenoticethatsubstitutionis unlikelytohelp,since x and eŠ xarenotcloselyrelated.Ontheother hand,theintegrandisaproduct,andoneoftheterms, x ,ismade simplerbydierentiation.Therefore,wechoosethetechniqueofintegrationbyparts,with x = u and v= eŠ x.Then u=1,while v = Š eŠ x,andweget xeŠ xdx = Š xeŠ xŠ eŠ xdx = Š xeŠ x+ eŠ x=(1 Š x ) eŠ x. 45.3.Radicals.Ifthetwomostgeneralmethods(substitutionandintegrationbyparts)arenothelpful,thenitisquitepossiblethatthereis arootsignintheintegrand.Inthatcase,therearetwospeci“cmethodsthatwecantry,reversesubstitutionandrationalizingsubstitution. Theeasiestwaytoknowwhentouseeachofthesetwomethodsisto remembertherelativelyfewcasesinwhichreversesubstitutionworks directly.Aswehaveseen,thesearetheintegralsinvolving 1 r2+ x2, r2Š x2,and x2Š r2.Iftheintegranddoesnotcontainanyof thesefunctions,thenitmaybesimplertousearationalizingsubstitution,suchasincomputingtheintegral x x +4dx .Theexercisesatthe endofthissectionwillaskthereadertodecidewhichmethodtouse forafewspeci“cexamples.45.4.IfEverythingElseFails.Ifnoneofourmethodswork,thenitmay bethatanunexpectedtransformationoftheintegrandmayhelp,at leastwithrelatingtheintegraltoonethatisnotquiteaschallenging tocompute. Example 7.24 Compute sin4x tan2xcos22 xdx .

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487.METHODSOFINTEGRATION Solution: Weusethetrigonometricidentitytan x =sin x/ cos x ,to rewritetheintegrand.Weget sin4x tan2x cos22 xdx = sin2x cos2x cos22 xdx = 1 4 sin22 x cos22 xdx = 1 16 sin24 xdx, whichiseasytointegratewiththemethodwelearnedforpowersof trigonometricfunctions. Itisimportanttopointoutthatsometimesthereisindeed nosolution ;thatis,thereexistelementaryfunctions f suchthatthereisno elementaryfunction F ( x )satisfying F( x )= f ( x ).Examplesofsuch functions f include ex2, ex/x ,and1 ln x.45.5.Exercises.Inalloftheexercisesbelow,computetheintegral. (1) x2eŠ 5 xdx (2) x x2Š 1dx (3) x +1 x3+1dx (4) esin xcos xdx (5) ex exŠ 1dx (6) 1 x2+4dx (7) 1 x +4dx (8) 1 x ln xdx (9) 1 x ln x ln(ln x )dx .46.IntegrationUsingTablesandSoftwarePackages46.1.TablesofIntegrals.TablesofintegralscanbefoundinmanycalculustextbooksandontheInternet.Thewebsite www.integraltable.com isagoodexample,andwewilluseitasareferencein thissubsection.(Whenwesaythetableofintegrals,Žwemeanthat table.) Nomatterhowextensiveatableofintegralsis,itcannotcontain all integrals.Itisthereforeimportanttoknowhowtousethesetables tocomputeintegralsthatarenotcontainedinthetablesinthesame form. Theeasiestcaseiswhentheintegraltobecomputedisa special case ofamoregeneralintegralthatisinthetable.

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46.INTEGRATIONUSINGTABLESANDSOFTWAREPACKAGES49 Example 7.25 Usethetableofintegralstocompute ln( x2+9) dx Solution: Lookingatthetableofintegralsfoundatthe website www.integral-table.com ,we“ndthattheintegrandisaspecialcase ofintegral(45),with a =3.Usingtheformulagiveninthetablefor thegeneralcasewith a =3,wegettheresult ln( x2+9) dx = x ln( x2+9)+6tanŠ 1 x 3 Š 2 x + C. Sometimes,wehavetoresorttointegrationbysubstitutiontobe abletousethetableofintegrals. Example 7.26 Usethetableofintegralstocompute 9 Š 4 x2dx Solution: Takingalookatthetableofintegrals,we“ndthatformula (30)providesaformulafor a2Š x2dx .Inordertobeabletouse thatformula,weset y =2 x ,whichimplies dy/dx =2.Therefore, 9 Š 4 x2dx = 1 2 9 Š y2dy. Nowwecanapplyformula(30)fromthetableofintegrals,toget 1 2 9 Š y2dy = y 4 9 Š y2+ 9 2 tanŠ 1 y 9 Š y2 + C = x 2 9 Š 4 x2+ 9 2 tanŠ 1 2 x 9 Š 4 x2 + C. Sometimesweneedtocarryoutsomealgebraicmanipulationbefore wecanusethetechniqueofsubstitutionsinconnectionwithusingthe tableofintegrals. Example 7.27 Usethetableofintegralstocompute x2+2 x +1 x2+2 x +10dx Solution: Thereisnointegralinthetableofintegralsthatwouldimmediatelystandoutasonethatisverysimilartothisone.Thecrucial observationisthatthesubstitution y = x +1signi“cantlysimpli“esour integrand.Thatsubstitutionleadstotheintegral y2 y2+32dy ,which, inturn,canbedirectlyfoundinthetableofintegralsasitem(36). Substituting x backintotheobtainedformula,weget x2+2 x +1 x2+2 x +10 dx = x2+2 x +10 2 Š 9 2 ln x +1+ x2+2 x +10 + C.

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507.METHODSOFINTEGRATION 46.2.SoftwarePackages.ComputersoftwarepackagessuchasMaple andMathematicaareveryusefultoolsofintegration.Thesepackages willcomputethede“niteorinde“niteintegralsofalargeclassoffunctions,thentheywillpresenttheresultsinaformthatisusually,but notalways,intheformtheuserexpected.Inthissection,weshowa fewexamplesoftheseunexpectedresultsandexplainhowtointerpret them. Tostartwithaverybasicexample,type int(x2+3x,x); intoMaple.Wegettheanswer1 3x3+3 2x2.Thisisthecorrectanswer havingconstantterm0 .Experimentingwithotherfunctions,wenote thatMaplealwaysanswersinthisway,thatis,withouttheconstant C attheend.Itisimportantnottoforgetthisifwewillbeusingthe obtainedfunctioninsomefurthercomputation. Mapledoesnotalwaysprovidethe simplestform fortheintegral thatitcomputes.Forinstance,ifweaskMapletocomputetheindefiniteintegral x ( x2+3)6dx ,wegettheoutput (7.16) 1 14 x14+ 3 2 x12+ 27 2 x10+ 135 2 x8+ 405 2 x6+ 729 2 x4+ 729 2 x2. However,itisveryeasytocompute x ( x2+3)6dx byhand,using thesubstitution u = x2.Thatsubstitutionleadstothesamesolution,butinamuchsimplerform,namely,1 14( x2+3)7.Ifwewant toverifythatthisresultindeedagreeswiththeonegivenbyMaple anddisplayedin(7.16),wecanaskMapleto expand theexpression H =1 14( x2+3)7usingthe expand command.Weseethattheexpandedexpressionindeedagreeswiththeonegivenin(7.16),upto theconstanttermsattheend. Thereareothercommandslike expand that,areusefulifwewant totransformtheoutputofanintegrationsoftwarepackage.Thecommands rationalize and simplify areexamplesofthese. Thereisoftenmorethanonewaytoexpressintegralsinvolving hyperbolicfunctions.Astrikingexampleisthefollowing.Ifweask Mapletocompute 1 1 Š x2dx bytyping int(1/(1-x2),x); MaplereturnstheanswertanhŠ 1x .Thisisverysurprising,sinceitis notdiculttointegratetheintegrandasarationalfunction,withno

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46.INTEGRATIONUSINGTABLESANDSOFTWAREPACKAGES51 hyperbolicfunctionsinvolved.Indeed, 1 1 Š x2= 0 5 x +1 Š 0 5 x Š 1 andso (7.17) 1 1 Š x2dx = 1 2 ln( x +1) Š 1 2 ln( x Š 1)=ln x +1 x Š 1 TheresultobtainedbyMapleactually agrees withtheresultgiven in(7.17),evenifthatisnotobvious.Indeed,if y =tanhŠ 1x ,then, byde“nition, x =( eyŠ eŠ y) / ( ey+ eŠ y).Solvingthisequationfor y isnotcompletelytrivial,butattheend,ityields y =ln x +1 x Š 1. ( Hint: Multiplyboththenumeratorandthedenominatorby eytoget x =( e2 yŠ 1) / ( e2 y+1).) Finally,whencomputingde“niteintegrals,Maplesometimesanswersbyusingtheacronymsofsomerarefunctions.Forinstance,if wewanttocompute 1 0sin( x2) dx bytyping int(exp(sin(x2)),x=0..1). thenwegettheanswer (1/2)*FresnelS(sqrt(2)/sqrt(Pi))*sqrt(2)*sqrt(Pi). Here FresnelS referstotheFresnelsineintegral,aconceptbeyond thescopeofthisbook.Ifwesimplywanttoknowanumericalvalue for 1 0ex2dx ,wecantype evalf(int(sin(x2),x=0..1)) instead.Mapleoutputs0.4596976941astheanswer.46.3.Exercises.(1)Usethetableofintegralstocompute x ln(3 x +5) dx (2)Usethetableofintegralstocompute x cos 2Š x dx (3)Usethetableofintegralstocompute x +7 x+4 x +5dx (4)UsetheTableofIntegralstocompute x2 x6+16dx (5)Useyourfavoritesoftwarepackagetocompute xexdx x2exdx ,and x3exdx .Doyouseeapattern?Trytoguess what x4exis,thenverifyyourguessbyusingyoursoftware packageagain. (6)Let f ( x )=1 1 Š x2andlet g ( x )= 1 1 Š x2.Clearly, f ( x )= g ( x )forall x wherethesefunctionsarede“ned.Compute theintegralofbothfunctionswithMaple.Iftheresultsseem dierent,showthattheyareinfactequal.

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527.METHODSOFINTEGRATION 47.ApproximateIntegration Sometimesitisnotpossibleto“ndtheexactvalueofade“nite integral b af ( x ) dx .Itcouldhappenthatwecannot“ndtheantiderivativeof f ( x )orthattheantiderivativeof f ( x )isnotanelementary function.Oritcouldhappenthat f itselfisnotgivenbyaformula, butinstead, f isgivenbyitsgraph,whichisplottedbyacomputer program.Inthiscase,weresorttomethodsof approximateintegration .47.1.BasicApproximationMethods.Thekeyobservationbehindthe approximationmethodsisthefactthatif f ( x ) 0for x [ a,b ],then b af ( x ) dx isequaltotheareabelowthegraphofthefunction f on thatinterval.Moreprecisely, b af ( x ) dx isequaltotheareaofthe domainborderedbythehorizontalaxis,theverticallines x = a and x = b ,andthegraphof f Inordertoestimatetheareaofthisdomain D ,wecut D intosmall verticalstrips.Todoso,wechooserealnumbers a = x0
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47.APPROXIMATEINTEGRATION53 (iii)Themidpointapproximationshows b af ( x ) dx ni =1( xiŠ xi Š 1) f ( xi) Astheaboveformulassuggest,itwillbeparticularlyeasytowork withtheseformulasifthepoints x1,x2,...,xn Š 1splittheinterval[ a,b ] into equal parts,sinceinthatcase xiŠ xi Š 1=( b Š a ) /n forall i Example 7.28 Findtheapproximatevalueof 1 0ex2dx Solution: Letususetheleft-endpointmethodwith n =4and x1, x2, and x3splitting[0 1]intofourequalparts.Thatmeansthat x1=1 / 4, x2=1 / 2,and x3=3 / 4.Then(7.18)implies 1 0ex2dx 1 4 e0+ e1 / 16+ e1 / 4+ e9 / 16 1 2759 since xiŠ xi Š 1=1 / 4forall i Ifweusetheright-endpointmethod,withthesamesetofpoints xi,weget 1 0ex2dx 1 4 e1 / 16+ e1 / 4+ e9 / 16+ e 1 7055 Itisnotsurprisingthatthesecondmethodyieldsthelargerresult, sincetheintegrand ex2isan increasingfunction ,so f ( xi) >f ( xi Š 1). Figure7.4. Left-endpointmethod.

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547.METHODSOFINTEGRATION Figure7.5. Right-endpointmethod. Furthermore,foreachpoint x [ xi Š 1,xi],wehave f ( xi Š 1) f ( x ) f ( xi).Sotheleft-endpointmethod underestimates theareaofeach strip Si,whiletheright-endpointmethod overestimates it.Therefore, theareaof D „andhencethecorrectvalueof 1 0ex2dx „isbetween thetwovaluesof1.2759and1.7055computedabove. Replacingthevalueof n =4bysomelargernumberwillresultin amorepreciseapproximation(andmorework).Usingthemidpoint methodwillresultinanapproximation A thatisclosertotheactual valueoftheintegral 1 0ex2dx ,butitisnotcompletelyobviousfrom whichside A approximates 1 0ex2dx ,thatis,whether A< 1 0ex2dx or A> 1 0ex2dx .47.2.MoreAdvancedApproximationMethods. 47.2.1.TrapezoidMethod.Ifthedierencebetween f ( xi Š 1)and f ( xi) islarge,thenestimatingtheareaof Sibyusingrectanglescouldlead tolargeerrors.Amorere“nedapproachistoestimatetheareaof Sibycomputingtheareaofthe trapezoid whoseverticesarethepoints ( xi Š 1, 0),( xi, 0), f ( xi),and f ( xi Š 1).Weknowthattheareaofthis trapezoidistheaveragelengthofitsparallelsidestimesthedistance ofthoseparallelsidesfromeachother,thatis,( f ( xi)+ f ( xi Š 1))( xiŠ xi Š 1) 2.

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47.APPROXIMATEINTEGRATION55 Figure7.6. Trapezoidmethod. Summingoverallpossiblevaluesof i ,wegetanestimateforallthe areaof D ,thatis,for b af ( x ) dx .Indeed,weobtaintheformula b af ( x ) dx =ni =1( f ( xi)+ f ( xi Š 1))( xiŠ xi Š 1) 2 Inparticular,ifthe xiarechosensothattheysplittheinterval[ a,b ] into n equalparts,thenthelastdisplayedequationsimpli“esto b af ( x ) dx = b Š a 2 nni =1( f ( xi)+ f ( xi Š 1)) = b Š a 2 n ( f ( a )+2 f ( x1)+2 f ( x2)+ +2 f ( xn Š 1)+ f ( b )) Notethat f ( x0)= f ( a )and f ( xn)= f ( b )occuronlyonceinthe suminthelastlinesince a and b areeachpartofonlyoneofthe intervals[ xi Š 1,xi]. Example 7.29 Usethetrapezoidmethodwith n =4 to“ndthe approximatevalueof 1 0ex2dx Solution: Wewillhave x1=1 / 4, x2=1 / 2,and x3=3 / 4,justasin Example7.28.Thisyields 1 0ex2dx = 1 8 e0+2 e1 / 16+2 e1 / 4+2 e9 / 16+ e =1 4907

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567.METHODSOFINTEGRATION Figure7.7. Trapezoidmethod. Thealertreadermayhavenoticedthattheresultweobtainedis preciselytheaverageoftheleft-endpointandright-endpointapproximationsweobtainedforthesameintegralintheprevioussection.(This meansthatitisabetterapproximationthanatleastoneofthetwo earlierones.)Thisisnotanaccident,andinExercise47.4.5,thereader willbeaskedtoprovethat,undercertainconditions,thisphenomenon willalwaysoccur.47.2.2.SimpsonsMethod.Asimilarmethodis Simpsonsmethod ,in whichweuseparabolasinsteadofstraightlinesforapproximation.For simplicity,letusnowassumethatthepoints xisplittheinterval[ a,b ] into n equalparts,thatis, xiŠ xi Š 1=( b Š a ) /n .Inordertosimplify thenotation,letusset yi= f ( xi). Foranyinteger i [1 ,n Š 1],considerthepoints( xi Š 1,yi Š 1),( xi,yi), and( xi +1,yi +1).Thereisexactlyoneparabola pioftheform y = Ax2+ Bx + C thatcontainsthesethreepoints.Itcanthenbeproved thattheareaunderthatparabola„moreprecisely,theareaofthe domain Piborderedbythehorizontalaxis,theverticallines x = xi Š 1and x = xi +1and pi„isequalto (7.19) b Š a 3 n ( yi Š 1+4 yi+ yi +1) Ifwesummed(7.19)overallpossiblevaluesof i ,wewouldnotgeta goodestimate,sincemostpointsofthedomainunderthecurvewould

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47.APPROXIMATEINTEGRATION57 bepartof two ofthe Pi.Forinstance,apointwithahorizontalcoordinatebetween xiand xi +1ispartofboth Piand Pi +1.Therefore, wesumthelastdisplayedequationoverall even valuesof i ,and,accordingly,westipulatethat n bean even number.Thisleadstothe followingestimate. Theorem 7.1(SimpsonsMethod) Let n bean even positiveinteger.Then b af ( x ) dx b Š a 3 n ( y0+4 y1+2 y2+4 y3+ +2 yn Š 2+4 yn Š 1+ yn) Example 7.30 UseSimpsonsmethodwith n =4 toapproximate 1 0sin( x2) dx Solution: Wehave y0=0, y1=sin(1 / 16), y2=sin(1 / 4), y3= sin(9 / 16),and y4=sin1.SoSimpsonsmethodyields 1 0sin x2 dx 1 12 (4sin(1 / 16)+2sin(1 / 4)+4sin(9 / 16)+sin1) 0 31 Figure7.8. Simpsonsmethod.

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587.METHODSOFINTEGRATION Notethatthisresultcon“rmsourintuitioninthatif x [0 1],then sin( x2) sin x ,andso 1 0sin x2 dx 1 0sin x =1 Š cos1 0 4597 .47.3.BoundsontheErrorTerm.Theerrorterm E ofanapproximation isthedierencebetweenthenumberobtainedbytheapproximation andtheactualvalueofthequantitythatwasapproximated.(Inthis section,thatactualvalueisthevalueof b af ( x ) dx .)Itgoeswithout sayingthatthesmallertheabsolutevalueoftheerrorterm,thebetter theapproximationis. The“eldof numericalanalysis studiestheerrortermsofapproximationmethods.Thetechniquesofthat“eldyieldvariousbounds onerrorterms.Wecollectedsomeoftheseboundsinthefollowing theorem. Theorem 7.2 Let f beatwice-dierentiablefunctionon [ a,b ] such that | f( x ) | M if x [ a,b ] .Thenthefollowingholdfortheapproximationmethodsusedtocompute b af ( x ) (a) If ETistheerrortermofthetrapezoidmethod,then | ET| M ( b Š a )3 12 n2. (b) If EMistheerrortermofthemidpointmethod,then | EM| M ( b Š a )3 24 n2. (c) If,forall x [ a,b ] ,thenumber f(4)( x ) isde“nedandisat mostaslargeastheconstant K ,and ESistheerrortermof Simpsonsmethod,then | ES| K ( b Š a )5 180 n4. Comparingtheformulasofparts(a)and(b)oftheprevioustheorem,wecanconcludethatthe worst-case scenarioofthemidpoint methodisbetterthantheworst-casescenarioofthetrapezoidmethod. This,ofcourse,doesnotmeanthatthemidpointmethodis always betterthanthetrapezoidmethod. Example 7.31 FindanupperboundfortheapproximationobtainedinExample7.29.

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48.IMPROPERINTEGRALS59 Solution: Weapplypart(a)ofTheorem7.2.Wehave f ( x )= ex2, so f( x )= ex2(4 x2+2).Thisisanincreasingfunctionon[0 1],soits maximumistakenat x =1,showingthat | f( x ) | 6 e .AsinExample 7.29,wechose n =4,sotheprevioustheoremyields | ET| M ( b Š a )3 12 n2= 6 e 12 16 0 085 47.4.Exercises.(1)Use n =4andthemidpointmethodto“ndtheapproximate valueof 1 0ex2dx (2)Use n =4andthetrapezoidmethodto“ndtheapproximate valueof 1 0eŠ x2dx (3)Use n =4andSimpsonsmethodto“ndtheapproximate valueof 1 0sin x2dx (4)Whatvalueof n shouldweuseineachoftheprecedingthree exercisestogetanerrortermthatislessthan10Š 6? (5)Letusassumethatthepoints x1,x2,...,xn Š 1usedforapproximateintegrationsplittheinterval[ a,b ]into n equalsegments. Provethattheresultobtainedbythetrapezoidmethodwillbe theaverageoftheleft-endpointmethodandtheright-endpoint method. (6)Howlargeistheerrortermoftheapproximationinexercise1 intheworstcase?48.ImproperIntegrals Inourstudiesofintegration,wehavenotdealtwithde“niteintegralsoverin“niteintervals,nordidweintegratefunctionsoveran intervalifthefunctionwasnotde“nedineverypointofthatinterval. Inthissection,wewillconsiderde“niteintegralsofthesekinds,which arecalled improperintegrals .48.1.In“niteIntervals.For“niteintervals,wehaveidenti“ed b af ( x ) dx withtheareaofthedomainlimitedbythegraphof f ,theverticallines x = a and x = b ,andthehorizontalaxis.Welearnedthat,bythefundamentaltheoremofcalculus,theequality b af ( x ) dx = F ( b ) Š F ( a ) holds,where F isanantiderivativeof f Nowletusconsidertheintegralof f overthe in“nite interval[ a, ). Recallingthat b af ( x ) dx isequaltoacertainarea,weintuitivelywant af ( x ) dx toequaltheareaofthedomainborderedbytheline x = a ,

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607.METHODSOFINTEGRATION Figure7.9. Areaunderthecurve y = f ( x )from x = a to x = b thehorizontalaxis,andthegraphof f .Notethatthisareamaybea “nitenumber,evenifitisnotsqueezedbetweentwoverticallines.One exampleofthisiswhen f ( x )=0if x>N forsomerealnumber N Timehascometoformallyde“ne af ( x ) dx Definition 7.1 Let f beafunction.Iftheintegral b af ( x ) dx existsforall b>a and limb b af ( x ) dx = L existsasa(“nite)real number,thenwesaythattheintegral af ( x ) dx is convergent ,and wewrite af ( x ) dx = L If limb b af ( x ) dx doesnotexistorisin“nite,thenwesaythat af ( x ) dx is divergent Notethatif F isanantiderivativeof f ,then limb b af ( x ) dx =limb ( F ( b ) Š F ( a )) (limb F ( b )) Š F ( a ) Therefore,theintegral af ( x ) dx isconvergentifandonlyiflimb F ( b )existsandis“nite. Example 7.32 Let f ( x )= xŠ 2.Compute 1f ( x ) dx .

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48.IMPROPERINTEGRALS61 Figure7.10. Areaunderthecurve y = f ( x )from1to Solution: Wehave 1xŠ 2dx =limb b 1xŠ 2dx =limb Š xŠ 1b 1=limb Š 1 b +1 =1 Inparticular, 1f ( x ) dx isconvergent. Encouragedbythesimplesolutionofthelastexample,wearegoing tocomputethemoregeneralintegral 1xrfor any realnumber r Example 7.33 Let f ( x )= xr.Compute 1f ( x ) dx Solution: Letus“rstassumethat r = Š 1.Thenwehave 1xrdx =limb b 1xrdx =limb 1 r +1 xr +1b 1. If r> Š 1,then r +1 > 0andlimx xr +1= ,sothelimitinthe lastdisplayedrowisin“nite,andhence 1xrdx isdivergent. If r< Š 1,then r +1 < 0andlimx xr +1=0,sothelimitinthe lastdisplayedrowisequalto1 r +1,andhence 1xrdx isconvergent.

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627.METHODSOFINTEGRATION If r = Š 1,thenweneedtocompute 1xrdx dierently,since,in thatcase, xadx =xa +1 a +1.Instead,wehave 1xŠ 1dx =limb b 1xŠ 1dx =limb [ln x ]b 1=limb ln b = Therefore, 1xŠ 1dx isdivergent. Notethattheresultsofthepreviousexampleprovethefollowing importanttheorem. Theorem 7.3 Let r bearealnumber. (i) If r Š 1 ,then 1xrdx isdivergent. (ii) If r< Š 1 ,then 1xrdx isconvergent. Thefollowingde“nitionisnotverysurprising.Itisthecounterpart ofDe“nition7.1. Definition 7.2 Let f beafunctionandlet b bearealnumber suchthat,forallrealnumbers a
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48.IMPROPERINTEGRALS63 Figure7.11. m Šf ( x ) dx isblue,while mf ( x ) dx isorange. Figure7.12. 0 ŠeŠ xdx Solution: Weset m =0andapplyDe“nition7.3.Wegetthat ŠeŠ xdx isconvergentifbothof 0 ŠeŠ xdx and 0eŠ xdx areconvergent.However, 0 ŠeŠ xdx =lima Š[ Š eŠ x]0 a=1+ isdivergentandthereforesois ŠeŠ x. Figure7.12showsthedomainwhoseareaisequalto 0 ŠeŠ xdx Thereadercouldaskhowweknewthatweneededtoselect0,and notsomeotherrealnumber,fortheroleof m ,thatis,tosplitthe realnumberlineintotwoparts.Theansweristhatwedidnot,and otherchoicesof m wouldhavegiventhesameresultsincetheintegrand

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647.METHODSOFINTEGRATION convergestoin“nityas x goestonegativein“nity.Wechose m =0 becauseitwasconvenienttodoso. Notethatallimproperintegralsdiscussedinthissectionarecalled Type1improperintegrals .48.2.VerticalAsymptotes.Sometimeswemaywanttocomputethe integralofafunction f ona“niteinterval[ a,b ]sothatinsomepoint c [ a,b ],thefunction f hasaverticalasymptote.Anexampleisthe function f ( x )=1 / ( x2Š 4)ontheinterval[1 3].Inthiscase,weuse thetechniqueoflimitstoformallyde“ne b af ( x ) dx ,aswedidinthe previoussection. Definition 7.4 Let f beafunctionthatiscontinuouson [ a,b ] exceptforonepoint c [ a,b ] .Thenweset (7.20) c af ( x ) dx =limt cŠt af ( x ) dx and (7.21) b cf ( x ) dx =limt c+b tf ( x ) dx. Furthermore,ifbothofthetwolimitsdisplayedaboveexistandare “nite,weset b af ( x ) dx = c af ( x ) dx + b cf ( x ) dx. Notethatiftheonlypoint c inwhich f isnotcontinuousisone oftheendpointsof[ a,b ],thenweonlyhavetocomputeoneof(7.20) and(7.21),sincetheotherintegralistakenoveratrivialintervaland ishencezero. Example 7.35 Compute 1 0xŠ 1 / 2dx Solution: Astheonlypointin[0 1]inwhich f ( x )= xŠ 1 / 2isnot continuousis0,weuseformula(7.21)with c =0and b =1. Weget 1 0xŠ 1 / 2dx =limt 0+1 txŠ 1 / 2dx =limt 0+[2 x1 / 2]1 t=2 Š limt 0+t1 / 2=2 Š 0 =2 .

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48.IMPROPERINTEGRALS65 Figure7.13. 1 0xŠ 1 / 2dx Sotheintegral 1 0xŠ 1 / 2dx isconvergent. Example 7.36 Compute 4 Š 1xŠ 2dx Solution: WeapplyDe“nition7.4sincetheinterval[ Š 1 4]hasone point, c =0,wheretheintegrandisnotcontinuous.Therefore, 4 Š 1xŠ 2dx = 0 Š 1xŠ 2dx + 4 0xŠ 2dx =limt 0Št Š 1xŠ 2dx +limt 0+4 txŠ 2dx =limt 0Š[ Š xŠ 1]t Š 1+limt 0+[ Š xŠ 1]4 t= + = Sotheintegralinquestionisdivergent. Figure7.14showsthedomainwhoseareaisequalto 4 0xŠ 2dx and thecorrectwayofbreakingthatintervaluptotwoparts. Notethatwewouldhavereachedthe wrong conclusionifwehad disregardedthefactthat xŠ 2isnotcontinuousat x =0andtriedto applythefundamentaltheoremofcalculus.Indeed,inthatcase,we wouldhaveobtainedthe wrong result:[ Š xŠ 1]4 Š 1=Š 1 4Š 1= Š5 4.This resultisincorrect,andtheincorrectstepwastoapplythefundamental theoremofcalculusforafunctionthatisnotcontinuousintheentire intervalofintegration. Theintegralsthatwehavediscussedinthissectionarecalled Type 2improperintegrals .

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667.METHODSOFINTEGRATION Figure7.14. 4 Š 1xŠ 2dx .48.3.FurtherRemarks. 48.3.1.ImproperIntegralsofMixedType.Therearesomeintegralsthat areimproperfortworeasons.Theyaretakenoveranin“niteinterval, andthatintervalcontainsapointinwhichthefunctionisnotcontinuous.Inthatcase,wesplituptheintervalofintegrationsothatnow wehavetwointegrals,oneofwhichisofType1andtheotherofwhich isType2. Example 7.37 Compute 0 1 ( x Š 2)2dx Solution: Webreakuptheinterval[0 )totheunionofthetwo intervals[0 2]and[2 ),getting 01 ( x Š 2)2dx = 3 01 ( x Š 2)2dx + 31 ( x Š 2)2dx. The“rsttermontheright-handsideisanimproperintegralofType 2,andthesecondtermontheright-handsideisanimproperintegral ofType1.Wecancomputebothbythemethodspresentedearlierin thissection. 48.3.2.ComparisonTest.Comparisontestsforimproperintegralswork verysimilarlytothoseforproperintegrals. Theorem 7.4 Letusassumethat,forall x a ,thechainof inequalities 0 f ( x ) g ( x ) holds. (i) If af ( x ) dx isdivergent,thensois ag ( x ) dx (ii) if ag ( x ) dx isconvergent,thensois af ( x ) dx Example 7.38 Showthat 3 1 x2ln xdx isconvergent.

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48.IMPROPERINTEGRALS67 Figure7.15. 0 1 ( x Š 2)2dx Figure7.16. 3 1 x2ln xdx Solution: If x 3,thenln x> 1,so x2ln x>x2,andtherefore theintegrandislessthan1 x2.SeeFigure7.16foranillustration.On theotherhand,weknowthat 31 /x2dx isconvergent,soourclaim followsfromthecomparisontest. 48.4.Exercises.(1)Is 1sin xdx convergentordivergent? (2)Is 0 5xŠ 1 5dx convergent? (3)Is 3 1 x Š 2dx convergent? (4)Is ŠxŠ 2dx convergent? (5)Is 0xeŠ xdx convergent? (6)Is ŠxeŠ x2dx convergent? (7)Is 0 1 ex+ xdx convergent?

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CHAPTER8 SequencesandSeries 49.In“niteSequences Asequencecanbethoughtofasanorderedlistofnumbers a1,a2, ...,an,an +1,... .Thesubscript n indicatesthepositionofanumber anin thesequence;forexample, a1isthe“rstelement, anisthe n thelement, andsoon. Definition 8.1(Sequence) A sequence isafunction f de“nedon thesetofallpositiveintegers;thatis,itisarulethatassignsanumber toeachpositiveinteger.If f ( n )= anfor n =1 2 ,... ,itiscustomary todenotetherangeof f bythesymbol { an} or { an} 1. Soasequencecanbede“nedbyspecifyingtherule an= f ( n )to calculatethe n thtermfromaninteger n .Forexample, an= n n +1 n n +1 1= 1 2 2 3 3 4 ,... an= ( Š 1)n n ( Š 1)n n 1= Š 1 1 2 Š 1 3 1 4 ,... an= qn{ qn Š 1} 0= { 1 ,q,q2,q3,... } Sequencescanalsobede“nedrecursively,thatis,byarelationthatallowsusto“nd anif am, m
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708.SEQUENCESANDSERIES Figure8.1. Setofpointsonthegraph y = x/ ( x + 1)correspondingtointegervalues x = n .Forlarge x x/ ( x +1)approaches1frombelow,andhence n/ ( n +1)= 1 / (1+1 /n ) 1as n .Thedierence1 Š n/ ( n +1)= 1 / ( n +1)canbemadesmallerthanany(small)number > 0for all n>N andsomeinteger N decreaseswithincreasing n andhencecanbemadesmallerthanany preassignedpositivenumber forall n>N ,where N dependson Forexample,put =10Š 2.Thenthecondition1 Š an< impliesthat 1 / ( n +1) < or1 / Š 1 99.If =10Š 4,then1 Š an< 10Š 4forall n>N =9999.In otherwords,nomatterhowsmall is,thereisonlya “nite numberof elementsofthesequencethatlieoutsidetheinterval(1 Š 1+ ).In thiscase,thesequenceissaidtoconvergetothelimitvalue1. Definition 8.2(LimitofaSequence) Asequence { an} hasthe limit a if,forevery > 0 ,thereisacorrespondinginteger N suchthat | anŠ a | < forall n>N .Inthiscase,thesequenceissaidtobe convergent ,andonewrites limn an= a or an a as n Ifasequencehasnolimit,itiscalled divergent Onecansaythat asequence { an} convergestoanumber a ifand onlyifeveryopenintervalcontaining a hasallbut“nitelymanyofthe elementsof { an} Theorem 8.1(UniquenessoftheLimit) Thelimitofaconvergent sequenceisunique: limn an= a andlimn an= a= a = a.

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49.INFINITESEQUENCES71 Figure8.2. De“nitionofthelimitofasequence.The dotsindicatenumericalvalues an(verticalaxis).The integer n increasesfromlefttoright(horizontalaxis). Theconvergenceof antoanumber a meansthat,for anysmall > 0,thereisaninteger N suchthat all the numbers an, n>N ,lieintheinterval( a Š ,a + ). Itisclearthat N dependson .Generally,asmaller requiresalarger N Proof. Fix > 0.Then,bythede“nitionofthelimit,thereare numbers N and Nsuchthat | anŠ a | if n>N and | anŠ a| if n>N.Hence,bothinequalitiesholdfor n> max( N,N)andforall such n : 0 | a Š a| = | a Š an+ anŠ a|| anŠ a | + | anŠ a| < 2 ; thatis,the nonnegative number | a Š a| issmallerthan any preassigned positivenumber,whichmeansthat | a Š a| =0or a = a. Sinceasequenceisafunctionde“nedonallpositiveintegers,there isagreatdealofsimilaritybetweentheasymptoticbehaviorofafunction f ( x )as x andasequence an= f ( n ). Theorem 8.2(LimitsofSequencesandFunctions) Let f bea functionon (0 ) .Supposethat limx f ( x )= a .If an= f ( n ) where n isaninteger,then limn an= a Thevalidityofthetheoremfollowsimmediatelyfromthede“nition ofthelimitlimx f ( x )= a (i.e.,given > 0,thereisacorresponding number M suchthat | f ( x ) Š a | < forall x>M )bynotingthatthe rangeof f ( x )containsthesequence an= f ( n ). Example 8.1 Findthelimitofthesequence an=ln n/n ifitexists orshowthatthesequenceisdivergent.

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728.SEQUENCESANDSERIES Solution: Considerthefunction f ( x )=ln x/x suchthat an= f ( n ) forallpositiveintegers.Hence, limn ln n n =limx ln x x =limx 1 /x 1 =0 wheretheindeterminateform arisingfromln x/x as x has beenresolvedbymeansoflHospitalsrule.NotethatlHospitalsrule appliesnottosequencesbuttofunctionsofarealvariable. Followingtheanalogybetweenthelimitsofsequencesandfunctions,onecanselectaparticularclassofdivergentsequences. Definition 8.3(In“niteLimits) Thelimit limn an= means that,foreverypositivenumber M ,thereisacorrespondinginteger N suchthat an>M forall n>N .Similarly,thelimit limn an= Š meansthat,foreverynegativenumber M ,thereisacorresponding integer N suchthat anN Example 8.2 Analyzetheconvergenceofthesequence an=1 /np, where p isreal. Solution: Put f ( x )=1 /xpfor x> 0.Then an= f ( n )andtherefore limn 1 np=limx 1 xp= 0if p> 0 1if p =0 if p< 0 Example 8.3 Analyzetheconvergenceofthesequence an= qn, n =0 1 ,... ,where q isreal. Solution: Suppose q> 0.Put f ( x )= qx= ex ln q.Fromtheproperties oftheexponentialfunction,itfollowsthat eax if a =ln q> 0, eax=1if a =ln q =0,and eax 0if a =ln q< 0.Therefore, an if q> 1, an=1 1if q =1,and an 0if0
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49.INFINITESEQUENCES73 Thus, limn qn= 0if q ( Š 1 1) 1if q =1 if q> 1 andthesequencedoesnotconvergeif q Š 1. 49.2.Subsequences.Givenasequence { an} ,considerasequence { nk} ofpositiveintegerssuchthat n1 0and an> 0 Thesqueezetheoremalsoappliestosequences. Theorem 8.3(SqueezeTheorem) If cn an bnfor n>N and limn bn=limn cn= a ,then limn an= a ,where a canalsobe Example 8.4 Findthelimitof an=sin( / n ) Solution: Since Š x sin x x if x 0,onehas cn= Š / n an / n = bn,where cn 0and bn 0as n .Bythesqueeze theorem,sin( / n ) 0as n Theorem 8.4 If limn | an| =0 ,then limn an=0 .

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748.SEQUENCESANDSERIES Figure8.3. Thesqueezetheorem.Thedotsindicatenumericalvalues(verticalaxis)ofthesequences bn(blue), cn(black),and an(red).Theinteger n increases fromlefttoright(horizontalaxis).Thesequences bnand cnconvergetoanumber a .Thismeansthatthedierences | bnŠ a | and | cnŠ a | canbemadearbitrarilysmallfor all n N andsomeinteger N .Since cn an bn,the dierence | anŠ a | isalsoarbitrarilysmallforall n N Bythede“nitionofthelimit,thesequence anmustconvergeto a ,too. Thistheoremfollowsdirectlyfromthede“nitionofthelimitofa sequencewhere a =0. Theorem 8.5 If an a as n andthefunction f iscontinuousat a ,then limn f ( an)= f ( a ) Thistheoremassertsthatifacontinuousfunctionisappliedtothe termsofaconvergentsequence,theresultisalsoconvergent. Proof. Thecontinuityof f at a meansthatlimx af ( x )= f ( a ) or,bythede“nitionofthislimit,forany > 0,thereisacorresponding > 0suchthat | f ( x ) Š f ( a ) | < whenever | x Š a | < .Havingfound such ,put = and,bythede“nitionofthelimitlimn an= a foranysuch > 0,thereisacorrespondinginteger N suchthat | anŠ a | <= if n>N .Therefore,forany > 0,onecan“nda correspondinginteger N suchthat | f ( an) Š f ( a ) | < forall n>N whichmeansthatlimn f ( an)= f ( a ).

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49.INFINITESEQUENCES75 Example 8.5 Findthelimitofthesequence an=exp(1 /n2) Solution: Considerthesequence bn=1 /n2.Then limn bn=limx 1 x2=0 Put f ( x )= eŠ x.Then an= f ( bn).Bycontinuityoftheexponential function, limn an=exp( Š limn bn)= e0=1 49.4.Exercises.(1)Findaformulaforthegeneralterm anofthesequences: { an} = 1 Š 1 3 1 5 Š 1 7 1 9 Š 1 11 ,... { an} = 1 1 2 1 4 1 8 1 16 ,... { an} = Š 1 4 2 7 Š 3 10 4 13 Š 5 16 ,... In(2)…(13),determinewhetherthesequenceconvergesordiverges.Ifitconverges,“ndthelimit. (2) an=2n. (3) an=2nŠ ( Š 1)n2n. (4) an=(3 Š 5 n2) / (1+ n2). (5) an=tan[ n/ (2+4 n )]. (6) an=sin2[ ( n2+2) / (2 n2+5)]. (7) an=ln( an ) / ln( bn ),where a and b arepositivenumbers. (8) an= npeŠ n,where p isreal. (9) an= n cos(1 /n ). (10) an= ( n3+1) / (8 n3+4 n2+2 n +1). (11) an=(ln n )p/n ,where p> 0. (12) an=tanŠ 1( n2). (13) an= n2+ n Š n .

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768.SEQUENCESANDSERIES 50.SpecialSequences Theorem 8.6(SpecialSequences) Let p and q berealnumbers. limn n p =1if p> 0 (8.1) limn n n =1 (8.2) limn nq pn=0if p> 1 (8.3) limn n nn=0 (8.4) limn qn n =0 (8.5) Proof. (8.1) .If p> 1,put an=n p Š 1.Then an> 0and,bythebinomial theorem, 1+ nan (1+ an)n= p. Notethatalltermsin(1+ an)n=1+ nan+ n ( n Š 1) an/ 2+ + nan Š 1 n+ an narepositive.So,byretainingonlythe“rsttwoterms,asmallernumber isobtained,thatis,1+ nan (1+ an)n.Itfollowsfromthisinequality that 0 1. (8.2) .Put an=n n Š 1.Then an 0and,bythebinomialtheorem, n =(1+ an)n n ( n Š 1) 2 a2 n. Hence,for n 2, 0 an 2 n Š 1 Bythesqueezetheorem, an 0orn n = an+1 1as n 0. (8.3) .Considerthefunction f ( x )= xqeŠ cx,where c> 0.Bythe asymptoticpropertyoftheexponentialfunction, f ( x ) 0as x forany q ;theexponentialgrowsfasterthananypowerfunction(which

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50.SPECIALSEQUENCES77 hasbeenprovedinCalculusI).Since an= f ( n )for c =ln p> 0if p> 1,oneconcludesthat limn nq pn=limn nqeŠ n ln p=limx xqeŠ x ln p=0 (8.4) .Thefollowinginequalityholds: an= n nn= 1 2 3 n n n n n = 1 n 2 n 3 n n n 1 n 0 0,thenthereisapositiveinteger k suchthat k Š 1 q 1 ,ifitexists orshowthatthesequencediverges. Solution: limn n nq=limn (n n )q=(limn n n )q=1q=1 by(8.2)andthebasiclimitlaws. 50.1.MonotonicSequences.Definition 8.4(MonotonicSequences) Asequence anissaidto be monotonicallyincreasingif an an +1, monotonicallydecreasingif an an +1forall n =1 2 ,... Theclassof monotonic sequencesconsistsoftheincreasingandthe decreasingsequences. Example 8.7 Showthatthesequence an= n/ ( n2+1) ismonotonicallydecreasing.

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788.SEQUENCESANDSERIES Figure8.4. Thesequenceontheleftismonotonically increasing,andthesequenceontherightismonotonicallydecreasing. Solution: Theinequality an an +1mustbeestablished.Itisequivalenttothefollowinginequalitiesobtainedbycross-multiplication: n +1 ( n +1)2+1 n n2+1 ( n +1)( n2+1) n [( n +1)2+1] n3+ n2+ n +1 n3+2 n2+2 n 1 n2+ n. Thelatterinequalityistruefor n 1.Therefore, an +1 an(infact, thestrictinequality an +1 0, a Š isnotanupperboundof S .Theleastupperboundiscalledthe supremum of S anddenoted sup S .Naturally,sup S M foranyupperbound M of S .If S has alowerbound m ,thenitalsohasthe greatestlowerbound ,denoted

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50.SPECIALSEQUENCES79 Figure8.5. Aboundedsequence.Thedotsindicate numericalvaluesof an(verticalaxis).Theinteger n increasesfromlefttoright(horizontalaxis).Allthenumbers anlieintheinterval: m an M inf S (the in“mum of S ).Thenumberinf S isalowerboundof S such thatinf S + isnotalowerboundof S foranypositive > 0;that is, m inf S foranylowerboundof S .Thecompletenessaxiomisan expressionofthefactthatthereisnogaporholeintherealnumber line. Theorem 8.7(MonotonicSequenceTheorem) Suppose { an} is monotonic.Then { an} convergesifandonlyifitisbounded. Proof. Suppose an an +1(theproofisanalogousintheother case).Let S betherangeof { an} .If { an} isbounded,let a bethe leastupperboundof S (itexistsbythecompletenessaxiom).Then an a forall n 1.Forevery > 0,thereisaninteger N suchthat a Š
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808.SEQUENCESANDSERIES Figure8.6. Monotonicsequencetheorem.Abounded monotonicsequencewithnumericalvaluesindicatedby dots(verticalaxis).Theinteger n increasesfromleftto right.If S =supn{ an} istheleastupperboundofall an, then,foranynumber > 0, S Š isnotanupperbound ofthesequence.Since anincreasesmonotonically,there isaninteger N suchthat all thenumbers an, n>N ,are greaterthan S Š andhencelieintheinterval( S Š ,S ). Thismeansthat anconvergesto S ismathematicalinduction.Thestatementistruefor n =1.Suppose thatthestatementistruefor n = k ,thenonehastoprovethatthe statementisalsotruefor n = k +1.Iftheproofgoesthrough,then startingwith n =1,onecanestablishthestatementfor n =2, n =3, ansoon.Thisisthebasicideaofmathematicalinduction.Usingthe recurrencerelation, ak +1>ak= 1 2 ( ak +1+3) > 1 2 ( ak+3)= ak +2>ak +1. Thus,thesequenceisindeedmonotonicallyincreasing.Ifithappensto bebounded,thenitconverges.Again,mathematicalinductionturns outtobehelpful.The“rsttermssuggestthat an< 3.Thisistruefor n =1.Supposetheinequalityistruefor n = k .Letustrytoprove thatthisassumptionimpliesthattheinequalityholdsfor n = k +1. Usingtherecurrencerelation, ak< 3= 1 2 ( ak+3) < 1 2 (3+3)= ak +1< 3 Thus,thesequenceismonotonicandboundedandhenceconverges.If thesequence anconvergesto a ,thensodoesthesequence an + kforany integer k (inthede“nitionofthelimit,change N to N + k toprove this).Sincetheexistenceofthelimithasbeenestablished,onecan

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50.SPECIALSEQUENCES81 takethelimitofbothsidesoftherecurrencerelation limn an +1= 1 2 (limn an+3)= a = 1 2 ( a +3)= a =3 Thus, an 3as n Example 8.9 Investigatetheconvergenceofthesequencede“ned bytherecurrencerelation a1= 2 and an +1= 2+ an. Solution: The“rstfewtermsofthesequencesuggestthatthesequenceisincreasing: a1= 2
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828.SEQUENCESANDSERIES (10) an=sin( qn ) /n ,where q isreal. In(11)…(15),“ndthelimitofthesequenceorshowthatitdoesnot exist. (11) a1=1and an +1=4 Š an. (12) a1=1and an +1=1 / (1+ an). (13) a1=1and an +1=3 Š 1 /an. (14) a1=2and an +1=1 / (3 Š an). (15) a1=1and an +1=1+1 / (1+ an). (16)Thesizeofanundisturbed“shpopulationhasbeenmodeled bytheformula pn +1= bpn/ ( a + pn),where pnisthe“shpopulationafter n yearsand a and b arepositiveconstantsthat dependonthespeciesandtheenvironment.Supposethat p0> 0.Showthat pn +1< ( b/a ) pn.Thenprovethat pn 0 if a>b ;thatis,thepopulationdiesout.Finally,showthat pn b Š a if b>a Hint: Showthat pnisincreasingandbounded,0 b Š a ,then pnisdecreasingandbounded, pn>b Š a .51.Series51.1.BasicDe“nitionsandNotation.Withasequence { an} ,onecan associateasequence { sn} ,where sn=nk =1ak= a1+ a2+ + an. Thesymbol (8.6)n =1an= a1+ a2+ a3+ iscalledan in“niteseries ,orjusta series .Thenumbers snarecalled the partialsums oftheseries(8.6).Thelimitsofsummationareoften omittedtodenoteaseries;thatis,thesymbol analsostandsforan in“niteseries.If { sn} convergesto s ,thentheseriesissaidto converge andonewritesn =1an= s orlimn nk =1ak= s. Thenumber s iscalledthe sumoftheseries .Ifthesequenceofpartial sums { sn} diverges,theseriesissaidtodiverge.Itshouldbeunderstoodthat s is thelimitofasequenceofsums ,anditisnotobtained merelybyaddition.

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51.SERIES83 Forexample,thesequenceofpartialsumsfortheseries ( Š 1)nis s1= Š 1, s2= Š 1+1=0, s3= s2Š 1= Š 1,or,generally, sn=(( Š 1)nŠ 1) / 2.Thissequencedivergesasithastwosubsequences s2 n=0and s2 n Š 1= Š 1,whichconvergetodierentnumbers,0 = Š 1. Ifonesimplyusesaddition,dierentvaluesforthesumoftheseries maybeobtained:n =1an=( a1+ a2)+( a3+ a4)+( a5+ a6)+ =0+0+ =0 ,n =1an=( a1+ a2+ a3)+( a4+ a5+ a6)+ = Š 1 Š 1 Š = Š ,n =1an= a1+( a2+ a3)+( a4+ a5)+ = Š 1+0+0+ = Š 1 Generally,bygroupingtermsinthesumindierentways(according totheassociativityofaddition),thesumisfoundtobe any integer! Thereaderisadvisedtoverifythis.Thus,theadditionrulescannot generallybeappliedtoevaluatethesumofaseries.51.2.GeometricSeries.Takeapieceofropeoflength1m.Cutitin half.Keeponehalfandcuttheotherhalfintwopiecesofequallength. Keepdoingthis,thatis,keepingonehalfandcuttingtheotherhalfin twoequal-lengthpieces.Thetotallengthoftheretainedpiecesis 1 2 + 1 4 + 1 8 + = 1 2 1+ 1 2 + 1 4 + = 1 2n =01 2n. Thisseriesmustconverge.Thepartialsum snhereisthetotallengthof retainedpieces.Thesequence { sn} ismonotonicallyincreasing(after eachcutpieceofropeisadded)andboundedbythetotallength1.So itconverges.Fromthegeometry,itisalsoclearthat1 Š sn=1 / 2n, where n isthenumberofcuts,andhence sn 1asonewouldexpect (thetotallengthoftherope).Soitisconcludedn =01 2n=2 Thisseriesisanexampleofthe geometricseries : 1+ q + q2+ q3+ =n =0qn,

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848.SEQUENCESANDSERIES where q isanumber.Thegeometricseriesdoesnotconvergeforany valueof q Theorem 8.8(ConvergenceofaGeometricSeries) Ageometric series n =1qnconvergesif | q | < 1 ,and,inthiscase,n =0qn= 1 1 Š q | q | < 1 andtheseriesdivergesotherwise. Proof. If q =1,thesequenceofpartialsumsobviouslydiverges. If q =1,onehas sn=1+ q + q2+ + qn Š 1= qsn= q + q2+ q3+ + qn. Subtractingtheseequations,oneinfers snŠ qsn=1 Š qn= sn= 1 Š qn 1 Š q Therefore, limn sn=limn 1 Š qn 1 Š q = 1 1 Š q Š 1 1 Š q limn qn. Ithasbeenfoundthatthesequence an= qnconvergesonlyif | q | < 1, and,inthiscase, qn 0as n .If | q | 1,thegeometricseries diverges. Example 8.10 Analyzetheconvergenceoftheseries 4 Š8 3+16 9Š32 27+ Solution: Theseriescanbewrittenintheform4 q0+4 q1+4 q2+4 q3+ ,where q = Š 2 / 3.Soitspartialsumsarefourtimesthepartial sumofthegeometricserieswith q = Š 2 / 3.Therefore,n =14 Š 2 3 n=4n =1 Š 2 3 n= 4 1 Š ( Š2 3) = 12 5 Whenrealnumbersarepresentedindecimalfrom,oneoftenencountersasituationwhenanumberhasarepeatedpatternofdecimal places.Take,forexample,thenumber1 2131313 ... ;thatis,thecombination13repeatsitselfinalldecimalplacesstartinginthesecond decimalplace. Example 8.11 Isthenumber 1 2131313 ... rationalorirrational? Ifitisrational,writeitasaratioofintegers.

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51.SERIES85 Solution: Byde“nitionofthedecimalrepresentation, 1 2131313 ... =1 2+ 13 103+ 13 105+ 13 107+ =1 2+ 13 103 n =0 1 100 n=1 2+ 13 103100 99 = 12 10 + 13 990 = 1201 990 51.3.NecessaryConditionforaSeriestoConverge.Thefollowingtheoremfollowsfromthelimitlawsappliedtothesequencesofpartial sums. Theorem 8.9(PropertiesofSeries) Supposethattheseries anand bnareconvergentandtheirsumsare s and t ,respectively.Let c beanumber.Thentheseries ( an+ bn) and canconvergeand ( an+ bn)= an+ bn= s + t, ( can)= c an= cs. Indeed,if { sn} and { tn} arethesequencesofpartialsumsofthe series anand bn,respectively,thenthepartialsumsoftheseries ( an+ bn)and ( can)are sn+ tnand csn,respectively.Bythelimit laws, sn+ tn s + t and csn cs Notethattheconvergenceoftheseries ( an+ bn)does not imply theconvergenceof anand bn.Forexample,put an=1and bn= Š 1.Theseries ( an+ bn)= 0=0,while an= 1and bn= ( Š 1)diverge,andtheequality ( an+ bn)= an+ bnbecomesmeaningless(0= Š Ž).Thisshowsthattherulesof algebrafor“nitesumsare not generallyapplicabletoseries.Only seriesfromaspecialclassof absolutelyconvergent series,discussedlater, behaveprettymuchas“nitesums. Itisclearthateverytheoremaboutsequencescanbestatedinterms ofseriesbyputting a1= s1and an= snŠ sn Š 1for n> 1andvice versa.Inparticular,iftheseriesconverges,thatis, sn s as n thenonecantakethelimitonbothsidesofthisrecurrencerelation andconcludethatlimn an=limn ( snŠ sn Š 1)= s Š s =0;thatis, foraconvergentseries an,thesequence { an} necessarilyconverges to0. Theorem 8.10(NecessaryConditionforaSeriestoConverge) If theseries anconverges,then limn an=0 Theconverseisnotgenerallytrue;thatis,theconditionlimn an= 0isnotsucientforaseriestoconverge.However,itcanstillbeused asatestfordivergenceofaseries.

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868.SEQUENCESANDSERIES Corollary 8.1(TestforDivergenceofaSeries) Ifthelimit limn andoesnotexistorif limn an =0 ,thentheseries andiverges. Example 8.12 Showthattheseries n3/ (3 n3+1) diverges. Solution: limn an=limn n3 3 n3+1 =limn 1 3+1 n3= 1 3 =0 sotheseriesdiverges. Ifthenecessaryconditionissatis“ed,theseriesmayconvergeor diverge.Thesequenceofpartialsumshastobeanalyzed. Example 8.13 Findthesumoftheseries n =1 1 n ( n +1)ifitexists orshowthatitdoesnotexist. Solution: Thenecessaryconditionforconvergenceisevidentlysatis“ed.Sothesequenceofpartialsumshastobeanalyzedforconvergence: sn=nk =11 k ( k +1) =nk =1 1 k Š 1 k +1 = 1 Š 1 2 + 1 2 Š 1 3 + 1 3 Š 1 4 + + 1 n Š 1 n +1 =1 Š 1 n +1 1as n Sothesequence { sn} convergesto1andhencen =11 n ( n +1) =1 Thisexampleisaparticularcaseofa telescopicseries Theorem 8.11(ConvergenceofaTelescopicSeries) Atelescopic series n =1( anŠ an +1) convergesif limn an= a ,and,inthiscase,n =1( anŠ an +1)= a1Š a, Theproofisanalogoustotheaboveexampleandbasedonthefact thatthesequenceofpartialsumsofatelescopicseries sn= a1Š an +1convergesto a1Š a .Thedetailsarelefttothereaderasanexercise.

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52.SERIESOFNONNEGATIVETERMS87 51.4.Exercises.In(1)…(3),determinewhetherthegeometricseriesconvergesordiverges. (1)n =0n 3n +1(2)n =1en 3n Š 1(3)n =0( Š 5)n 32 nIn(4)…(9),determinewhethertheseriesconvergesordiverges.Ifit converges,“nditssum.Here p isapositivenumber, p> 0. (4)n =1k2 k2+ k +1 (5)n =12 Š 3n 5n(6)n =1n p (7)n =1(sin p )n(8)n =1en np(9)n =1 eŠ 2 n+ 4 n ( n +1) In(10)…(12),determinewhethertheseriesconvergesordivergesbyexpressingitasatelescopicseries.Findthesumoftheseriesifitexists. (10)n =21 n2Š 1 (11)n =13 n2+3 n +3 (12)n =1ln n n +1 In(13)and(14),expressthenumberasaratioofintegers (13)1 23232323 .... (14)1 53525252 .... In(15)…(17),“ndthevaluesof x forwhichtheseriesconverges.Find thesumoftheseriesforthosevaluesof x (15)n =1xn 2n(16)n =1sinnx 3n(17)n =1( x Š 5)nIn(18)and(19),solvetheequation. (18)n =2(1+ x )Š n=3(19)n =0enx=952.SeriesofNonnegativeTerms Inmanyapplications,thetermsofaseriesdecreasemonotonically. Itappearsthatthereisarelationbetweenconvergenceofsuchseries andconvergenceofimproperintegralsoveraninterval[1 ).This relationallowsonetoestablisha necessaryandsucientcondition for seriesofnonnegativetermstoconverge.

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888.SEQUENCESANDSERIES 52.1.TheIntegralTest.Suppose f ( x )isapositive,continuous,monotonicallydecreasingfunctionon[1 )suchthat f ( x ) 0.Suppose alsothattheimproperintegral 1f ( x ) dx =lima a 0f ( x ) dx = Ifexists.Thevalue Ifistheareaunderthegraph y = f ( x )overthe interval[1 ).Considertheseries n =1f ( n ).Thenecessaryconditionforconvergenceisful“lledas f ( n ) 0as n .Toinvestigate theconvergenceoftheseries,onehastoanalyzetheconvergenceofits partialsums: sn=nk =1f ( k )= f (1)+ f (2)+ + f ( n ) Sincethefunction f ( x )monotonicallydecreasesandiscontinuouson everyinterval[ k,k +1],itattainsitsminimalandmaximalvalues f ( k +1) f ( x ) f ( k )onthisintervalandtherefore (8.7) f ( k +1) k +1 kf ( x ) dx f ( k ) Thisinequalityleadstothefollowingupperandlowerestimatesof thepartialsums: sn f (1)+ 2 1f ( x ) dx + + n n Š 1f ( x ) dx = f (1)+ n 1f ( x ) dx, sn 2 1f ( x ) dx + 3 2f ( x ) dx + + n +1 nf ( x ) dx = n +1 1f ( x ) dx, sothat (8.8) n +1 1f ( x ) dx sn f (1)+ n 1f ( x )forall n 1 Thisinequalityshowsthatthefollowingtheoremholds. Figure8.7. Integraltest.Anillustrationofinequality(8.7).

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52.SERIESOFNONNEGATIVETERMS89 Theorem 8.12(IntegralTest) Suppose f isacontinuous,positive,decreasingfunctionon [1 ) andlet an= f ( n ) .Thentheseries n =1anconvergesifandonlyiftheimproperintegral 1f ( x ) dx converges.Inotherwords, 1f ( x ) dx converges= n =1f ( n )converges 1f ( x ) dx diverges= n =1f ( n )diverges Proof. Iftheimproperintegralconvergestoanumber If,then by(8.8)thesequenceofpartialsumsisbounded, sn f (1)+ If,and monotonicallyincreases, sn sn+ f ( n +1)= sn +1.Therefore,itis convergent.Iftheimproperintegraldiverges,then,foranynumber M> 0,thereisaninteger N suchthat n +1 1f ( x ) dx M forall n>N .Bytheleftinequalityof(8.8), M snforall n>N ;thatis, { sn} isamonotonicallyincreasing,unboundedsequenceandhenceit diverges. Remark. Supposethat an= f ( n ),where f ( x )isafunctionon [1 ),suchthatitiscontinuous,positive,anddecreasingon[ N, ), where N 1isaninteger.Then (8.9)n =1anconverges Nf ( x ) dx converges; thatis,theintegraltestappliesevenifthesequence anbecomesmonotonicallydecreasingonlyfor n N 1.Thisiseasytounderstand byisolatingthe“rst N Š 1termsintheseriesn =1an= a1+ a2+ + aN Š 1+n = Nan= a1+ a2+ + aN Š 1+n =1bn, where bn= aN + n Š 1.Convergenceof bnimpliesconvergenceof anandviceversaastheydierbya number .Put bn= g ( n ),where g ( x )= f ( x + N Š 1),whichisacontinuous,positive,decreasingfunction on[1 ),and 1g ( x ) dx = 1f ( x + N Š 1) dx = Nf ( u ) du bychangingtheintegrationvariable u = x + N Š 1.

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908.SEQUENCESANDSERIES 52.2.SpecialSeriesofNonnegativeTerms.Theorem 8.13 The p-seriesn =11 npconvergesif p> 1 anddivergesif p 1 Proof. If p 0,theseriesdivergesbecausethenecessaryconditionforconvergenceisnotful“lled, an if p< 0and an=1 =0if p =0.For p> 0,considerthefunction f ( x )= xŠ p,whichispositive, continuous,anddecreasingon[1 ),and a 1dx xp= 1 p Š 1 1 Š1 ap Š 1 if p =1 ln a if p =1 So,bytheintegraltest,theseriesconvergesif p> 1becausethe improperintegraldivergesif0

1(the limit a existsonlyif p> 1). Notethattheseries nŠ pdivergesforall0

1; thatis,thisseries de“nes afunctionon(1 ).Thisfunctioniscalled Riemannszetafunction Example 8.14 Investigatetheconvergenceoftheseries n =1( n +2)Š 3 / 2. Solution: Theseriescanbewrittenasn =11 ( n +2)3 / 2=n =31 n3 / 2= Š 1 Š 1 23 / 2+n =11 n3 / 2. Thelatterseriesisa p -seriesthatconvergesfor p =3 / 2 > 1. Theorem 8.14 Theseriesn =21 n (ln n )pconvergesif p> 1 ,anditdivergesif p 1 Proof. Considerthefunction g ( x )= x (ln x )pfor x> 1.Itsderivativereads g( x )=(ln x )p Š 1( p +ln x ).If p 0,then g( x ) > 0for all x> 1and g ( x )increases,whileitsreciprocal f ( x )=1 /g ( x )should

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52.SERIESOFNONNEGATIVETERMS91 decrease.If p< 0,then g( x ) > 0forall x>eŠ pandhence g ( x )increases,while f ( x )=1 /g ( x )decreasesif x>eŠ p> 1.Thus,forany p thereisaninteger N suchthatthefunction f ( x )=1 / [ x (ln x )p]iscontinuous,positive,anddecreaseson[ N, ).Bytheintegraltest(8.9), theseriesinquestionconvergesifandonlyiftheimproperintegral Ndx x (ln x )p= ln Ndu upconverges,wheretheintegrationvariablehasbeenchanged, u =ln x du = dx/x .Thisintegraldivergesif p 1andconvergesif p> 1,and theconclusionofthetheoremfollows. 52.3.EstimateoftheSum.Ifapartialsum snisusedtoestimatethe sumofaconvergentseriesofnonnegativeterms f ( n ),howgood issuchanestimate?Theremainder s Š snhastobeinvestigatedto answerthisquestion. Corollary 8.2(EstimateofSums) Suppose f isacontinuous, positive,decreasingfunctionon [1 ) andlet an= f ( n ) .Iftheseries anconvergestoanumber s ,then n +1f ( x ) dx s Š sn nf ( x ) dx, where { sn} isthesequenceofpartialsums. Proof. The“rstinequalityisobtainedbytakingthelimit n in(8.8)withtheresult (8.10) 1f ( x ) dx n =1an f (1)+ 1f ( x ) dx, whichisalegitimateoperationbecause(8.8)holdsforall n andthe seriesconverges(andsodoestheimproperintegralbytheintegral test).Theremainderestimateisobtainedbysubtracting(8.8)from (8.10).Notethevalueoftheimproperintegraldoes not coincidewith thesum;itonlydeterminesaninterval(8.10)inwhichthesumofa serieslies. Example 8.15 Testtheseries n =1( n2+1)Š 1forconvergenceor divergence.Ifitconverges,estimateitssum. Solution: Put f ( x )=( x2+1)Š 1,whichisacontinuous,positive,decreasingfunctionon[1 ),suchthattheseriesinquestionis f ( n ).

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928.SEQUENCESANDSERIES Therefore,theintegraltestapplies,andtheseriesconvergesbecause 1dx x2+1 =lima tanŠ 1x a 1=lima tanŠ 1a Š 4 = 2 Š 4 = 4 By(8.10),itssumliesintheinterval 4 s f (1)+ 4=1 2+ 4. Example 8.16 Testtheseries n =1neŠ nforconvergenceordivergence.Ifitconverges,estimateitssum. Solution: Considerthefunction f ( x )= xeŠ x.Since f( x )= eŠ xŠ xeŠ x=(1 Š x ) eŠ x 0if x 1,thefunctiondecreaseson[1 ),and theintegraltestappliestoassessconvergenceoftheseries f ( n ): 1xeŠ xdx = Š 1xdeŠ x= Š lima xeŠ x a 1+ 1eŠ xdx = 1 e + 1 e = 2 e wheretheintegrationbypartshasbeenusedtoevaluatetheintegral. Theseriesconvergestoanumber s thatliesintheinterval2 eŠ 1 s f (1)+2 eŠ 1=3 eŠ 1. Example 8.17 EstimatevaluesofRiemannszetafunction ( p ) Howmanytermsdoesoneneedtoretaininthepartialsum sntoapproximate ( p ) correctto N decimalplaces? Solution: Riemannszetafunctionisde“nedbythesumoftheseries ( p )= n =1nŠ p.For p> 1, 1dx xp=lima x1 Š p 1 Š p a 1=lima a1 Š p 1 Š p Š 1 1 Š p = 1 p Š 1 Since f (1)=1,by(8.10), 1 p Š 1 ( p ) p p Š 1 ByCorollary8.2, 0 ( p +1) Š sn ndx xp +1= 1 pnp. If ( p +1), p> 0,istobeapproximatedby sncorrectto N decimal places,thentheremaindershouldbelessthan5 10Š N Š 1,whichyields theconditiononthenumberofterms:1 pnp< 5 10Š N Š 1or np> 10N +1/ (5 p )or n p 10N +1/ (5 p ).

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53.COMPARISONTESTS93 52.4.Exercises.In(1)…(9),determinewhethertheseriesconvergesordiverges. (1)n =11 n9 / 8(2)n =2(ln n )4 n (3)n =21 Š n ln n n2(4)n =1n2Š 2 n Š 5 3 n7 / 3(5)n =11 n2Š 4 n +5 (6)n =1n n4+1 (7)n =1e1 /n n2(8)n =12 n +1 n ( n +1) (9)n =1tanŠ 1n n2+1 In(10)…(14),determinethevaluesof p forwhichtheseriesisconvergent. (10)n =31 n ln n (ln(ln n ))p(11)n =1n (1+ n2)p(12)n =1npeŠ n(13)n =1pln n,p> 0(14)n =1 p n Š 1 n +1 (15)HowmanytermsoftheseriesinTheorem8.14wouldoneneedto addto“nditssumcorrectto N decimalplaces? (16)Showthatthesequence an=1+ 1 2 + 1 3 + + 1 n Š ln n converges.Thelimitlimn an= iscalledthe Eulernumber Hints :(1)Use(8.8)toshowthatif snisthepartialsumoftheharmonic series,then sn 1+ln n andhence an 1(i.e.,thesequence { an} is bounded). (2)Interpret anŠ an +1asadierenceofareastoshowthat { an} is monotonic.53.ComparisonTests Considertheseriesn =1an=n =11 n4+1 Thisserieshastermssmallerthanthecorrespondingtermsoftheconvergent p -series:n =1bn=n =11 n4

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948.SEQUENCESANDSERIES because an
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53.COMPARISONTESTS95 Solution: Onehas an= n +1 Š n = ( n +1 Š n )( n +1+ n ) n +1+ n = 1 n +1+ n 1 2 n + n = 1 1+ 2 1 n = bn. The p -series 1 / n divergesandsodoestheseriesinquestionbythe comparisontest. Theorem 8.16(LimitComparisonTest) Supposethat anand bnareserieswithpositiveterms.Let c =limn ( an/bn) € If c =0 and bnconverges,then anconverges. € If 0 N bythede“nitionofthelimit.Hence, anN .If bnconverges,then anconvergesbythecomparison test.If c (0 ),then,bythede“nitionofthelimit,foranynumber c>> 0,thereisaninteger N suchthat c Š an bn < m = c Š < an bnN. Therefore, mbnN. Bythecomparisontest,convergenceof bnimpliesconvergenceof anduetotheinequality an 0,thereisaninteger N suchthat an/bn>M when n>N .Theinequality an>Mbnshowsthatdivergenceof bnimpliesdivergenceof anbythecomparisontest. Itisoftenhelpfultoinvestigatetheasymptoticbehaviorof anas n toidentifyasuitable bninthelimitcomparisontest. Example 8.20 Testtheseries n =1(2 n3 / 2+ n ) / n6+ n4+1 for convergence. Solution: Letus“ndtheasymptoticbehaviorof anas n .For large n ,thetopoftheratiobehavesas 2 n3 / 2,whilethebottomof theratiobehavesas ( n6)1 / 2= n3.Therefore, an= 2 n3 / 2+ n n6+ n4+1 = 2 n3 / 2(1 Š1 2 n) n3 1+1 n2+1 n6 2 n3 / 2 n3=2 nŠ 3 / 2= bn

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968.SEQUENCESANDSERIES intheasymptoticregion n .Thisshowsthattheratio an/bnconvergesto c =1as n .Bythelimitcomparisontest,theseries anconvergesbecausethe p -series bn=2 nŠ 3 / 2converges. Example 8.21 Testtheseries n =1( n5+3n) / n3+5nforconvergence. Solution: Recallthatthepowerfunctionincreasesmoreslowlythan theexponentialfunction,thatis, npqŠ n 0as n forany q> 1 andany p .Hence,theasymptoticbehaviorof anis an= 3n(1+ n53Š n) 5n(1+ n35Š n) 3n 5n=(3 / 5)n= bn. Thisshowsthattheratio an/bnconvergesto c =1as n .By thelimitcomparisontest, anconvergesbecausethegeometricseries bnconverges( q =3 / 5 < 1). 53.1.EstimatingSums.Ifaseries anconvergesbycomparisonwith aseries bn,thenthesumof ancanbeestimatedbycomparing remaindersfortheseries bn.Indeed,put an= s and bn= t Let { tn} and { sn} bethesequencesofpartialsumsfor bnand an, respectively.Theremainderssatisfytheinequality: s Š sn= an +1+ an +2+ bn +1+ bn +2+ = t Š tn. Sotheaccuracyoftheapproximation s snisthesameorhigher thanthatoftheapproximation t tn.If,forexample,one“nds that n = N issucientfortheequality t = tNtobecorrecttoa speci“cnumberofdecimalplaces,then s = sNisalsocorrecttothat orevenahighernumberofdecimalplaces.Theremainderiseasyto estimatewhen bn= f ( n ),wherethefunction f issimpletointegrate, t Š tn nf ( x ) dx Example 8.22 Determinehowmanytermsareneededtoestimate thesumoftheseries n =1tanŠ 1( n2) / ( n3+1) correctto“vedecimal places. Solution: ThefunctiontanŠ 1x ismonotonicallyincreasingfor x> 0 approachingasymptoticallythevalue / 2.Therefore, an= tanŠ 1( n2) n3+1 2 1 n3+1 2 1 n3= bn. Hence, s Š sn t Š tn 2 ndx x3= 4 n2< 5 10Š 6 n> 2 5 103 396

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54.ALTERNATINGSERIES97 53.2.Exercises.In(1)…(12),determinewhethertheseriesconvergesordiverges. (1)n =1n n5 / 3+ n1 / 3+1 (2)n =2n (ln n )4 n2+1 (3)n =2cos2( n ) n2(4)n =11+( Š 1)n n3 / 2+1 (5)n =11 3 n3+ n +1 (6)n =11+2n n2+2n(7)n =1e1 / n n +1 (8)n =1sin2 1 n (9)n =1n nn(10)n =11 n1+1 /n(11)n =1(n n Š 1)n(12)n =1n2 n (13)Howmanytermsdoesoneneedinthepartialsumtoestimatethe sumoftheseries n =1sin3n/ ( n3+ n )upto“vedecimalplaces? (14)Considerasequence { an} ,where ancantakeanyvaluefromthe set { 0 1 2 ,...,p Š 1 } ,where p> 1isaninteger.Themeaningofthe representationofanumber0 .a1a3a3... withbase p isthat (8.11)0 .a1a3a3... = a1 p + a2 p2+ a3 p3+ When p =10,thedecimalsystemisobtained.Thebinaryrepresentationcorrespondsto p =2.TheMayaused p =20(thenumberof “ngersandtoes).TheBabyloniansused p =60.Showthattheseries (8.11)alwaysconverges. (15)Showthatif an> 0and anconverges,then ln(1+ an)converges,too. (16)Provethattheconvergenceof an,where an> 0,impliesthe convergenceof an/n (17)If anconvergesandifthesequence { bn} ismonotonicand bounded,provethat anbnconverges.54.AlternatingSeries Definition 8.6(AlternatingSeries) Let { bn} beasequenceof nonnegativeterms.Theseries ( Š 1)n Š 1bn= b1Š b2+ b3Š b4+ b5Š iscalledanalternatingseries.

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988.SEQUENCESANDSERIES Forexample,theseries (8.12)1 Š 1 2 + 1 3 Š 1 4 + 1 5 + =n =1( Š 1)n Š 1 n isanalternatingseries.Itiscalledthe alternatingharmonicseries Theorem 8.17(AlternatingSeriesTest) Ifasequenceofpositive terms { bn} ismonotonicallydecreasingand limn bn=0 ,thenthe alternatingseries ( Š 1)n Š 1bnconverges: ( i ) bn +1 bnforall n ( ii )limn bn=0 = n =1( Š 1)n Š 1bnconverges Proof. Theconvergenceofthesequenceofpartialsums { sn} is tobeestablished.Considerasubsequenceofevenpartialsums { s2 k} Onehas s2= b1Š b2 0, s4= s2+( b3Š b4) s2,and,ingeneral, s2 k= s2( k Š 1)+( b2 k Š 1Š b2 k) s2( k Š 1) s2( k Š 2) s2 0 bythemonotonicityofthesequence { bn} .Thus,thesubsequence { s2 k} ismonotonicallyincreasing.Byregroupingthetermsinadierentway, onecanseethat s2 k= b1Š ( b2Š b3) Š ( b4Š b5) ŠŠ ( b2 k Š 2Š b2 k Š 1) Š b2 k b1becauseallnumbersinparenthesesarenonnegativebyhypothesis(i), whichshowsthat { s2 k} isalsobounded.Therefore,itconvergesbythe monotonicsequencetheorem: limk s2 k= s. Forthesubsequenceofoddpartialsums s2 k +1= s2 k+ b2 k +1,oneinfers bythelimitlawsandhypothesis(ii)that limk s2 k +1=limk s2 k+limk b2 k +1= s +0= s. Theconvergenceoftwoparticularsubsequencesofasequencetothe samenumber s doesnotgenerallyguaranteethatthesequenceconvergesto s (allitssubsequencesshouldconvergeto s ).Byde“nition, thelimitsof { s2 k} and { s2 k +1} meanthat,givenanynumber > 0, therearepositiveintegers N1and N2suchthat | s2 kŠ s | < if k>N1and | s2 k +1Š s | < if k>N2.Put N =max(2 N1, 2 N2+1).Then | snŠ s | < forall n>N ,whichmeansthat sn s as n Bythistest,thealternatingharmonicseries(8.12)convergesbecausethesequence bn=1 /n ismonotonicallydecreasingandconverges to0.

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54.ALTERNATINGSERIES99 Figure8.8. Alternatingseriestest.Anillustrationof itsproofwheretwosubsequences, s2 kand s2 k Š 1,ofthe sequence snofpartialsumsareanalyzedforconvergence. Example 8.23 Testtheseries n =1sin( n/ 2) /n forconvergence. Solution: Onehassin( n/ 2)=1 0 Š 1 0 1 ,... for n =1 2 3 4 5 ,... respectively,or,ingeneral,forodd n =2 k Š 1,sin( n/ 2)=( Š 1)k Š 1, whileforeven n =2 k ,sin( n/ 2)=sin( k )=0.Thus,theseriesin questionisanalternatingseries:n =1sin( n/ 2) n =n =1( Š 1)n Š 1 2 n Š 1 =n =1( Š 1)n Š 1bn,bn= 1 2 n Š 1 Thesequence { bn} ismonotonicallydecreasingand bn 0as n Sotheseriesconvergesbythealternatingseriestest. Remark. ItshouldbenotedthatTheorem8.17providesonlya sucient conditionforanalternatingseriestoconverge.Sothereare convergentalternatingseriesthatdo not satisfythehypothesesofTheorem8.17.Forexample,thealternatingserieswith bn=sin2( n/q ) /n2, where q isaninteger,isconvergentbythecomparisontestbecause | ( Š 1)n +1bn| = bn 1 /n2andthe p -series 1 /n2converges(seethe nextsectionon absolutely conversingseries).However,thesequence { bn} is not monotonicallydecreasingbecause bn 0andithasazero

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1008.SEQUENCESANDSERIES subsequence bkq=0, k =1 2 ,... .So bnoscillatesbetweenthezero sequenceandthesequence1 /n2. Remark. Hypothesis(i)ofTheorem8.17maybeweakened (i) bn +1 bnforall n N forsomeinteger N 1.Indeed,n =1( Š 1)n Š 1bn= b1Š b2+ b3ŠŠ bN Š 1+n = N( Š 1)n Š 1bn= b1Š b2+ b3ŠŠ bN Š 1+( Š 1)N Š 1 n =1( Š 1)n Š 1cn, where cn= bn + N Š 1.Theseries ( Š 1)n Š 1bnand ( Š 1)n Š 1cndier byanumber,andthereforetheconvergenceofoneofthemimplies theconvergenceoftheother.Theseries ( Š 1)n Š 1cnconvergesby Theorem8.17as cn +1 cnforall n andlimn cn=limn bn + N Š 1= 0. Example 8.24 Testtheseries n =1( Š 1)n Š 1np/ ( n +1) forconvergenceif p< 1 Solution: Here bn= np/ ( n +1)and,for p< 1, limn bn=limn np n +1 =limn np Š 1 1+1 n=limn np Š 1=0 Sohypothesis(ii)ofTheorem8.17isful“lled.However,themonotonicityof { bn} isnotobvious.Toinvestigateit,considerthefunction f ( x )= xp/ ( x +1),where x 1.If f ( x )monotonicallydecreases,then sodoesthesequence bn= f ( n ).Thecondition f( x ) 0hastobe veri“ed: f( x )= pxp Š 1( x +1) Š xp ( x +1)2 0 ( p Š 1) xp+ pxp Š 1 0 p x (1 Š p ) If p 0,thisistrueas x 1.If0 N .Sotheseriesconverges forall p< 1. 54.1.EstimatingSumsofAlternatingSeries.Apartialsum snofany convergent alternatingseriescanbeusedasanapproximationofthe totalsum s ,butthisisnotofmuchuseunlesstheaccuracyofthe

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54.ALTERNATINGSERIES101 approximationisassessed.Thefollowingtheoremassertsthattheabsoluteerroroftheapproximation s sndoesnotexceedthevalueof bn +1. Theorem 8.18(AlternatingSeriesSumEstimation) If s = ( Š 1)n Š 1bnisthesumofanalternatingseriesthatsatis“es (i)0 bn +1 bnforall n and(ii)limn bn=0 then | s Š sn| bn +1. Proof. Intheproofofthealternatingseriestest,itwasfoundthat thesubsequence { s2 k} approachesthelimitvalue s frombelow, s2 k s Ontheotherhand,thesubsequence { s2 k Š 1} approachesthelimitvalue s fromabove.Indeed, s1= b1, s3= s1Š b2+ b3 s1because b3 b2, and,ingeneral, s2 k +1= s2 k Š 1Š b2 k+ b2 k +1 s2 k Š 1;thatis, { s2 k +1} ismonotonicallydecreasing.Thisshowsthatthesequenceofpartial sums snoscillatesaround s sothatthesum s alwaysliesbetweenany twoconsecutivepartialsums snand sn +1asdepictedinFigure8.8. Hence, | s Š sn|| sn +1Š sn| = bn +1. Example 8.25 Estimatethenumberoftermsinapartialsum snneededtoapproximatethesumofthealternatingharmonicseries correctto N decimalplaces. Solution: Here, bn=1 /n .Hence,theapproximation s sniscorrect to N decimalplacesiftheabsoluteerrordoesnotexceed5 10Š N Š 1: | s Š sn| bn +1< 5 10Š N Š 1or1 / ( n +1) < 5 10Š N Š 1or n> 0 2 (10NŠ 1). Remark. Ifthemonotonicitycondition bn +1 bnholdsonlyif n N ,theconclusionofTheorem8.18alsoholdsonlyif n N Indeed,inthenotationfromRemark2,put t = ( Š 1)n Š 1cn,where cn= bn + N Š 1.Let tnbeapartialsumfortheseries ( Š 1)n Š 1cn.Then s = sN Š 1+( Š 1)N Š 1t and sn= sN Š 1+( Š 1)N Š 1tn Š N +1for n N Therefore, | s Š sn| = | t Š tn Š N +1| cn Š N +2= bn +1forall n N.54.2.Exercises.In(1)…(15),determinewhethertheseriesconvergesordiverges(here p isreal).

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1028.SEQUENCESANDSERIES (1)n =1( Š 1)n ln( n +3) (2)n =1( Š 1)nn n3+1 (3)n =1cos( n/ 2) n4 / 5(4)n =1( Š 1)nn ( n3 / 2+1)2 / 3(5)n =2( Š 1)nn (ln n )p(6)n =1( Š 1)nn3 2n(7)n =2( Š 1)n(ln n )p n (8)n =1( Š 1)nsin n (9)n =1( Š 1)nn2 n4+1 (10)n =1( Š 1)n n1+1 /nŠ n (11)n =1( Š 1)n(n n Š 1)n(12)n =1( Š 1)nnn n (13)n =1( Š 1)n n + p (14)n =1( Š 1)n np(15)n =1( Š 1)n( n2+ n +1) (2 n +3)2In(16)and(17),“nd n forwhichtheapproximationbypartialsums s sniscorrectto N decimalplacesfortheseries. (16)n =1( Š 1)nn 10n(17)n =1( Š 1)nn1 / 3 n1 / 3+6 (18)Provethatthesumofthealternatingharmonicseriesisn =1( Š 1)n Š 1 n =ln2 Hint :Showthatapartialsumofthealternatingharmonicseriesis s2 n= h2 nŠ hn,where hn= an+ln n andthesequence { an} isde“ned inExercise52.4.16.Thenusetheresultofthelatterexercisetoprove that sn ln2as n .55.RatioandRootTests55.1.AbsolutelyConvergentSeries.Definition 8.7(AbsoluteConvergence) Aseries aniscalled absolutelyconvergent iftheseriesofabsolutevalues | an| isconvergent. Theabsoluteconvergenceisstrongerthanconvergence,meaning thatthereareconvergentseriesthatdonotconvergeabsolutely.For example,thealternatingharmonicseries an, an=( Š 1)n Š 1/n ,is convergent,butnotabsolutelyconvergentbecausetheseriesofabsolute values | an| =1 /n isnothingbuttheharmonicseries 1 /n ,whichis divergent(asa p Š serieswith p =1).Ontheotherhand,theabsolute convergenceimpliesconvergence.

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55.RATIOANDROOTTESTS103 Theorem 8.19(ConvergenceandAbsoluteConvergence) Every absolutelyconvergentseriesisconvergent. Proof. Foranysequence { an} ,thefollowinginequalityholds; 0 an+ | an| 2 | an| because | an| iseither anor Š an.Itshowsthattheseries bn,where bn= an+ | an| ,convergesbythecomparisontestbecause 2 | an| = 2 | an| convergesif anconvergesabsolutely.Hence,theseries an= bnŠ | an| convergesasthedierenceoftwoconvergent series. Example 8.26 Testtheseries [sin n Š 2cos(2 n )] /n3 / 2forabsoluteconvergence. Solution: Makinguseoftheinequality | A + B || A | + | B | andthe propertiesthat | sin x | 1and | cos x | 1,oneinfers | an| = | sin n Š 2cos(2 n ) | n3 / 2 | sin n | +2 | cos(2 n ) | n3 / 2 3 n3 / 2. Theseriesofabsolutevalues | an| con-vergesbycomparisonwiththe convergent p Š series3 nŠ 3 / 2(here p =3 / 2 > 1).Sotheseriesin questionconvergesabsolutely. Definition 8.8(ConditionalConvergence) Aseries aniscalled conditionallyconvergent ifitisconvergentbutnotabsolutelyconvergent. Thus,allconvergentseriesareseparatedintotwoclassesofconditionallyconvergentandabsolutelyconvergentseries.Thekeydifferencebetweenpropertiesofabsolutelyconvergentandconditionally convergentseriesisstudiedinthenextsection.55.2.RatioTest.Theorem 8.20(RatioTest) Givenaseries an,supposethe followinglimitexists: limn an +1 an = c, where c 0 or c = € If c< 1 ,then anconvergesabsolutely. € If c> 1 ,then andiverges. € If c =1 ,thenthetestgivesnoinformation.

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1048.SEQUENCESANDSERIES Proof. If c< 1,thentheexistenceofthelimitmeansthat,for any > 0,thereisaninteger N suchthat Š < an +1 an Š c< = an +1 an 0 smallenoughsothatthenumber q = c + < 1.Inparticular,put n = N + k Š 1,where k 2.Applyingtheinequality | an +1| 1,thenthereisaninteger N suchthat | an +1| / | an| > 1or | an +1| > | an| 0forall n N .Hence,thenecessaryconditionfora seriestoconverge, an 0as n doesnothold;thatis,theseries andiverges.If c =1,itissucient,togiveexamplesofaconvergent anddivergentseriesforwhich c =1.Considera p -series nŠ p.One has c =limn an +1 an =limn np ( n +1)p=limn 1 (1+1 /n )p=1 forany p .Buta p -seriesconvergesif p> 1anddivergesotherwise. Example 8.27 Findallvaluesof p and q forwhichtheseries n =1npqnconvergesabsolutely. Solution: Here an= npqn.Onehas c =limn an +1 an =limn ( n +1)p np| q |n +1 | q |n= | q | limn (1+1 /n )p 1 = | q | So,for | q | < 1andany p ,theseriesconvergesabsolutelybytheratio test.If q = 1,theratiotestisinconclusive,andthesecaseshavetobe studiedbydierentmeans.If | q | =1,then | an| = np= 1 /nŠ p, whichisa p -seriesthatconvergesif Š p> 1or p< Š 1.Thus,the seriesconvergesabsolutelyforall p if | q | < 1andfor p< Š 1if q = 1.Notethat,for Š 1 p< 0and q = Š 1,theseriesconditionally converges(i.e.,itisconvergentbutnotabsolutelyconvergent).Inthis case,itisaconvergentalternating p -series ( Š 1)n/nŠ p(seeExercise 54.2.14).

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55.RATIOANDROOTTESTS105 55.3.RootTest.Theorem 8.21(RootTest) Givenaseries an,supposethefollowinglimitexists: limn n | an| = c, where c 0 or c = € If c< 1 ,then anconvergesabsolutely. € If c> 1 ,then andiverges. € If c =1 ,thenthetestgivesnoinformation. Proof. If c< 1,then,asintheproofoftheratiotest,theexistence ofthelimitmeansthat,forany c 1,thenthereexistsaninteger N suchthatn | an| > 1 forall n N ,andhencethecondition an 0as n does nothold.Theseries andiverges.If c =1,considera p -series:n nŠ p=(n n )Š p 1Š p=1byTheorem8.6.Buta p -seriesconverges if p> 1anddivergesif p< 1.Sotheroottestisinconclusive. Example 8.28 Testtheconvergenceoftheseries an,where an= [(2 n2+5) / (3 n2+2)]n. Solution: Here | an| = an,andtheabsoluteconvergenceisequivalent totheconvergence.Onehas limn n | an| =limn 2 n2+5 3 n2+2 =limn 2+5 /n2 3+2 /n2= 2 3 < 1 Sotheseriesconverges. 55.4.OscillatoryBehaviorofSequencesintheRootandRatioTests.Considerasequencede“nedrecursivelyby a1=1and an +1=1 2(sin n ) an. Anattempttotesttheconvergenceof anbytheratiotestleadsto thesequence cn= | an +1| / | an| =1 2| sin n | whichdoesnotconvergeasit oscillatesbetween0and1 / 2.Similarly,thesequenceusedintheroot testmayalsoexhibitoscillatorybehaviorandbenonconvergent,for example, an=(1 2sin n )nsothat cn=n | an| =1 2| sin n | .Theratioand roottests,asstatedinTheorems8.20and8.21,assumethe existence ofthelimit cn c .Whatcanbesaidabouttheconvergenceofaseries whenthislimitdoesnotexist?

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1068.SEQUENCESANDSERIES Toanswerthisquestion,recallthat,intheproofoftheratioor roottest,theexistenceoflimn cn= c< 1hasbeenusedonlyto establishtheboundednessofthesequence cn q< 1forall n N whichis sucientfortheseries antoconverge .Buttheboundedness propertydoes not implytheconvergence!Evidently,theboundedness conditionholdsintheaboveexamples, cn=1 2| sin n |1 2< 1forall n .Similarly,theexistenceofthelimitvalue c> 1hasonlybeenused toshowthatn | an| 1or | an| 1for in“nitelymany n toconclude thatthesequence { an} cannotconvergeto0andhence andiverges. If | an +1| / | an| 1forall n N ,thenagain { an} cannotconvergeto0 (bytheproofoftheratiotest).Thus,theconvergenceof { cn} inthe rootorratiotestisnotreallynecessary. Theorem 8.22(RatioandRootTestsRe“ned) Givenaseries nan,put cn= | an +1| / | an| or cn=n | an| .Then cn q< 1forall n N = anconverges n | an| 1forin“nitelymany n| an +1| | an| 1forall n N = andiverges forsomeinteger N .55.5.WiderScopeoftheRootTest.Ifthelimitof | an +1| / | an| exists, thensodoesthelimitofn | an| and (8.14)limn n | an| =limn | an +1| | an| Theconverseisnottrue;thatis,theexistenceofthelimitofn | an| doesnotgenerallyimplytheexistenceofthelimitof | an +1| / | an| (the lattermayormaynotexist).Furthermore,ifthesequencen | an| does notconverge,neitherdoes | an +1| / | an| .Aproofoftheseassertionsis giveninmoreadvancedcalculuscourses.Thus,theratiotesthasthe samepredictingpowerastheroottestonlyif | an +1| / | an| converges. Ingeneral,theroottest(asinTheorem8.22)has wider scope, meaningthatwhenevertheratiotestshowsconvergence,theroottest does,too,andwhenevertheroottestisinconclusive,theratiotest is,too.Thesubtletytonotehereisthattheconverseofthelatter statementisnotgenerallytrue;thatis,theinconclusivenessoftheratio test doesnot implytheinconclusivenessoftheroottest.Theassertion canbeillustratedwiththefollowingexample.Consideraconvergent seriesobtainedfromthesumoftwogeometricseriesinwhichtheorder

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55.RATIOANDROOTTESTS107 ofsummationischanged:n =1an= 1 2 + 1 3 + 1 22+ 1 32+ 1 23+ 1 33+ = 1 2k =01 2k+ 1 3k =01 3k= 3 2 wherethesumofageometricserieshasbeenused(Theorem8.8).Now notethatif n =2 k iseven,then a2 k=(1 / 3)k,and a2 k Š 1=(1 / 2)kif n =2 k Š 1isodd.Takethesubsequenceofratiosforeven n =2 k c2 k= a2 k +1/a2 k=(2 / 3)k/ 9.Itconvergesto0as k .Ontheother hand,thesubsequenceofratiosforodd n =2 k Š 1diverges: c2 k Š 1= (3 / 2)k as k .Sothelimitof cndoesnotexist;moreover, theratiotest(asinTheorem8.22)failsmiserablybecause cnisnot evenbounded.Theseriesconvergesbytheroottest.Indeed, c2 k=2 k a2 k=1 / 2 < 1and c2 k Š 1=2 k Š 1 a2 k Š 1=1 / 3 < 1.Althoughthe sequence cndoesnotconverge(itoscillatesbetween1 / 3and1 / 2), itisbounded, cn q =1 / 2 < 1forall n ,andhencetheseries convergesbyTheorem8.22.AsimilarexampleisgiveninExercise 55.7.19.Thus, theratiotestissensitivetotheorderofsummation, whilethisisnotsofortheroottest .55.6.WhentheRatioTestIsInconclusive.Theorem 8.23(DeMorgansRatioTest) Let anbeaseriesin which | an +1| / | an| 1 as n .Theseriesconvergesabsolutelyif limn n an +1 an Š 1 = b< Š 1 Theproofofthistheoremislefttothereaderasanexercise(see Exercise55.7.18).Considertheasymptoticbehavioroftheratio cn= | an +1| / | an| as n .Thetheoremassertsthatif cnbehavesas cn 1+ b/n forlarge n (i.e.,neglectingtermsoforder1 /npwhere p> 1),thentheseries anconvergesif b< Š 1. Fora p -series,theratiotestisinconclusive(seetheproofofthe ratiotest).However,DeMorganstestresolvestheinconclusiveness. Indeed,forlarge n cn= np ( n +1)p= 1 (1+1 /n )p 1 Š p n wheretheasymptoticbehaviorhasbeenfoundfromthelinearization f ( x )=(1+ x )Š p f (0)+ f(0) x =1 Š px forsmall x =1 /n .So b = Š p andtheseriesconvergesif b< Š 1or p> 1. ThisillustratesabasictechnicaltricktoapplyingDeMorganstest. Supposethatthereisafunction f ( x )suchthat | an +1| / | an| = f (1 /n ).

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1088.SEQUENCESANDSERIES If f isdierentiableat x =0,thatis, f ( x ) f (0)+ f(0) x as x 0, then | an +1| / | an| = f (1 /n ) f (0)+ f(0)(1 /n )=1+ f(0) /n, andtheseries anconvergesabsolutelyif f(0) < Š 1.Notethatthe property f (0)=1followsfromtheinconclusivenessoftheratiotest.55.7.Exercises.In(1)…(15),determinewhethertheseriesisabsolutelyconvergent,conditionallyconvergent,ordivergent(here p isreal). (1)n =1( Š 1)n 3 n (2)n =1n2(2 3)n(3)n =1n pn,p =0 (4)n =1pn n (5)n =2( Š 1)n21 /n np(6)n =1( Š 1)nnp +3 n (7)n =2( Š 1)n(ln n )p n (8)n =1( Š 1)nn 4 n3+1 (9)n =1n nn(10)n =1( Š 1)n (ln n )p(11)n =1 2 n3+ n 3 n3+5 n(12)n =1 1+ 1 n n2(13)n =1np (ln n )n(14)n =12 4 (2 n ) n (15)n =1pnn 5 8 (3 n +2) (16)Forwhichintegers p> 0istheseries n =1( n !)2/ ( pn )!convergent? (17)(Estimatingsums).Givenaseries anwithpositiveterms,put cn= an +1/an.Supposethat cn c< 1,thatis,theseriesconverges, an= s .Let snbeapartialsum.Provethat s Š sn an +1 1 Š cn +1if { cn} isadecreasingsequence,and s Š sn an +1 1 Š c if { cn} isanincreasingsequence. Hint :Usethegeometricseriesasintheproofoftheratiotestto estimatetheremainder s Š sn= an +1+ an +2+ (18)ProveDeMorgansratiotest.

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56.REARRANGEMENTS109 Hints :Comparetheseries | an| withtheconvergent p -series bn, where bn= A/npand p = Š (1 Š b ) / 2 > 1if b< Š 1.Showthat n ( bn +1/bnŠ 1) b as n .Next,showthat,bychoosingthe constant A onecanalwaysmake | an|
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1108.SEQUENCESANDSERIES Takingthelimit n inthisequality,one“nds s= s + s/ 2=3 s/ 2, where s and sarethesumsof(8.15)and(8.16),respectively.Thus, arearrangementoftheserieshaschangeditssum!Thisfactisnot speci“ctotheexampleconsideredbutinherentinallconditionally convergentseries.Termsofaconditionallyconvergentseriesoccur withdierentsigns(positiveandnegative).Byregroupingpositive andnegativeterms,itwillbeprovedthatthesumofaconditionally convergentseriescanbemade anynumber or .Theanalysisbegins bystudyingthepropertiesofsumsofpositiveandnegativetermsofa conditionallyconvergentseries. Givenanumber x ,put x=( x | x | ) / 2.Thenumber x+= x if x> 0and x+=0otherwise.Similarly, xŠ= x if x< 0and xŠ=0 otherwise. Lemma 8.1 Givenaseries an,considertwoseries a+ nand aŠ n,where a n=( an| an| ) / 2 (theseriesofpositiveandnegative terms).Then (i) If anconvergesabsolutely,then a+ nand aŠ nconverge. (ii) If anisconditionallyconvergent,then a+ nand aŠ ndiverge. Proof. Let an= s< and | an| = t ,where t< if anconvergesabsolutelyand t = ifitisconditionallyconvergent.Let s nbepartialsumsof a n, snbepartialsumsof an,and tnbe partialsumsof | an| .Since sn s and tn t as n ,oneinfers that a+ nŠ aŠ n= ana+ n+ aŠ n= | an| = s+ nŠ sŠ n= sns+ n+ sŠ n= tn= s+Š sŠ= s s++ sŠ= t, where sarethelimitsof s n.If anconvergesabsolutely,then t< andhence s=( t s ) / 2;thatis,bothseries a nconverge.If anisconditionallyconvergent,then t = ,andbothsequences s ndiverge. Theorem 8.24(RiemannsRearrangementTheorem) Let anbeaseriesthatconverges,butnotabsolutely.Then,forany c thatisa realnumberor ,thereexistsarearrangement a nwhosesequence ofpartialsums { s n} convergesto c Proof. Let p1,p2,.... denotenonnegativetermsof anintheorderinwhichtheyoccur,andlet q1,q2,... denotenegativetermsof anintheorderinwhichtheyoccur.InthenotationofLemma8.1,the series pnand a+ naswellas qnand aŠ nmayonlydierby zeroterms(ifsome an=0).Sotheseries pnand qndiverge. Considerthefollowingrearrangement.Givenanumber c ,take“rst

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56.REARRANGEMENTS111 k1terms pn,suchthatthenumber c liesbetweenthepartialsums sk1= p1+ p2+ + pk1and sk1Š 1;thatis, k1isde“nedbythecondition sk1Š pk1c>sk1+ m1or | c Š sk1+ m1| < | qm1| .Thiscan alwaysbedonebecausepartialsumsof pncanbelargerthanany number,whilepartialsumsof qncanbesmallerthananynumber owingtothedivergenceoftheseseries.So s1 sn sk1, 1 n k1, where | c Š sk1| c ,andtake m2nextterms qn,where m2isthesmallest integerforwhich sk1+ m1+ k1+ m2 0,thereisaninteger N such

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1128.SEQUENCESANDSERIES that | t Š tn| < forall n>N .Therefore,nk = N +1| ak| = | tnŠ tN +1| = | tnŠ t + t Š tN +1| | tnŠ t | + | t Š tN +1| < 2 So,bytaking N largeenough,thesumof any numberofterms | ak| k>N ,canbemadesmallerthan any preassignedpositivenumber. Let snand s nbepartialsumsof ananditsrearrangement a n. Onecantake n>N largeenoughsuchthat s ncontains a1, a2,..., aN(i.e.,theintegers1 2 ,...,N areinthesetofintegers k1,k2,...,knin thenotationsofDe“nition8.9).Thenthedierence | s nŠ sn| contains onlyterms | ak| with k>N (theterms a1, a2,..., aNarecancelled). Therefore, | s nŠ sn| < 2 forall n>N .If s n sand sn s ,then | sŠ s | < 2 ,whichshowsthat s= s because > 0isarbitrary. Thus, anabsolutelyconvergentseriesismuchlikea“nitesum.The sumdoesnotdependontheorderinwhichthesummationiscarried out.Incontrast,thesumofaconditionallyconvergentseriesdepends onthesummationorder.Thisisthecharacteristicdierencebetween thesetwoclassesofconvergentseries.56.1.StrategyforTestingSeries.Itwouldnotbewisetoapplytestsfor convergenceinaspeci“corderto“ndonethat“nallyworks.Instead, aproperstrategy,aswithintegration,istoclassifytheseriesaccording toitsform.Oneshouldalsokeepinmindthataconclusionaboutthe convergenceofaseriescanbereachedindierentways. 1.Specialseries .Aseries ancoincideswith(orisacombinationoforisequivalentto)specialseriessuchasa p -series,alternating p -series,geometricseries,telescopicseries,andsoon.Theirconvergencepropertiesareknown. 2.Seriessimilartospecialones .Ifaseries anhasaform thatissimilartooneofthespecialseries,thenoneofthecomparison testsshouldbeconsidered.Forexample,if anisarationaloralgebraicfunction(containsrootsofpolynomials),thentheseriesshould becomparedwitha p -series. 3.Necessaryconditionforconvergence .Itisisalwayseasier tocheckthecondition an 0as n thanitistoinvestigatethe series anforconvergence.Iftheconditiondoesnothold,theseries diverges. 4.Alternatingseries .If an=( Š 1)nbn, bn 0,thenthealternatingseriestestisanobviouspossibility.

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56.REARRANGEMENTS113 5.Ratioandroottests .Absoluteconvergenceimpliesconvergence.So,iftheratioorroottestshowsconvergence,thentheseriesin questionconvergesabsolutely.Ifthesetestsshowdivergence,thenthe seriesinquestionmaystillconvergebutnotabsolutely,andafurther investigationisrequired.Theroottestisconvenientforseriesofthe form ( bn)n.Theratiotestisconvenientwhen aninvolvesthefactorial n !orsimilarproductsofintegers.Theroottesthasawiderscope, butitismorediculttouse.Theratiotestisofteninconclusiveif anisarationaloralgebraicfunction( cn= | an +1| / | an| 1).Inthiscase, theasymptoticbehaviorof cnisrathereasyto“nd, cn 1+ b/n as n ,andthenuseDeMorganstest. 6.Seriesofnonnegativeterms .If an= f ( n ) 0andthe integral 1f ( x ) dx iseasytoevaluate,thentheintegraltestiseective. Also,itcanbeusedincombinationwiththecomparisontest: an f ( n )and 1f ( x ) dx convergesandsois an,or f ( n ) anand 1f ( x ) dx divergesandsois an. Example 8.29 Testtheseries ( n +1) / ( n2+ n +1) forconvergence. Solution: Forlarge n ,theleadingtermsofthetopandbottomof theratioare n and n2,respectively.So an 1 /n asymptoticallyfor large n .Theseriesresemblestheharmonicseries,whichdiverges.Itis natural,then,totrytoprovethedivergenceoftheseriesbycomparing itwiththeharmonicseries: n +1 n2+ n +1 > n n2+ n +1 n n2+ n2+ n2= 1 2 n Thus,theseriesindeeddivergesbycomparisonwiththeharmonicseries. Example 8.30 Testtheseries 3n/ (2 4 6 (2 n )) forconvergence. Solution: Eachterm aninvolvesafactorial-likeproductofintegers, whichsuggeststheuseoftheratiotest: an +1 an= 3n +1 2 4 (2 n ) (2 n +2) 2 4 (2 n ) 3n= 3 2 n +2 0 So,theseriesconverges. Example 8.31 Testtheseries sin( n2) eŠ n3 / 2forconvergence. Solution: Onehas | an| = | sin( n2) | eŠ n3 / 2 eŠ n3 / 2 eŠ n.

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1148.SEQUENCESANDSERIES Theseries eŠ nconvergesbytheintegraltest: 1eŠ xdx =1 /e< Hence,theseriesinquestionconvergesabsolutely.Alternatively,the convergenceof bn,where bn= eŠ n3 / 2,canbeestablishedbytheroot test:n bn= eŠ n1 / 2 0 < 1as n Example 8.32 Put hn=1+1 / 2+ +1 /n (apartialsumof theharmonicseries).Investigatetheconvergenceof npeŠ qhn,where p = q Solution: Theratiotestisinconclusive: an +1 an= ( n +1)peŠ qhn +1 npeŠ qhn= ( n +1)p npeŠ q ( hn +1Š hn)= (1+1 n)p 1 eŠq n +1 1 ToapplyDeMorganstest,theasymptoticbehaviorof an +1/anhas tobeinvestigated.Put f ( x )=(1+ x )pexp( Š qx/ (1+ x ))sothat an +1/an= f (1 /n ).Usingthelinearizationnear x =0,(1+ x )p 1+ px andexp( Š qx/ (1+ x )) 1 Š qx/ (1+ x ) 1 Š qx ,theasymptoticbehavior isobtained: f ( x ) (1+ px )(1 Š qx ) 1+( p Š q ) x = an +1 an 1+ p Š q n Thus,theseriesconvergesif p Š q< Š 1or p 0(12)n =1(n p2Š 1)n(13)n =1n enp(14)n =1npn Š 1 pnŠ (1 Š 1 /n )n, | p | < 1(15)n =1(n p Š 1) ,p 0

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57.POWERSERIES115 57.PowerSeries Definition 8.10( Power Series) Givenasequence { cn} ,theseriesn =0cnxn= c0+ c1x + c2x2+ c3x3+ iscalleda powerseries inthevariable x .Thenumbers cnarecalled the coecients oftheseries. Ingeneral,theserieswillconvergeordiverge,dependingonthe choiceof x .Thepowerseriesalwaysconvergesfor x =0tothenumber c0. Example 8.33 Forwhatvaluesof x doesthepowerseries n =0xn/n converge? Solution: Bytheroottest,n | xn| n = | x | n n | x | as n Sotheseriesconvergesforall Š 1 1 or x< Š 1.Theroottestisinconclusivefor x = 1.Thesevalues havetobeinvestigatedbydierentmeans.For x =1,thepowerseries becomestheharmonicseries 1 /n ,whichisdivergent.For x = Š 1, thepowerseriesbecomesthealternatingharmonicseries ( Š 1)n/n whichisconvergent.Thus,thepowerseriesconvergesif x [ Š 1 1) anddivergesotherwise. Givenanumber a ,considerapowerseriesinthevariable y = x Š a :n =0cnyn=n =0cn( x Š a )n. Itisalsocalleda powerseriescenteredat a ora powerseriesabout a Let S bethesetofallvaluesof x forwhichapowerseriesin x convergesandlet Sabethesetofallvaluesof x forwhichthecorresponding powerseriesin( x Š a )converges.Whatistherelationbetween S and Sa?Sincetheseriesareobtainedfromoneanotherbymerelyshifting thevalueofthevariablebyanumber a x x Š a ,theset Sais thereforeobtainedbyaddingthenumber a toeveryelementof S : (8.18) x Sa x Š a S = Sa= { x | x Š a S } Forexample,theseries n =0( x Š 2)n/n convergesif x Š 2 [ Š 1 1) or x [1 3)anddivergesotherwisebyExample8.33.Thus,theproblemof“ndingtheset Saisequivalenttotheproblemof“ndingthe set S .

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1168.SEQUENCESANDSERIES 57.1.PowerSeriesasaFunction.Supposethatapowerseriesin x convergesonaset S .Thenitde“nesafunctionon S : f ( x )= n =0cnxn,x S. Theset S iscalledthe domain ofsuchafunction.Functionsde“ned bypowerseriesaremostcommoninapplications.Manyofthemhave specialnotations(likeelementaryfunctionssin,cos,exp,etc.).Their propertiesarewellstudied.Inwhatfollows,itwillbeshownthat familiarelementaryfunctionssuchassin x ,cos x ,andexp x ,etccan alsoberepresentedaspowerseries. Example 8.34 FindthedomainoftheBesselfunctionoforder 0 thatisde“nedbythepowerseries J0( x )= n =0( Š 1)n 22 n( n !)2x2 n, where,bycommonconvention, 0!=1 Solution: Since an= cnx2 ncontainsthefactorial,theratiotestis moreconvenient: | an +1| | an| = x2| cn +1| | cn| = x222 n( n !)2 22( n +1)(( n +1)!)2= x2 22( n +1)2 0 as n .Sotheseriesconvergesforall x Valuesofafunctionde“nedbyapowerseriescanbeestimatedby partialsumsthatarepolynomialsinthevariable x : f ( x ) fn( x )=nk =0ckxk= c0+ c1x + c2x2+ + cnxn. Thus,partialsumsde“nea sequenceofpolynomialsthatconvergesto thefunction on S fn( x ) f ( x )forall x S .Theaccuracyof theapproximationisdeterminedbytheremainder Rn( x )= f ( x ) Š fn( x ).TheaccuracyassessmentisdiscussedinSection8.59.Sincethe remainder Rn( x )isafunctionon S ,theerroroftheapproximationis not generallyuniform;thatis,itdependson x .57.2.RadiusofConvergence.Theset S onwhichapowerseriesis convergentisanimportantcharacteristicanditspropertieshavetobe studied. Lemma 8.2(Propertiesofa Power Series) (i) .Ifapowerseries cnxnconvergeswhen x = b =0 ,thenitconvergeswhenever | x | < | b | .

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57.POWERSERIES117 (ii) .Ifapowerseries cnxndivergeswhen x = d =0 ,thenitdiverges whenever | x | > | d | Proof. If cnbnconverges,then,bythenecessaryconditionfor convergence, cnbn 0as n .Thismeans,inparticular,that,for =1,thereexistsaninteger N suchthat | cnbn| < =1forall n>N Thus,for n>N | cnxn| = cnbnxn bn = | cnbn| x b n< x b n. whichshowsthattheseries cnxnconvergesbycomparisonwiththe geometricseries qn,where q = x/b and | x/b | < 1or | x | < | b | Supposethat cndndiverges.If x isanynumbersuchthat | x | > | d | then cnxncannotconvergebecause,bypart(i)ofthelemma,the convergenceof cnxnimpliestheconvergenceof cndn.Therefore, cnxndiverges. Thislemmaallowsustoestablishthefollowingdescriptionofthe set S Theorem 8.26(ConvergencePropertiesofa Power Series) Fora powerseries cnxn,thereareonlythreepossibilities: (i) Theseriesconvergesonlywhen x =0 (ii) Theseriesconvergesforall x (iii) Thereisapositivenumber R suchthattheseriesconvergesif | x | R Proof. Supposethatneithercase1norcase2istrue.Thenthere arenumbers b =0and d =0suchthatthepowerseriesconvergesfor x = b anddivergesfor x = d .ByLemma8.2,thesetofconvergence S liesintheinterval | x || d | forall x S .Thisshowsthat | d | isan upperboundfortheset S .Bythecompletenessaxiom, S hasaleast upperbound R =sup S .If | x | >R ,then x S ,and cnxndiverges. If | x | | x | .Since b S cnxnconvergesby Lemma8.2. Theorem8.26showsthatapowerseriesconvergesina single open interval( Š R,R )anddivergesoutsidethisinterval.Theset S mayor maynotincludethepoints x = R .Thisquestionrequiresaspecial investigationjustlikeinExample8.33.Sothenumber R ischaracteristicforconvergencepropertiesofapowerseries. Definition 8.11(RadiusofConvergence) The radiusofconvergence ofapowerseries cnxnisapositivenumber R> 0 suchthat

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1188.SEQUENCESANDSERIES theseriesconvergesintheopeninterval ( Š R,R ) anddivergesoutside it.Apowerseriesissaidtohavea zeroradiusofconvergence R =0 ifitconvergesonlywhen x =0 .Apowerseriesissaidtohavean in“niteradiusofconvergence R = ,ifitconvergesforallvaluesof x Theratioorroottestcanbeusedtodeterminetheradiusofconvergence. Corollary 8.3(RadiusofConvergenceofa Power Series) Given apowerseries cnxn, iflimn | cn +1| | cn| = = R = 1 iflimn n | cn| = = R = 1 where R =0 if = and R = if =0 Proof. Put an= cnxnintheratiotest(Theorem8.20).Then | an +1| / | an| = | x || cn +1| / | cn|| x | .Theseriesconvergesif | x | < 1, whichshowsthat R =1 / .Similarly,usingtheroottest(Theorem 8.21),n | an| = | x |n | cn|| x | < 1,whichshowsthat R =1 / Remark. IfthesequencesinCorollary8.3donotconverge,then Theorem8.22shouldbeused,where an= cnxn. Oncetheradiusofconvergencehasbeenfoundand0 0 Solution: | cn +1| | cn| = qn +1 n +2 n +1 qn= q n +1 n +2 = q 1+1 /n 1+2 /n q = Therefore, R =1 / =1 /q .If x = Š 1 /q ,then cnxn=( Š 1)n/ n +1= ( Š 1)nbn.Thesequence bnconverges monotonically to0sothat ( Š 1)nbnconvergesbythealternatingseriestest.If x = Š 1 /q ,then cnxn= 1 / n +1 > 1 / 2 n n 1.The p -series 1 /n1 / 2diverges( p =1 / 2 < 1)sothat 1 / n +1divergesbythecomparisontest.Thus,the intervalofconvergenceis S =[ Š 1 /q, 1 /q ).

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57.POWERSERIES119 Example 8.36 Findtheradiusofconvergenceandtheintervalof convergenceofthepowerseries n2( x +1)n/qn,where q> 0 Solution: Put y = x +1.If S istheintervalofconvergenceof cnyn, where cn= n2/qn,thentheintervalofconvergenceinquestionisobtainedbyadding Š 1toallnumbersin S accordingtotherule(8.18). ByCorollary8.3,n | cn| = 1 qn n2= 1 q (n n )2 1 q = So R =1 / = q .If y = q ,then cnyn= n2,andtheseries n2diverges ( an= n2doesnotconvergeto0).If y = Š q ,then cnxn=( Š 1)nn2, andtheseriesdivergesbecause an=( Š 1)nn2,doesnotconvergeto0. Theseriesconvergesonlyif | y | = | x +1 | 0 (6)n =0 n ( x +1)n(7)n =1( Š 1)nn2+1 n3+3 ( x Š 1)n(8)n =1(4 x +1)n n2(9)n =1xn 1 3 5 (2 n Š 1) (10) .n =1n2x2 n 2 4 (2 n ) (11)n =0( n !)k ( kn )! xn,k> 0(integer) (12)n =04n n ( x +3)n(13)Let p
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1208.SEQUENCESANDSERIES Finditsdomain. (15)Afunction f isde“nedbythepowerseries f ( x )= p + qx + px2+ qx3+ px4+ qx5+ ; thatis,itscoecients c2 k= p and c2 k Š 1= q ,where p and q arereal. Findthedomainof f andanexplicitexpressionof f ( x )(thesumof theseries). (16)If f ( x )= cnxn,where cn +4= cnforall n 0,“ndthedomain of f andaformulafor f ( x ). (17) Power series cnxnand bnxnhavetheradiiofconvergence R1and R2,respectively.Whatistheradiusofconvergenceof ( cn+ bn) xn? (18)Supposethattheradiusofconvergenceof cnxnis R .Whatis theradiusofconvergenceof cnxkn,where k> 0isaninteger?58.RepresentationofFunctionsasPowerSeries Considerapowerseries 1 Š x2+ x4Š x6+ x8+ =n =0( Š 1)nx2 n. Itisageometricserieswith q = Š x2,andthereforeitconvergesfor all | q | = x2< 1or x ( Š 1 1).Usingtheformulaforthesumofa geometricseries,oneinfersthat 1 1+ x2=1 Š x2+ x4+ =n =0( Š 1)nx2 nforall Š 1 0 ,anddeterminetheintervalofitsvalidity. Solution: Put y = x Š a .Thefunctioncanberewritteninaform thatresemblesthesumofageometricseries: 1 x = 1 a (1+ y/a ) = 1 an =0 Š y a n=n =0( Š 1)n an +1( x Š a )n,x (0 2 a ) .

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58.REPRESENTATIONOFFUNCTIONSASPOWERSERIES121 Thegeometricseriesconvergesif | q | = |Š y/a | = | y | /a< 1,andhence thisrepresentationisvalidonlyif Š a 0 ,thenthefunction f de“nedby f ( x )= c0+ c1( x Š a )+ c2( x Š a )2+ =n =0cn( x Š a )nisdierentiable(andthereforecontinuous)ontheinterval ( a Š R,a + R ) and f( x )= c1+2 c2( x Š a )+3 c3( x Š a )2+ =n =1ncn( x Š a )n Š 1, f ( x ) dx = C + c0( x Š a )+ c1( x Š a )2 2 + = C +n =0cn( x Š a )n +1 n +1 .

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1228.SEQUENCESANDSERIES Theradiiofconvergenceofthesepowerseriesareboth R Thus,for powerseries ,thedierentiationorintegrationandthe summationcanbecarriedoutinanyorder: d dx cn( x Š a )n= d dx [ cn( x Š a )n] cn( x Š a )n dx = [ cn( x Š a )n] dx. Remark. Theorem8.27statesthattheradiusofconvergenceof apowerseriesdoesnotchangeafterdierentiationorintegrationof theseries.Thisdoesnotmeanthatthe intervalofconvergence does notchange.Itmayhappenthattheoriginalseriesconvergesatan endpoint,whereasthedierentiatedseriesdivergesthere. Example 8.38 Findtheintervalsofconvergencefor f f,and fif f ( x )= n =1xn/n2. Solution: Here cn=1 /n2andhencen | cn| =1 /n n2=(1 /n n )2 1= .Sotheradiusofconvergenceis R =1 / =1.For x = 1, theseriesisa p -series 1 /n2thatconverges( p =2 > 1).Thus, f ( x ) isde“nedontheclosedinterval x [ Š 1 1].ByTheorem8.27,the derivatives f( x )= n =1xn Š 1/n and f( x )= n =2( n Š 1) xn Š 2/n have thesameradiusofconvergence R =1.For x = Š 1,theseries f( Š 1)= ( Š 1)n Š 1/n isthealternatingharmonicseriesthatconverges,whereas theseries f( Š 1)= ( Š 1)n( n Š 1) /n divergesbecausethesequenceof itstermsdoesnotconvergeto0: | ( Š 1)n( n Š 1) /n | =1 Š 1 /n 1 =0. For x =1,theseries f(1)= 1 /n istheharmonicseriesandhence diverges.Theseries f(1)= ( n Š 1) /n alsodiverges(( n Š 1) /n does notconvergeto0).Thus,theintervalsofconvergencefor f f,and fare,respectively,[ Š 1 1],[ Š 1 1),and( Š 1 1). Theterm-by-termintegrationofapowerseriescanbeusedtoobtainapowerseriesrepresentationofantiderivatives. Example 8.39 Findapowerseriesrepresentationfor tanŠ 1x Solution: tanŠ 1x = dx 1+ x2= n =0( Š x2)n dx = C +n =0( Š 1)n +1x2 n +1 2 n +1 SincetanŠ 10=0,theintegrationconstant C satis“esthecondition 0= C +0or C =0.Thegeometricserieswith q = Š x2convergesif | q | < 1.Hence,theradiusofconvergenceoftheseriesfortanŠ 1x is R =1(thepowerseriesrepresentationisvalidfor x ( Š 1 1)).

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58.REPRESENTATIONOFFUNCTIONSASPOWERSERIES123 Inparticular,thenumber1 / 3islessthantheradiusofconvergenceofthepowerseriesfortanŠ 1x .SothenumbertanŠ 1(1 / 3)= / 6canbewrittenasthenumericalseriesbysubstituting x =1 / 3 intothepowerseriesfortanŠ 1x .Thisleadstothefollowingrepresentationofthenumber : =2 3n =0( Š 1)n (2 n +1)3n.58.2.PowerSeriesandDifferentialEquations.Apowerseriesrepresentationisoftenusedtosolve dierentialequations .Therelationbetween afunction f ( x ),itsargument x ,anditsderivatives f( x ), f( x ),and sooniscalledadierentialequation.Afunction f ( x )thatsatis“esa dierentialequationisgenerallydicultto“ndinaclosedform.A powerseriesrepresentationturnsouttobehelpful.Sinceinthisrepresentationafunctionisde“nedbyasequence { cn} f ( x )= cnxn,and soareitsderivatives f( k )( x ),adierentialequationimposesconditions on cnthataresolvedrecursively. Example 8.40 Findapowerseriesrepresentationofthesolution oftheequation f( x )= f ( x ) anddetermineitsradiusofconvergence. Solution: Put f ( x )= cnxnandhence f( x )= ncnxn Š 1.Then theequation f= f gives c1+2 c2x +3 c3x2+4 c4x3+ = c0+ c1x + c2x2+ c3x3+ Bymatchingthecoecientsatthemonomialterms1, x x2, x3,and soon,one“nds: c0= c1c2= c1 2 ,c3= c2 3 ,...,cn= cn Š 1 n Usingthelatterrelationrecursively: cn= 1 n cn Š 1= 1 n ( n Š 1) cn Š 2= 1 n ( n Š 1)( n Š 2) cn Š 3= = c0 n So f ( x )= c0 n =0xn/n !,where c0isaconstant(theequationissatis“edforanychoiceof c0).Bytheratiotest,theseriesconvergesforall x (so R = ).Indeed, cn=1 /n !and cn +1/cn=1 / ( n +1) 0= andhence R =1 / = Forthissimpledierentialequation,itisnotdicultto“nd f ( x )= c0exbyrecallingthepropertiesoftheexponentialfunction:( ex)= ex. Thecondition f (0)= e0=1determinestheconstant c0=1.Thus,

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1248.SEQUENCESANDSERIES theexponentialfunctionhasthefollowingpowerseriesrepresentation: (8.19) ex=1+ x 1! + x2 2! + x3 3! + =n =0xn n Theseriesconvergesontheentirerealline.Inparticular,thenumber e hasthefollowingseriesrepresentation: e =1+ 1 1! + 1 2! + 1 3! + =n =01 n .58.3.ApproximationofDe“niteIntegrals.Ifaninde“niteintegralof f ( x )isdiculttoobtain,thentheevaluationoftheintegral b af ( x ) dx posesaproblem.Apowerseriesrepresentationoersasimplewayto approximatethevalueoftheintegral.Supposethat f ( x )= cnxnfor Š R
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58.REPRESENTATIONOFFUNCTIONSASPOWERSERIES125 Adirectcalculationshowsthat b7 1 32 10Š 5and b8 1 46 10Š 6. So n =7issucienttoapproximatetheintegralwiththerequired accuracy. 58.4.Exercises.In(1)…(3),“ndapowerseriesrepresentationforthefunctionanddeterminetheintervalofconvergence. (1) f ( x )= 1 1 Š x4(2) f ( x )= x 3 x2+2 (3) f ( x )= x +1 2 x2Š x Š 1 In(4)…(6),usedierentiationto“ndapowerseriesrepresentationfor thefunctionanddeterminetheintervalofconvergence. (4) f ( x )= 1 (1+ x )2(5) f ( x )= x3 (1 Š 4 x2)2(6) f ( x )= 1 (1+ x4)3In(7)…(9),useintegrationto“ndapowerseriesrepresentationforthe functionanddeterminetheradiusofconvergence. (7) f ( x )=ln(1+ x )(8) f ( x )=ln 1+ x2 1 Š x2 (9) f ( x )=tanŠ 1(3 x ) In(10)…(12),“ndapowerseriesrepresentationfortheinde“niteintegralanddeterminetheradiusofconvergence. (10) ln(1 Š x ) x dx (11) exŠ 1 Š x x2dx (12) tanŠ 1( x2) dx (13)Findapowerseriesrepresentationforsin x andcos x usingthe dierentialequation f+ f =0.Determinetheintervalofconvergence. (14)ShowthattheBesselfunctionoforder0de“nedinExample8.34 satis“esthedierentialequation: x2J 0( x )+ xJ 0( x )+ x2J0( x )=0 In(15)…(17),usedierentiationorintegrationto“ndthesumofthe series. (15)n =1nxn Š 1(16)n =2n ( n Š 1) xn Š 2(17)n =0xn +1 n +1 In(18)…(20),howmanytermsdoesoneneedinapowerseriesapproximationtoevaluatetheintegralwiththeabsoluteerrornotexceeding 10Š 6? (18) 1 0dx 1+ x8(19) 1 0eŠ xŠ 1 x dx (20) 1 / 2 0ln(1+ x4) dx

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1268.SEQUENCESANDSERIES (21)Findtheradiusofconvergenceofthehypergeometricseries: 1+ ab 1! c x + a ( a +1) b ( b +1) 2! c ( c +1) x2+ a ( a +1)( a +2) b ( b +1)( b +2) 3! c ( c +1)( c +2) x3+ where a b ,and c arereals.UseDeMorganstesttodeterminethe intervalofconvergence.59.TaylorSeries59.1.RealAnalyticFunctions.Supposeafunction f isrepresentedby apowerseries( R> 0): f ( x )= c0+ c1( x Š a )+ c2( x Š a )2+ =n =0cn( x Š a )n, | x Š a | 0 ,thenitscoecientsare cn= f( n )( a ) n Definition 8.12(RealAnalyticFunctions) Afunction f onan openinterval I issaidtobeanalyticif,forany a I ,ithasapower seriesrepresentation f ( x )= cn( x Š a )nthatconvergesinsomeopen interval ( a Š ,a + ) I ,where > 0 .

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59.TAYLORSERIES127 Theclassofanalyticfunctionsplaysasigni“cantroleinapplications.Theirpropertiesarediscussednext. Theorem 8.29( Power SeriesRepresentationofAnalyticFunctions) Afunction f thatisanalyticonanopeninterval I hasthe powerseriesrepresentation (8.20) f ( x )=n =0f( n )( a ) n ( x Š a )nforany a I thatconvergesinanopensubintervalof I thatincludes a ThistheoremfollowsfromDe“nition8.12andTheorem8.28. InExample8.37itwasfoundthat (8.21) 1 x =n =0( Š 1)n an +1( x Š a )n,x (0 2 a ) Thisshowsthatthefunction f ( x )=1 /x isanalyticforall x> 0 because a canbeanypositivenumber;thatis,thefunctionhasa powerseriesrepresentationthatconvergesinanopensubintervalof (0 )containingany a> 0.Similarly,theanalyticityof f ( x )=1 /x canbeestablishedforall x< 0. Itisimportanttoemphasizethatapowerseriesforananalytic functiondoes not necessarilyconvergeontheentiredomainofthe function.Butananalyticfunctioncan always berepresentedbya convergentpowerseriesinaneighborhoodofeverypointofitsdomain. Equation(8.21)illustratesthepoint. Theorem 8.30(PropertiesofAnalyticFunctions) (i) Thesumsandproductsofanalyticfunctionsareanalytic. (ii) Thereciprocal 1 /f ofananalyticfunction f isanalyticif f is nowherezero. (iii) Thecomposition f ( g ( x )) ofanalyticfunctions f and g isanalytic. (iv) .Analyticfunctionsaredierentiablein“nitelymanytimes. Aproofofproperties(i)…(iii)isgiveninmoreadvancedcalculus courses.Property(iv)followsfromTheorem8.27. Itsconverseis notgenerallytrue;thatis,therearefunctionsthataredierentiable in“nitelymanytimesatapoint,buttheycannotberepresentedbya powerseriesthatconvergesinanopenintervalthatincludesthispoint Asanexample,considerthefunction f ( x )= eŠ 1 /x2if x =0and f (0)=0 .

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1288.SEQUENCESANDSERIES Thefunctioniscontinuousat x =0becauselimx 0eŠ 1 /x2=limu eŠ u=0= f (0).Itisdierentiableat x =0because f(0)=limx 0f ( x ) Š f (0) x =limx 0eŠ 1 /x2 x =0 The“rstequalityisthede“nitionof f(0).Thelastlimitisestablished byinvestigatingtheleftandrightlimits x 0withthehelpofthe substitution x =1 /u as x 0;theleftandrightlimits coincidebecause ueŠ u2 0as u (theexponentialfunction decreasesfasterthananypowerfunction).Inasimilarfashion,itcan beprovedthat f( n )(0)=0forall n (seeExercise59.5.24).Thus, f ( x ) hasnopowerseriesrepresentation cnxninaneighborhoodof x =0 because,ifitdid,then,byTheorem8.29thefunctionshouldhavebeen identically0insomeinterval( Š ), > 0,(as f( n )(0)=0forall n ), whichisnottrue( f ( x ) =0forall x =0).Hence,thefunctionisnot analyticat x =0.59.2.TaylorandMaclaurinSeries.Definition 8.13(TaylorandMaclaurinSeries) Theseriesin (8.20)iscalledthe Taylorseries ofafunction f at a (orabout a orcenteredat a ).ThespecialcaseoftheTaylorserieswhen a =0 is calledthe Maclaurinseries ofafunction f TheTaylorseriesoftheexponentialfunction exabout x =0is givenby(8.19).Theseriesconvergesforall x ;thatis,itsradiusof convergenceis R = Trigonometricfunctions. ConsidertheMaclaurinseriesof f ( x )= sin x .Onehas f( x )=(sin x )=cos x and f( x )=(cos x )= Š sin x Hence, f(2 n )( x )=( Š 1)nsin x,f(2 n +1)( x )=( Š 1)n Š 1cos x, and f(2 n )(0)=0, f(2 n +1)(0)=( Š 1)n Š 1.So sin x =n =0( Š 1)n Š 1x2 n +1 (2 n +1)! = x Š x3 3! + x5 5! + ,R = Bytheratiotest, | cn +1| / | cn| =(2 n +1)! / (2 n +3)!=1 / [(2 n +2)(2 n + 3)] 0= ,andtheradiusofconvergenceis R =1 / = .The seriesconvergesontheentirerealline. TheMaclaurinseriesfor f ( x )=cos x isobtainedbydierentiating cos x =(sin x )=n =0( Š 1)n Š 1x2 n (2 n )! =1 Š x2 2! + x4 4! + ,R = .

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59.TAYLORSERIES129 ByTheorem8.27italsoconvergesontheentirerealline. Binomialseries. Let f ( x )=(1+ x )p,where p isanyrealnumber. Itsderivativesare f( x )= p (1+ x )p Š 1, f( x )= p ( p Š 1)(1+ x )p Š 2,and, ingeneral, f( n )( x )= p ( p Š 1) ( p Š n +1)(1+ x )p Š n. TheMaclaurinseriesfor(1+ x )piscalledthe binomialseries .The traditionalnotationforitscoecientsis cn= f( n )(0) n = p ( p Š 1) ( p Š n +1) n = p n Thesenumbersarecalledthe binomialcoecients .Thebinomialseries anditsradiusofconvergenceare (1+ x )p=n =0 p n xn=1+ p 1! x + p ( p Š 1) 2! x2+ ,R =1 Thecoecientssatisfytherecurrencerelation cn +1= cn( p Š n ) / ( n +1). Therefore,bytheratiotest, | cn +1| / | cn| = | p Š n | / ( n +1)= | 1 Š p/n | / (1+ 1 /n ) 1= as n .Hence, R =1 / =1.59.3.TaylorSeriesofAnalyticFunctions.Everyanalyticfunctionina neighborhoodofanypointisrepresentedbytheTaylorseriesabout thatpoint.IftheTaylorseriesconvergesontheentirerealline,then thefunctionisanalyticeverywhere.Inparticular,theexponential exandtrigonometricfunctionssin x andcos x areanalyticeverywhere. Moreover,thepropertiesofanalyticfunctionsstatedinTheorem8.30 allowsustoadd,multiply,andmakeacompositionoftheTaylorseries(onthecommonintervalsoftheirconvergence)justlikeordinary sumstoobtaintheTaylorseriesrepresentationofthesums,products, andcompositionsofanalyticfunctions.Theseareextremelyuseful propertiesinapplications. Example 8.42 Find“rstfourtermsoftheTaylorseriesforthe function f ( x )=exp(tanŠ 1x ) about x =0 Solution: Calculationofthederivativesofsuchafunctionisrather tedious.Instead,notethat exandtanŠ 1x arebothanalyticinaneighborhoodof x =0.SothecompositionoftheirTaylorseries(see(8.19) andExample8.39)givesthesought-afterTaylorseries.Onlymonomials1, x x2,and x3havetoberetainedwhencalculatingthecomposition.Thisimpliesthatitissucienttoretaintwoleadingtermsin

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1308.SEQUENCESANDSERIES theTaylorseriestanŠ 1x = x Š x3/ 3+ andfourleadingtermsin theTaylorseries(8.19)oftheexponentialfunction: etanŠ 1x=1+tanŠ 1x + 1 2 tanŠ 1x 2+ 1 6 tanŠ 1( x ) 3+ =1+ x Š x3 3 + + 1 2 x Š x3 3 + 2+ 1 6 x Š x3 3 + 3+ =1+ x Š x3 3 + 1 2 x2+ 1 6 x3+ =1+ x + 1 2 x2Š 1 6 x3+ 59.4.ApproximationsbyTaylorPolynomials.Ananalyticfunction f can beapproximatedbya“nitesumoftheTaylorseries: f ( x ) nk =0f( k )( a ) k ( x Š a )k= Tn( x ) Thepolynomial Tn( x )iscalledthe Taylorpolynomial about a .The convergenceoftheTaylorseriesguaranteesthattheremainderconvergesto0: Rn( x )= f ( x ) Š Tn( x ) 0as n forall | x Š a |
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59.TAYLORSERIES131 Sincethe( n +1)thderivativeofapolynomialofdegree n vanishesand hence g( n +1)( t )= f( n +1)( t ) Š n M ,theproofwillbecompleteifonecan showthat g( n +1)( )=0forsome between x and a (thelatterwould implythat M = f( n +1)( ) /n !).Bythede“nitionofTaylorpolynomials, f( k )( a )= T( k ) n( a )for k =0 1 ,...,n ,andhence g ( a )= g( a )= = g( n )( a )=0.Thefunction g ( t )isdierentiableand g ( x )= g ( a )=0 bythechoiceof M ;therefore,byRollestheorem,thereisanumber t1between x and a suchthat g( t1)=0.Similarly,thefunction g( t ) isdierentiableand g( t1)= g( a )=0;hence,thereisanumber t2between t1and a suchthat g( t2)=0.After n +1stepsofthis procedure,onearrivesattheconclusionthat g( n +1)( tn +1)=0forsome number tn +1= between tnand a ,thatis,between x and a Corollary 8.4(TaylorsInequality) If | f( n +1)( x )| Mnfor | x Š a | d
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1328.SEQUENCESANDSERIES Figure8.9. Anillustrationofanapproximationof f ( x )=sin x (thedashedredcurve)byitsTaylorpolynomialsat x =0(thesolidbluecurve).As n increases, Tn( x )approaches f ( x )=sin x .Theapproximationbecomesbetterinalargerintervalforalarger n inaccordancewiththeanalysisofExample8.43.59.5.Exercises.In(1)…(5),“ndtheMaclaurinseriesforthefunctionandtheradiusof convergence. (1)ln(1+ x )(2)tan x (3)sinh x (4)cosh x (5) x6+2 x5Š x3+ x Š 3 In(6)…(9),“ndtheTaylorseriesforthefunctionabout a andtheradius ofconvergence. (6)cos x,a = (7)1 / x,a =4 (8)sin x,a = / 2(9)(1+ x )2 / 3,a =7 In(10)…(13),useMaclaurinseriesforbasicfunctionsto“ndtheMaclaurinseriesforthefunction. (10) x cos( x2/ 2)(11) x 3 1+ x4(12) x Š sin x x3(13) x2tanŠ 1( x2)

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59.TAYLORSERIES133 In(14)…(17),usetheproductsandcompositionoftheMaclaurinseries forbasicfunctionsto“ndthe“rstthreenon-vanishingtermsofthe Maclaurinseriesforthefunction. (14)sin( (cos x ))(15) esin x(16)tanŠ 1x ln(1+ x )(17)ln(cos x ) (18)Findthe“rst“venonvanishingtermsoftheMaclaurinseriesfor f ( x )= ex/ cos x Hint :Put f ( x )= c0+ c1x + + c4x4+ ,thenusetheproductof theMaclaurinseriesto“ndthecoecientsfrom ex= f ( x )cos x In(19)…(21),“ndthedegreeofaTaylorpolynomialtoapproximate theintegrandsothattheerrorofapproximatingtheintegraldoesnot exceed10Š 4. (19) 1 / 2 0tanŠ 1( x2) dx (20) 1 0exŠ 1 x dx (21) 1 / 2 0(1+ x4)1 / 4dx In(21)…(23),“ndthesumoftheseries. (21)n =0( Š 1)n2 n 62 n(2 n )! (22)n =0( Š 1)n3n 2nn (23)1 Š ln2+ (ln2)2 2! Š (ln2)3 3! + (24)(i)Forthefunction f ( x )= eŠ 1 /x2if x =0and f (0)=0,show that f( n )(0)=0forall n andhence f cannotberepresentedasapower seriesnear0.(ii)Let f ( x )= eŠ 1 /xif x> 0and f ( x )=0if x 0.Is thisfunctionanalyticeverywhere?

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CHAPTER9 FurtherApplicationsofIntegration 60.ArcLength60.1.TheLengthofaCurve.Wehaveseenvariousapplicationsofintegrationtothecomputationoftheareaofadomainandtothecomputationofthevolumeofasolid.Itisperhapsmoresurprisingthat wecanalsouseintegrationtocomputethelengthofacurvebetween twogivenpoints.Thismaysoundcounterintuitiveat“rst,sincein theapplicationswehaveseensofar,integrationwasusedtocompute someparameterofanobjectthatexistedina higherdimension than thefunctionthatwasbeingintegrated. Let f beafunctionsothat,ontheinterval[ a,b ],thederivative fof f existsandisacontinuousfunction.Wewouldliketoknowthe lengthofthecurveof f ,startingatthepoint A =( a,f ( a ))andending atthepoint B =( b,f ( b )). Intuitively,wecanimaginethatwelayaropeoverthegraphof f betweenthetwoendpoints,mark A and B ontherope,thenstraighten thatropeout,andmeasurethedistancebetweenthem. Amoreformalde“nition,whichisusefulintheactualcomputation ofthelengthofthecurve,isthefollowing.Cuttheinterval[ a,b ] into n equalparts,usingpoints a = x0
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1369.FURTHERAPPLICATIONSOFINTEGRATION Figure9.1. Arclengthasalimit. Letusnowreturnto(9.1)inordertocomputelimn Kn.Let ( b Š a ) /n =( xiŠ xi Š 1) /n = x .Notethatthen | Pi Š 1Pi| = ( xiŠ xi Š 1)2+( yiŠ yi Š 1)2= ( x )2+( x )2( yiŠ yi Š 1)2 ( xiŠ xi Š 1)2= x 1+ ( yiŠ yi Š 1)2 ( xiŠ xi Š 1)2. Nowobservethatsince fiscontinuous,theintermediatevalue theoremimpliesthatthereisarealnumber x i [ xi Š 1,xi]suchthatyiŠ yi Š 1 xiŠ xi Š 1= f( x i).Hence,thepreviouschainofequalitiesyields | Pi Š 1Pi| = x 1+ f( x i)2. Summingoverall i ,weget Kn=ni =1 x 1+ f( x i)2. As n goestoin“nity,theleft-handside,byde“nition,convergestothe lengthofthecurveof f between A and B ,whiletheright-handside, beingaRiemannsum,convergesto b a 1+ f( x )2dx .Hence,wehave provedthefollowingtheorem. Theorem 9.1 If fisacontinuousfunctionontheinterval [ a,b ] thenthelengthofthegraphof f ( x ) fromthepoint ( a,f ( a )) tothepoint

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60.ARCLENGTH137 Figure9.2. Thecurveof f ( x )=2 3x3 / 2. ( b,f ( b )) isequalto L = b a 1+ f( x )2dx. Example 9.1 Findthelengthofthecurveof f ( x )=2 3x3 / 2from (0 0) to (1 2 / 3) .SeeFigure9.2foranillustration. Solution: Wehave f( x )= x ,so fisacontinuousfunctionon[0 1], andthereforeTheorem9.1applies.Usingthattheorem,weobtain L = 1 0 1+( x )2dx = 1 0 1+ xdx = 2 3 (1+ x )3 / 21 0= 4 2 3 Š 2 3 Notethattheresultisremarkablyclosetothelengthofthe straight line thatconnectsthetwopointsinquestion,whichis 13 / 3. Wecanuseournewtechniquetoverifyaclassicformula. Example 9.2 UseTheorem9.1tocomputethecircumferenceofa circleofradius1. Solution: Letusplacethecenteroftheunitcircleattheorigin.Then theboundaryofthecircleisthesetofpointssatisfying x2+ y2=1.

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1389.FURTHERAPPLICATIONSOFINTEGRATION Figure9.3. Onequarteroftheunitcircle. WewanttouseTheorem9.1,soweneedapartofthecirclewhere thatsatis“estheverticallinetest(so y isafunctionof f )andwhere thetangentlinetothecircleisneververtical(sothat f( x )exists). Forinstance,wecanchoosethequarterofthecirclethatstartsinthe point Š 2 2 andendsinthepoint Š 2 2 .SeeFigure9.3foran illustration.Onthatpartofthecurve, f( x )= Šx 1 Š x2iscontinuous, soTheorem9.1implies L = 2 / 2 Š 2 / 2 1+ f( x )2dx = 2 / 2 Š 2 / 2 1+ x2 1 Š x2dx = 2 / 2 Š 2 / 21 1 Š x2dx =[sinŠ 1x ] 2 / 2 Š 2 / 2= / 2 Thisimpliesthatthecircumferenceofthefullcircleisfourtimesthis much,thatis,2

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61.SURFACEAREA139 60.2.Remarks.Recallthatinthe“rstparagraphofthissection,we discussedwhyitmayseemcounterintuitivethatintegrationplaysarole inthecomputationofarclengths.Nowwecanseethatthepurported contradictionexplainedthereisresolvedbythefactthattheintegrand inTheorem9.1contains f,not f Comparedtootherformulaswelearnedinourearlierstudiesof integration,itisrelativelyrarethattheformulagivenbyTheorem9.1 canbeexplicitlycomputed,since b a 1+ f( x )2dx isoftendicult tohandle.Therefore,wemustoftenresorttoapproximateintegration whilecomputingarclengths.60.3.Exercises.(1)Findthelengthofthecurve f ( x )= x2/ 2betweenthepoints givenby x =0and x =1. (2)Findthelengthofthecurve f ( x )= x3+3 4 xbetweenthepoints givenby x =2and x =4. (3)Findthelengthofthecurve f ( x )=ln(cos x )betweenthe pointsgivenby x =0and x = / 4. (4)ProvethatTheorem9.1providesthecorrectvalueforthearc lengthof f when f isa linearfunction (5)Useamethodofapproximateintegrationtoestimatethelength ofthecurveof f ( x )= exasfrom(0 1)to(1 ,e ). (6)Useamethodofapproximateintegrationtoestimatethelength of f ( x )=sin x from(0 0)to( 0).61.SurfaceArea61.1.TheDe“nitionofSurfaceArea.Inthelastsection,wede“nedthe lengthofacurve,anddeducedaformulaforthecomputationofthat length.Letusnowtakeacurve,sayofafunction f ( x )= y ,where x [ a,b ]and fiscontinuouson[ a,b ].Letus rotate thiscurvearound thehorizontalaxis,asshowninFigure9.4.Whatistheareaofthe obtained surfaceofrevolution ? Thede“nitionoftheareainquestion,anditscomputation,willbe quitesimilartowhatwehavediscussedintheprevioussectionforthe arclength. Cuttheinterval[ a,b ]into n equalparts,usingpoints a = x0
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1409.FURTHERAPPLICATIONSOFINTEGRATION Figure9.4. Asurfaceobtainedbyrotatingacurve. Figure9.5. Approximatingasurfaceofrevolution. resultsinthelateralsurface Sn,iofatruncatedconewithslantheight liandradii yi Š 1and yi.SeeFigure9.5foranillustration.Itisthen notdiculttoprovethattheareaof Sn,iisequalto (9.2) A ( Sn,i)= ( yi Š 1+ yi) li. As n goestoin“nity,thesumoftheareasofthesurfaces Sn,iapproximateswhatweintuitivelythinkofastheareaofthesurfaceobtained byrotatingthecurve.

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61.SURFACEAREA141 Infact,itcanbeprovedthatthelimit (9.3) S ( A )=limn ni =1Sn,i=limn ni =1 ( yi Š 1+ yi) liexists.Wede“ne thislimit tobetheareaofthesurfaceofrevolution obtainedwhenthecurveof f isrotatedaroundthehorizontalaxis, with x [ a,b ]. Inordertocomputethissurfacearea,recallfromthelastsection thatthereexistsarealnumber x i [ xi Š 1,xi]suchthat li= | Pi Š 1Pi| = x 1+ f( x i)2.Alsonotethat,since f iscontinuous,smallchanges in x leadtosmallchangesin f ( x )= y ,soif n islargeenough,then f ( x i) f ( xi)= yiand f ( x i) f ( xi Š 1)= yi Š 1.Therefore,(9.3) implies (9.4) S ( A )=limn ni =12 f ( x i) x 1+ f( x i)2, where x =( b Š a ) /n .Nownoticethatthelastexpressionobtained for S ( A )isthelimitofaRiemannsum,thatis,anintegral(ofthe function2 f ( x ) 1+ f( x )2).Thismeansthatweprovedthefollowing theorem. Theorem 9.2 Let f beafunctionthat fiscontinuousonthe interval [ a,b ] .Let S bethesurfaceobtainedbyrotatingthecurve y = f ( x ) ,where x [ a,b ] ,aroundthehorizontalaxis. Thentheareaof S is A ( S )= b a2 f ( x ) 1+ f( x )2dx. Example 9.3 Computethesurfaceareaofasphereofradius r Solution: Suchaspherecanbeobtainedbyrotatingthesemicircle givenbytheequation f ( x )= r2Š x2aroundthehorizontalaxis.See Figure9.6foranillustration.

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1429.FURTHERAPPLICATIONSOFINTEGRATION Figure9.6. Asphereasasurfaceofrevolution. Theorem9.2thenyields A ( S )= r Š r2 r2Š x2 1+ x2 r2Š x2dx = r Š r2 r2Š x2 r2 r2Š x2dx = r Š r2 rdx =[2 rx ]r Š r=4 r2. 61.2.Variations.Ifwerotateourcurvearoundthe verticalaxis instead ofthehorizontalaxis,thenmostofthepreviousargumentremains valid.Theonlydierenceisthatwhenthepoint Pi=( xi,yi)isrotated, itisrotatedinacircleofradius xi,not yi.Thisleadstothefollowing theorem. Theorem 9.3 Let f beafunctionsuchthat fiscontinuouson theinterval [ a,b ] .Let S bethesurfaceobtainedbyrotatingthecurve y = f ( x ) ,where x [ a,b ] ,aroundtheverticalaxis. Thentheareaof S is A ( S )= b a2 x 1+ f( x )2dx. Notethatthe f ( x )termintheintegrandofTheorem9.2isreplaced by x .

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61.SURFACEAREA143 Example 9.4 Rotatethecurvegivenby y = f ( x )= x2/ 2 ,with x [0 1] ,aroundtheverticalaxis.Findtheareaoftheobtainedsurface. Figure9.7. Thecurveof y = x2/ 2andthesurface obtainedbyitsrotation. Solution: Theorem9.3implies A ( S )= 1 02 x 1+ x2dx =2 1 3 ( x2+1)3 / 21 0=2 2 2 Š 1 3 0 6095 NotethatanotherwayofwritingtheresultofTheorem9.2is A ( S )= b a2 f ( x ) 1+ dy dx 2dx. Byinterchangingtherolesof x and y ,thisimpliesthat,forcurvesgiven byanequation g ( y )= x ,thefollowingholds. Theorem 9.4 Let g beafunctionsuchthat giscontinuouson theinterval [ a,b ] .Let S bethesurfaceobtainedbyrotatingthecurve x = g ( y ) ,where y [ a,b ] ,aroundtheverticalaxis.

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1449.FURTHERAPPLICATIONSOFINTEGRATION Thentheareaof S isgivenby A ( S )= b a2 g ( y ) 1+ dx dy 2dy. Whilethesurfaceareaofaconecanbecomputedbyelementary methods,itiselucidatingtocomputeitwithournewmethodandsee thattheresultiswhatweexpectittobe. Example 9.5 Findthesurfaceareaofarightconeofbaseradius R andheight h Solution: Thebasecircleoftheconehasarea R2 .Inordertocomputethelateralsurface,notethatthelateralsurfacecanbeobtainedby rotatingthelinesegmentgivenbytheequation x = ŠRy h+ R = g ( y ), where y [0 ,h ],aroundtheverticalaxis.Therefore,Theorem9.4 applies,andforthelateralsurfacearea,ityields A ( S )= h 02 Š Ry h + R 1+ R2 h2dy =2 R R2+ h2 h2 Š y2 2 h + y h 0= R R2+ h2= Rs, where s = R2+ h2isthe slantheight ofthecone. Sothetotalsurfaceareaoftheconeisthesumoftheareaofits baseplustheareaofitslateralsurface,thatis, R2 + Rs = R ( R + s ). Theorem9.4alsohasaversionthatappliestocurvesgivenasfunctionsof y thatarerotatedaroundthehorizontalaxis. Theorem 9.5 Let g beafunctionsuchthat giscontinuouson theinterval [ a,b ] .Let S bethesurfaceobtainedbyrotatingthecurve x = g ( y ) ,where y [ a,b ] ,aroundthehorizontalaxis. Thentheareaof S isgivenby A ( S )= b a2 y 1+ dx dy 2dy. Notethatforthecomputationofsomesurfaceareas,wewillhave achoiceoftwotheoremsdiscussedinthissection.Thereaderisencouragedtodescribethecurveswhoserotationsleadtosuchsurfaces.

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62.APPLICATIONSTOPHYSICSANDENGINEERING145 61.3.Exercises.(1)Computethesurfaceareaobtainedbyrotatingthecurve y = ex,for x [0 1],aroundthehorizontalaxis. (2)Rotatethecurveof y = f ( x )= x ,where x [0 1],around theverticalaxis.Findthesurfacearea. (3)Rotatethecurveof y = f ( x )=tan x ,where x [0 ,/ 4], aroundtheverticalaxis.Findthesurfacearea. (4)Rotatethecurveof y = f ( x )= x3,where x [0 1],around thehorizontalaxis.Findthesurfacearea. (5)Rotatethecurveof x = g ( y )=2 3( y +1)3 / 2,where y [2 3], aroundthehorizontalaxis.Findthesurfacearea. (6)SolveExample9.5usingTheorem9.3.62.ApplicationstoPhysicsandEngineering62.1.CenterofMass. 62.1.1.One-DimensionalSystems.Letusassumethatwehavetwo objectsofmass m1and m2placedonthelineofrealnumbers,at points x1and x2,respectively.Wewantto“ndthepoint xgsuchthat ifweplaceafulcrumundertheinterval[ x1,x2]at xg,theobjectsatthe endpointsoftheintervalwillbalance.SeeFigure9.8foranillustration. Weassumethattheinterval[ x1,x2],orthestickrepresentingit,has negligiblemass. If m1= m2,thenweclearlyhave xg=( x1+ x2) / 2.Otherwise, wemakeuseofthewell-knownfactofphysicsthattheintervalwill balanceifthe moments onthetwosidesofthefulcrumareequal,that is,when (9.5) m1( xgŠ x1)= m2( x2Š xg) holds.Solving(9.5)for xg,weget (9.6) xg= m1x1+ m2x2 m1+ m2. Figure9.8. Centerofmass.

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1469.FURTHERAPPLICATIONSOFINTEGRATION Thepoint xgofthereallineiscalledthe centerofmass or centerof gravity ofthesystemdescribedabove,thatis,thesystemofanobject ofmass m1at x1andanobjectofmass m2at x2.The moment ofan objectwithrespecttoapoint P isthemassoftheobjecttimesthe distanceoftheobjectfrom P .Inparticular,intheabovesystem,the twoobjectshadmoments m1x1and m2x2withrespecttotheorigin. Sothetotalsystemhadmoment m1x1+ m2x2.Notethatifwereplace thetwoobjectsbyasimpleobjectofmass m1+ m2placedat xg,then themomentofthesystemabouttheorigindoesnotchange.Thisisan importantpropertythatonlythecenterofmasshas,andthereforewe repeatit.Ifweconcentratethetotalmassofthesystematthecenter ofmass,themomentofthesystemwithrespecttotheoriginwillnot change. Ifweconsiderasystemof k distinctobjectsofmass m1,m2,...,mkplacedatpoints x1,x2,...,xkalongthehorizontalaxis,thenwecanuse ananalogousargumenttoshowthatthecenterofmassofthesystem isat (9.7) xg= k i =1mixi k i =1mi.62.2.Two-DimensionalSystems. 62.2.1.DiscreteTwo-DimensionalSystems.Letusnowconsiderthe moregeneralcasewhenthe k objectsofmass m1,m2,...,mkareplaced inpoints( x1,y1) ( x2,y2) ,..., ( xk,yk)ofthe plane .Wewouldliketo “ndthecenterofmass( xg,yg)ofthissystem.Inotherwords,we assumethataplateofnegligiblemassisplacedunderoursystem,and wewantto“ndthepoint( xg,yg)withthepropertythatifweplacea fulcrumundertheplateatthatpoint,theplatewillbalance. Usingmethodssimilartotheone-dimensionalcase,itcanbeproved thattheplatewillbalanceifthefulcrumisplacedat( xg,yg)with (9.8) xg= k i =1mixi k i =1miand yg= k i =1miyi k i =1mi. Thiscorrespondstotheintuitivelyappropriateconceptthatthe platewillbalanceifitbalancesbothhorizontallyŽandvertically.Ž Thesum Mx= k i =1miyiiscalledthe momentofthesystemwith respecttothe x axis .Thisnameisduetothefactifwetriedto balancethesystemonthe x axis,thelargerthenumber Mx,themore wouldtheweightsofthesystemrotatetheplate.Similarly,thesum

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62.APPLICATIONSTOPHYSICSANDENGINEERING147 My= k i =1mixiiscalledthe momentofthesystemwithrespecttothe y axis .62.2.2.SymmetryLines.Nowletusconsiderthe continuous versionof theproblem.Let P beaplateandletustryto“ndthecenterofmass of P .(Wenolongerassumethatthemassoftheplateisnegligible;in fact,thatmassistheobjectofourstudynow.)Letusassume,forthe restofthischapter,thatthemassof P isuniformlydistributedover P .Letusalsoassumethatthe density ofthematerialofwhich P is madeis1.Thatis,themassofaunitsquarewithin P is1. Sometimeswecan“ndthecenterofmassof P withoutcomputation.A symmetryline of P isastraightline t suchthattheimageof P whenre”ectedthrough t is P itself.Thatimpliesthatthetwoparts intowhich t cuts P arecongruent,andtheplatebalancesontheline t .Consequently,thecenterofmass C of P mustbeon t ,sinceifwe concentratetheentiremassof P in C ,itstillhastobalanceontheline P Theargumentofthepreviousparagraphshowsthatthecenterof massof P mustbeon every symmetrylineof P .Soif P hasmore thanonesymmetryline,thenthesesymmetrylinesmustallintersect inonepoint,namely,inthecenterofmassof P .Inthiscase,weobtain thecenterofmassof P astheintersectionofanytwosymmetrylines of P .Forexample,wecan“ndthecenterofmassofacircle,ellipse, rectangle,orrhombusinthisway.62.2.3.AFormulaforContinuousTwo-DimensionalSystems.Letuskeep theconditionsfromtheprevioussectionandletusimposethenewconditionthat P isadomainunderacurveŽ;thatis,thebordersof P are theverticallines x = a and x = b ,thehorizontalaxis,andthegraph ofthecontinuousfunction f ( x )= y Wewouldliketouseformula(9.8)to“ndtheapproximatelocation ofthecenterofmassof P .Letuscuttheinterval[ a,b ]into n equal parts,usingtheintermediatepoints a = x0
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1489.FURTHERAPPLICATIONSOFINTEGRATION theareaof Ri,thatis, xf ( x i).Sowehave Mx( n )=ni =1miyi=ni =1 x f ( x i) f ( x i) / 2 Theright-handsideisaRiemannsum,so,as n goestoin“nity, itwillconvergetothecorrespondingintegral,whiletheleft-handside willconvergetothemoment Mxoftheoriginalplatewithrespectto the x axis. Thisyields (9.9) Mx= b a1 2 f ( x )2dx. Asimilarargumentusinghorizontalstripesinsteadofverticalones showsthat (9.10) My= b axf ( x ) dx. Finally,nowthatthemomentsof P areknown,itisstraightforward tocomputethecoordinatesofthecenterofmassof P .Indeed,the centerofmassistheuniquepoint( xg,yg)withthepropertythatif theentiremass A = b af ( x ) dx of P isplacedinthatpoint,thenthe momentsofthisone-objectsystemareidenticaltothemomentsof P Inotherwords, Mx= ygA and My= xgA .Therefore, xg= My/A and yg= Mx/A ,whichmeansthatformulas(9.9)and(9.10)implythe followingtheorem. Theorem 9.6 Let f beafunctionsuchthat f ( x ) 0 if x [ a,b ] Let D beadomainwhosebordersaretheverticallines x = a and x = b thehorizontalaxis,andthecurveofthefunction f ( x )= y .Let A ( D ) denotetheareaof D Let xgand ygbethecoordinatesofthecenterofmassof D .Then wehave xg= b axf ( x ) dx A ( D ) and yg= b a 1 2[ f ( x )]2dx A ( D ) Example 9.6 Findthecoordinatesofthecenterofmassofthe quarteroftheunitcirclethatisinthenortheasternquadrant. Notethatifweaskedthesamequestionforthe entire unitcircle, theanswerwouldobviouslybethat xg= yg=0,sincethecenterof gravityofanydomainmustbeonallsymmetrylines.Ifweaskedthe samequestionforthehalfoftheunitcirclethatisinthenorthern

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62.APPLICATIONSTOPHYSICSANDENGINEERING149 half-plane,then xg=0wouldclearlyhold,sincetheverticalaxisisa symmetrylineofthatsemicircle. Solution: (ofExample9.6):Notethatthedomain D inquestionhas asymmetryline,namely,thelinedeterminedbytheequation x = y Sothecenterofgravityof D isonthatline,thatis, xg= yg.Therefore, itsucestocomputeoneof xgand yg.Wehave f ( x )= 1 Š x2= y so ygissomewhateasiertocompute.Theorem9.6yields yg= 1 0 1 2 (1 Š x2) dx / 4 = 2 x Š x3 3 1 0= 2 2 3 = 4 3 Sothecenterofgravityofthequarteroftheunitcircleinthenortheasternquadrantisat(4 3 ,4 3 ).SeeFigure9.9foranillustration. Notethat4 / (3 ) 0 424.Thismakesperfectsensesincethis showsthatthecenterofgravityof D isclosertothehorizontalaxis (thebottomofthequartercircle)thantothe y =1line(thetopof Figure9.9. Thecenterofmassofaquarteroftheunitcircle.

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1509.FURTHERAPPLICATIONSOFINTEGRATION thequartercircle).Thatisreasonable,sincethebottomof D iswider thanthetopof D ,soitconstitutesalargerportionofthetotalweight of D thanthetopof D Example 9.7 Findthecenterofgravityofthedomain D whose bordersaretheverticallines x =0 and x =1 ,thehorizontalaxis,and thegraphofthefunction f ( x )= x2= y Solution: Thedomain D inquestiondoesnothaveasymmetryline, sowemustuseTheorem9.6tocomputebothof xgand yg.Wehave xg= b axf ( x ) dx A ( D ) = 1 0x x2dx 1 0x2dx = [ x4/ 4]1 0 [ x3/ 3]1 0= 3 4 and yg= b a 1 2[ f ( x )]2dx A ( D ) = 1 0( x4/ 2) dx 1 / 3 = [ x5/ 10]1 0 1 / 3 = 3 10 Sothecenterofgravityof D isat(0 75 0 3).Thisagreeswithour intuition,sincethebottomof D islargerthanitstop,andtheleft-hand sideof D issmallerthantheright-handsideof D .SeeFigure9.10for anillustration. 62.3.Exercises.(1)Findthecenterofmassoftheunitsemicirclethatliesinthe northernhalf-plane. (2)Findthecenterofmassoftheplatewhosebordersarethelines x =0and x = / 2,thegraphofthefunction f ( x )=sin y andthehorizontalaxis.

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63.APPLICATIONSTOECONOMICSANDTHELIFESCIENCES151 (3)Findthecenterofmassoftheplatewhosebordersarethe verticallines x =1and x =2,thehorizontalaxis,andthe graphofthefunction f ( x )=ln x (4)Findthecenterofmassoftheplatewhosebordersarethe verticallines x =1and x =2,thehorizontalaxis,andthe graphofthefunction f ( x )= ex. (5)Findthecenterofmassofthetrapezoidwhoseverticesareat (0 0),(15 0),(1 1),and(8 1). (6)Anobjectconsistsoftwosquares.The“rstisthesquare withvertices(0 0),(0 2),(2 0),and(2 2),andtheotheris thesquarewithvertices(0 2),(1 2),(0 3),and(1 3).The densityofthematerialofthesmallsquareistwicethedensity ofthematerialofthelargesquare.Whereisthecenterofmass ofthisobject?63.ApplicationstoEconomicsandtheLifeSciences63.1.ConsumerSurplus.Letusconsidertheproblemofpricingsome merchandisewhosevalueishighlysubjective;thatis,itisworthmore tosomecustomersthantoothers.Examplesofthiscouldbetickets forvarioussportingevents,airlineticketstovacationdestinations,or popularbooks. Let p ( x )bethe demand functionofthiscommodity.Thatis, p ( x ) isthepricethatwillresultinselling x unitsofthecommodity.Lower pricesusuallyleadtohighersalestherefore, p ( x )isusuallya decreasing function asillustratedinFigure9.11. Theareaunderthegraphof p representsthetotalrevenuethe companycouldpossiblyhave,ifitmanagedtochargeeachcustomer Figure9.10. Centerofmassoftheareade“nedinExample9.7.

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1529.FURTHERAPPLICATIONSOFINTEGRATION Figure9.11. Thedemandfunction p ( x )ofacommodity. themaximumpricethatthatcustomeriswillingtopay.Indeed,ifthe highestamountanyoneiswillingtopayforoneunitis p ( x1),and x1customersarewillingtopaythatprice,thentherevenuecomingfrom thesemostenthusedcustomersis x1p ( x1),whichistheareaofthe domainunderthegraphof p thatisbetweentheverticallines x =0 and y = x1.Wecouldcontinueinthisway,notingthatifthesecond highestpricethatsomecustomersarewillingtopayis p ( x2),andthere are x2Š x1peoplewhoarewillingtopaythisprice(notincludingthose whoarewillingtopayeven p ( x1)),thentherevenuefromthemwill be( x2Š x1) p ( x2).Thisistheareaofthedomainunderthegraphof p thatisbetweenthelines x = x1and x = x2,andsoon. Ifthesellerdecidestosetone“xedprice p ( z ),thenthesellerwill sell z items,foratotalrevenueof zp ( z )(theareaoftherectangle R borderedbythetwocoordinateaxesandthelines x = z and y = p ( z )). Thismeansthatthecustomerswhowouldhavepaidanevenhigher priceforthesegoodshavesavedmoney.Besideslosingthatpotential revenue,theselleralsolosesrevenuebynotgettinganypurchasesfrom customerswhowerewillingtopaysomeamount,butnot z ,forone unit. Let xnbethenumberofitemsthatthesellercansellatthelowest priceatwhichthesellerisstillwillingtoselltheseitems.Itisadirect consequenceoftheabovediscussionthatthetotalamountsavedbyall customerswhoboughttheitemat z dollarsisthearea underthecurve of p butabovetherectangle R ,thatis, (9.11)ni =1( p ( xi) Š z )( xiŠ xi Š 1) .

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63.APPLICATIONSTOECONOMICSANDTHELIFESCIENCES153 Ifthenumber n ofpricesatwhichvariouscustomersarewillingto buygoestoin“nity,thentheRiemannsumin(9.11)approachesthe de“niteintegral (9.12) CS = z 0( p ( x ) Š z ) dx. Ineconomics, CS iscalledthe customersurplus forthegivencommodity. Similarly,theintegral zp ( x ) dx istheamountofmissedrevenue,thatis,themoneythecompanycould havereceivedfrombuyerswhofoundtheproducttooexpensive.Note thatthisistheareaofthedomainunderthegraphof p ,butonthe rightof R Example 9.8 Ticketsforacertain”ightarenormallypricedat $ 300,andinanaveragemonth,500ticketsaresold.Researchshows that,forevery $ 10thatthepriceisreduced,thenumberofticketssold goesupby20.Findthedemandcurveandcomputetheconsumer surplusfortheseticketsifthepriceissetat $ 240. Solution: Iftheairlinewantstosell x tickets,thenthepricethatthe airlineneedstochargeis p ( x )=300 Š 10 x Š 500 20 =300 Š x Š 500 2 Indeed,inordertosell x Š 500extratickets,theairlineneedstodecrease itspriceby$10foreach20packofextratickets. Ifthepriceissetat z =$240,thenformula(9.12)showsthatthe customersurplusis CS = 240 0( p ( x ) Š 240) dx = 240 0 60 Š x Š 500 2 dx = 240 0 310 Š x 2 dx =60 000 Socustomerswouldsaveatotalof$60,000inanaveragemonthifthe priceoftheticketsweresetat$240.

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1549.FURTHERAPPLICATIONSOFINTEGRATION 63.2.SurvivalandRenewal.Example 9.9 Letusassumethattherearecurrently30,000people intheUnitedStateswhohaveacertainillness.Letusalsoassume thatweknowthatthefractionofthatpopulationwhowillstillhavethe illness t monthsfromnowisgivenbythefunction f ( t )= eŠ 0 05 t.We alsoknowthateverymonth1000newpatientswillgettheillness.How manypeopleintheUnitedStateswillhavetheillnessin20months? Solution: Clearly, f (20)= eŠ 1=0 368ofthepeoplewhocurrently havetheillnesswillstillhaveit20monthsfromnow.Nowwehave tocomputethenumberofpeoplewhowillgettheillness between now ( t =0)and20monthsfromnow( t =20)andwillstillbeill20months fromnow. Subdividetheinterval[0 20]into n equalsubintervalsusingthe points 0= t0
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63.APPLICATIONSTOECONOMICSANDTHELIFESCIENCES155 thenext20months.Tryto“ndanintuitiveexplanationforthatfact thatdoesnotinvolveintegration.63.3.Exercises.(1)Acountrycurrentlyhasapopulationof80millionpeopleand anaturalgrowthrateof1.5percent.Thenaturalgrowth g of thepopulationinagivenyeariscomputedasthedierence betweenthenumberofbirthsandthenumberofdeathsinthat year,whilethenaturalgrowthrateforthatyearis g divided bythesizeofthepopulationatthebeginningofthatyear. Letusassumethateachyear1.1millionpeopleemigrate fromthiscountry.Ifthecurrenttrendscontinue,howlarge willthepopulationofthiscountrybein20years? (2)Acountrycurrentlyhasapopulationof80millionpeopleand anaturalgrowthrateof Š 0 5percent.Letusassumethat eachyear0.35millionpeopleimmigratetothiscountry.Ifthe currenttrendscontinue,howlargewillthepopulationofthis countrybein20years? (3)Wedeposit$100,000intoabankaccount,whereitwillearn anannualinterestof5percent.Theinterestiscompounded continuously,soin t years,theoriginaldepositwillbeworth f ( t )=$100 000 1 05t.Eachyear,wedeposit$2000tothis sameaccountinacontinuousmanner.Whatwillouraccount balancebein15years? (4)Wedeposit$100,000intoabankaccount,whereitwillearn anannualinterestof5percent.Theinterestiscompounded continuously.Eachyear,wewithdrawatotalof$4000ina continuousmanner.Whatwillouraccountbalancebein20 years? (5)Ticketstoacertainsectionofthearenaforabasketballgame usuallycost$50.Thisresultsinthesalesof1000tickets.For everydollarthatthepriceisdropped,thenumberoftickets soldgoesup1percent.Findthedemandfunctionforthese ticketsandcomputetheconsumersurplusiftheticketsare soldat$40. (6)Let S ( x )bethesupplyfunctionforacertaincommodity.That is, S ( x )isthepricethatoneunitofthecommodityhastocost inordertoattractenoughsellerstoprovide x unitsforsale. Notethat S ( x )isanincreasingfunction,sinceahigher priceisneededtoattractmoresellers.

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1569.FURTHERAPPLICATIONSOFINTEGRATION Letusassumethattheunitsaresoldata“xedprice T = S ( t ).Thatmeansthatthesellerswhowouldbewillingto sellatalowerpricearemakingapro“t.Thetotalamountof thepro“tmadebyallsellersissometimescalledthe producer surplus .Provethattheproducersurplusforthiscommodity canbecomputedbytheformula t 0( T Š S ( x )) dx.64.Probability ThewordprobabilityŽisoftenusedininformalconversations,even ifitissometimesnotclearwhatthespeakermeansbythatword.It turnsoutthattherearetwodistinctconceptsofprobability.Thesetwo conceptscomplementeachotherinthattheyareapplicableindierent circumstances,anduseverydierentmethods.64.1.DiscreteProbability.Letussaythatwearetossingafaircoin fourtimes.Whatistheprobabilitythatwewillgetatleastthree heads? Thisisasituationinwhichthe event thatwestudy,thatis,the sequenceoffourcointosses,hasonlya “nitenumber ofoutcomes. Indeed,thereare24possibleoutcomes,sinceeachcointosshastwo outcomes(headsortails),andthecoinistossedfourtimes. Amongthese16possibleoutcomes,“veare favorable outcomes, namely, HHHH HHHT HHTH HTHH ,and THHH .Furthermore,eachsingleoutcome(favorableornot)isequallylikelytooccur, sincethecoinisfair,andtheresultofeachcointossisequallylikely tobeheadsortails. Inthissituation,thatis,whenthenumberofallpossibleoutcomes is“nite,andeachoutcomeisequallylikelytooccur,de“ne (9.13)Probabilityofevent= Numberoffavorableoutcomes Numberofalloutcomes which,inourexample,showsthattheprobabilityofgettingatleast threeheadsis5 / 16. Probabilitiesde“nedbyformula(9.13)arecalled discreteprobabilities .Theformulaisapplicableonlywhenthenumberofpossible outcomesis“nite.Ifwewanttoapplythisformulaincomplicated situations,weneedadvancedtechniquestocountthenumberofall

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64.PROBABILITY157 outcomesandthenumberoffavorableoutcomes.Thefascinatingdisciplinestudyingthosetechniquesiscalled enumerativecombinatorics andwillnotbediscussedinthisbook.64.2.ContinuousProbability.LetussaywewanttoknowtheprobabilitythatduringthenextcalendaryearthecityofGainesville,Florida, willhavemorethan40inchesofprecipitation,orwewanttoknowthe probabilitythatGainesvillewillhavelessthan50inchesofprecipitation,orthatGainesvillewillhaveatleast42butatmost48inchesof precipitation. Inthiscase,formula(9.13)isnotapplicable,sinceboththenumber offavorableoutcomesandalloutcomesisin“nite.Indeed,theamount ofprecipitationinGainesvillenextyearcanbeanynonnegativereal number.Furthermore,notalloutcomesareequallylikely.Receiving verylittleorverymuchprecipitationisfarlesslikelythanreceiving closetotheusualamount.Weneedatotallydierentapproach.Our approach,whiledierentfromtheoneintheprevioussection,shares someofthemostimportantfeaturesofthatapproach.Forinstance, theprobabilityofaneventthatiscertaintohappenwillbe1,while theprobabilityofaneventthatneverhappenswillbe0. Similarsituationsoccurwhenwewanttoknowtheprobabilitythat acertaindevicewillworkformorethan t years,orthatarandomly selectedpersonweighsmorethan p poundsbutlessthan q pounds,or thatthebloodpressureofarandomlyselectedpersonisbelowagiven value.Thequantitiesmentionedherearecalled randomvariables Wewouldliketode“netheprobability F ( a )thattheamount X ofprecipitationinGainesvillenextyearwillbeatmost a inches.This probabilitywillsometimesbedenotedby P ( X a ). Whilewedonotyetknowhowtocompute F ( a ),weknowthatthe function F hastosatisfythefollowingrequirements: (1)Wewillhave F ( a1) F ( a2)if a1
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1589.FURTHERAPPLICATIONSOFINTEGRATION Figure9.12. Theprobability P ( a X b )asanarea. Itcanbeprovedthatif F hasalltheseproperties,thenthereexists auniquefunction f : R R thathasthefollowingproperties: (a)Forall a R ,wehave a Šf ( x ) dx = F ( a )= P ( X a ). (b)Theequality Šf ( x ) dx =1holds. (c)Forallrealnumbers x ,theinequality f ( x ) 0holds. If f istheuniquefunctiondescribedbythethreepropertiesabove, then f iscalledthe probabilitydensityfunction ,orsimply densityfunction ,ofthecontinuousrandomvariable X Notethatproperty(a)aboveimpliesthat,forallrealnumbers a
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64.PROBABILITY159 Figure9.13. Thegraphofthefunction f inExample9.10. Example 9.10 Letthecontinuousrandomvariable X havedensity function f ( x )= 0 if x, 6 x (1 Š x ) if 0 x 1 0 if x> 1 Verifythat f isindeedadensityfunctionandcomputetheprobability P (0 3 X 0 6) Solution: Inordertoseethat f isindeedadensityfunction,we mustverifythatitsde“niteintegral,takenovertheentirelineofreal numbers,isequalto1.Thisisnotdicult,since f ( x )=0outsidethe interval[0 1].Thisleadsto Šf ( x ) dx =6 1 0 x Š x2 dx =6 x2 2 Š x3 3 1 0=6 1 6 =1 So f isindeedavaliddensityfunction. Wecanuseformula(9.14)tocomputetherequestedprobability. Weget P (0 3 X 0 6)= 0 6 0 3f ( x ) dx =6 0 6 0 3x Š x2dx =6 x2 2 Š x3 3 0 6 0 3=6(0 108 Š 0 036) =0 432 .

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1609.FURTHERAPPLICATIONSOFINTEGRATION Figure9.14. =1(red), =1 5(blue), =2(orange). 64.2.1.ExponentialDistribution.Considerthefollowingdensityfunction.Let beapositiverealnumberandlet f ( x )= 0if x< 0 eŠ xif0 x. (9.15) Weseethat f isadecreasingfunctionontheinterval[0 ).Figure 9.14showshowthespeedatwhich f decreasesdependsontheparameter Itturnsoutthatthisdensityfunctionisaveryfrequentlyoccurring one.Therefore,ithasaname.Itiscalledthe exponentialdensityfunction withparameter .Usingtherightconstant andundertheright circumstances,itcanbeusedinmanyscenarios,typicallyconnected to waitingtimes .Forinstance,itcouldbeusedtomeasuretheprobabilitythat,givenastartingmoment,agivencellphonewillringinless than t minutes,orthat,atagivenlocation,itwillstartraininginless than h hours,orthat,givenarandomstore,acustomerwillenterin s seconds.Theexponentialdensityfunctionwillgiveagoodapproximationtocomputetheseprobabilitiesifthementionedprocessestake placeataroughlyconstantrate.Thatis,weshouldchooseapartof thedaywhenthatgivencellphonereceivescallsatroughlyconstant frequency,aseasonwhenitrainsatthatlocationatroughlyconstant timeperiods,oratimeofdaywhencustomersenterthatstoreata roughlyconstantrate. Example 9.11 Theprobabilitythatacertainkindofnewrefrigeratorwillneedamajorrepairin x yearsisgivenbytheexponential

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64.PROBABILITY161 densityfunctionwithparameter =1 / 9 .Whatistheprobabilitythat anewrefrigeratorwill not needamajorrepairfor10years? Solution: First,wecomputetheprobabilitythattherefrigerator will needamajorrepairin10years.Let X denotethenumberofyears passingbeforethe“rstmajorrepairisneeded.Thenthatprobabilityis P ( X< 10)= 10 Šf ( x ) dx = 10 01 9 eŠ x/ 9dx =[ Š eŠ x/ 9]10 0=1 Š eŠ 10 / 9=0 671 Therefore,theprobabilitythattherefrigeratorwill not needamajor repairin10yearsis P ( X 10)=1 Š 0 671=0 329. 64.2.2.Mean.Ifwewanttocomputetheaverageweightofapersonselectedfromagivenpopulationof n people,wecansimplytake theweights a1,a2,...,anofthosepeopleandcomputetheirarithmetic mean,oraverage,thatis,therealnumber A = a1+ a2+ + an n Thiscouldtakeaverylongtimeif n isaverylargenumber.Ifthedata aregiveninamoreorganizedform,wemaybeabletosavesometime. Inparticular,ifweknowthatthereare b1peopleinthepopulation whoseweightis x1,thereare b2peoplewhoseweightis x2,andsoon, thenwecancomputetheaverageweightofthepopulationas (9.16) A = b1x1+ b2x2+ + bkxk b1+ b2+ + bk, sincethisfractionisthetotalweightofthepopulationdividedbythe numberofpeopleinthepopulation. Nownotethat pi= bi/ ( b1+ b2+ + bk)isjusttheprobabilitythat arandomlyselectedpersonofthispopulationhasweight xi.Therefore, (9.16)isequivalentto (9.17) p1x1+ p2x2+ + pkxk. Theoretically,theweightofapersoncantakein“nitelymanyvaluessincethemeasuringscalecanbealwaysbemoreprecise.Itisnot diculttoprovethat,as k goestoin“nity,thesumin(9.17)willturn

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1629.FURTHERAPPLICATIONSOFINTEGRATION intoaRiemannsum,andtheweights xiwillbemeasuredbyacontinuousrandomvariable X ,andtheprobabilities piwillbeexpressibleby thede“niteintegralsofadensityfunction.Thisleadstothefollowing de“nition. Definition 9.1 Let f bethedensityfunctionofthecontinuous randomvariable X .Thenthevalueof ( X )= Štf ( t ) dt iscalledthe averagevalue or mean or expectedvalue of X Example 9.12 Let X bethecontinuousrandomvariablewhose densityfunctionistheexponentialdensityfunctionwithparameter thatwede“nedin(9.15).Then ( X )=1 / Solution: UsingDe“nition9.1,wehave ( X )= Štf ( t ) dt = 0teŠ tdt = Š teŠ tŠ 1 eŠ t 0= 1 Inotherwords,theparameterandthemeanofanexponentialdistributionarereciprocalsofeachother.Inviewofthis,wecanreformulatetheresultofExample9.11asfollows.Iftheaveragetimebeforea newrefrigeratorneedsamajorrepairis9years,thentheprobability thatarefrigeratorwillnotneedamajorrepairfor10yearsis0.329.64.2.3.NormalDistribution.Let bearealnumberandlet bepositiverealnumber.Considerthedensityfunction f ( x )= 1 2 exp ( x Š )2 2 2 Thedistributionde“nedbythisdensityfunctioniscalledthe normaldistribution withparameters and .Thisdistributionisdenoted by N ( ).Inparticular,if =0and =1,thentheobtained distribution N (0 1)iscalledthe standardnormaldistribution Plottingthegraphof f forvariousvaluesof and ,weseethat thegraphhasabellcurve;itshighestpointisreachedwhen x = ,

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64.PROBABILITY163 Figure9.15. =1(red), =1 5(orange), =2 (green),and =2 5(blue). anditincreasesontheleftofthatanddecreasesontherightofthat. Thesmallerthevalueof ,thesteeperistheriseandfallofthegraph of f .SeeFigure9.15foranillustration. Itcanbeprovedthat ispreciselythemeanof N ( ).The constant iscalledthe standarddeviation of N ( ).Itmeasureshow spreadoutthevaluesofourvariable X are.(Theprecisede“nitionis that isthesquarerootofthemeanof( X Š )2. ) Manyscenariosaremodeledbyanormaldistribution,suchastest scores,athleticresults,orannualsnowfallatagivenlocation. Example 9.13 Inanaverageyear,Northtowngets10feetofsnow, withastandarddeviationof2feet.Whatistheprobabilitythat,ina randomyear,Northtowngetsbetween9and12feetofsnowifsnowfall ismodeledbyanormaldistribution? Solution: Let X denotethesnowfallinarandomyearinNorthtown. Weneedto“ndtheprobability P (9 X 12).Assnowfallismodeled byanormaldistribution,thegivenparametersimplythatthatmust bethedistribution N (10 2).Therefore,byformula(9.14),wehave P (9 X 12)= 12 91 2 exp ( x Š 10)2 8 =0 5326 wherethede“niteintegralhastobecomputedbysomeapproximation method(orasoftwarepackage)since eŠ x2hasnoantiderivativeamong elementaryfunctions.

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1649.FURTHERAPPLICATIONSOFINTEGRATION 64.3.Exercises.(1)Forwhichvalueof c will f ( x )= cx4beadensityfunctionon [0 1]? (2)Let X bearandomvariablewhosedensityfunctionis0outside theinterval[0 1]andsatis“es f ( x )=2 x for x [0 1].Prove that f isindeedadensityfunction,andcomputethemeanof X (3)Letussaythatthelifetimeofabicycletire(measuredin months)hasanexponentialdistributionwith = Š 7.What istheprobabilitythatatirewilllastbetween“veandeight months? (4)Theaveragescoreonanexamis100points.Inordertopass, astudentcannotbemorethan2standarddeviationsbelow theaverage.Ifthescoreshaveanormaldistributionwitha standarddeviationof6,howlargeafractionofthestudents willpasstheexam? (5)UsingtheconditionsofExample9.13,whatistheprobability thatNorthtownwillgetlessthan5feetofsnowinagiven year? (6)Let X betherandomvariablethatcountsthegoalsscored byanoensivesoccerplayerofacertaineliteleagueduring anentireseason.Anoensiveplayerisconsideredexceptional ifhethenumberofgoalshescoresexceedstheaverageofall oensiveplayersbyatleast3standarddeviations.Letussay that X hasdistribution N (33 3).Whatpercentageofoensive playersisconsideredexceptional?

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CHAPTER10 PlanarCurves 65.ParametricCurves Everypointinaplanecanbede“nedasanorderedpairofrealnumbers( x,y )calledthe rectangularorCartesiancoordinates .Agraphof afunction f isthesetpointsinaplanewhosecoordinatessatisfy thecondition y = f ( x ).Thegraphgivesasimpleexampleofaplanar curve.Moregenerally,aplanarcurvecanbede“nedasthesetofpoints whosecoordinatessatisfythecondition F ( x,y )=0calledthe Cartesianequation ofacurve.Inmanyinstances,anequation F ( x,y )=0 hasmultiplesolutionsforeverygiven x .Forexample,considerthe circleofunitradius: (10.1) x2+ y2=1= y = 1 Š x2,x [ Š 1 1] Thetwosolutionsrepresenttwosemicircles.Thegraph y = 1 Š x2isthesemicircleabovethe x axis,whilethegraph y = Š 1 Š x2is thesemicirclebelowthe x axis.Theunionofthetwographsisthefull circle.Thisexampleshowsade“ciencyindescribingplanarcurvesby thegraphofafunctionbecausethecurvescannotalwaysberepresented asthegraphofasinglefunction. Ontheotherhand,(10.1)admitsadierentsolution: (10.2) x2+ y2=1= x =cos t,y =sin t,t [0 2 ] whichimmediatelyfollowsfromthetrigonometricidentitycos2t + sin2t =1forallvaluesof t Thisrepresentationmeansthatapointofthecoordinateplaneis assignedtoeveryvalueof t [0 2 ]bytherule( x,y )=(cos t, sin t ). Thecoordinatesofpointsofthecirclearefunctionsofathirdvariable calleda parameter .As t changes,thepoint(cos t, sin t )tracesoutthe circleofunitradiuscenteredattheoriginintheplane.Theparameter t hasasimplegeometricalinterpretation.Itistheanglecounted counterclockwisefromthepositive x axistoarayfromtheoriginon whichthepoint(cos t, sin t )lies.Thisobservationadmitsanatural generalization. Definition 10.1(Parametriccurves) Let x ( t ) and y ( t ) becontinuousfunctionson [ a,b ] .A parametriccurve inthecoordinateplaneis165

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16610.PLANARCURVES Figure10.1. Circle: x ( t )=cos t y ( t )=sin t Figure10.2. Parametriccurve.As t increasesfrom a to b ,thepoint( x ( t ) ,y ( t ))tracesoutacurveinthe xy plane. thesetofpointssatisfyingtheconditions,calledthe parametricequations x = x ( t ) ,y = y ( t ) ,t [ a,b ] Thepoints ( x ( a ) ,y ( a )) and ( x ( b ) ,y ( b )) arecalledthe initial and terminal pointsofthecurve,respectively. Thegraphofafunction f isaparticularexampleofaparametric curve: x = t y = f ( t ).

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65.PARAMETRICCURVES167 Figure10.3. Thespiral x = t cos t y = t sin t t [0 2 ].Thedistancefromtheorigin R = x2+ y2= t increaseslinearlyastheangle t ,countedcounterclockwisefromthepositive x axis,increasesfrom0to2 Parametriccurvesarecommonineverydaylife.Thepositionofa particleinaplaneisde“nedbyitsrectangularcoordinates( x,y )inthe plane.Whentheparticlemoves,itscoordinatesbecomefunctionsof time t sothattheparametriccurve x = x ( t ), y = y ( t )isthetrajectory oftheparticle.Theparticlemovesinaspeci“cdirectionalongits trajectory.Theparticlemayrepeatitstrajectory(orsomeportionsof it)multipletimes. Example 10.1 Sketchthecurvewiththeparametricequations x = t cos t y = t sin t t [0 2 ] Solution: Abasicapproachtovisualizetheshapeofaparametric curveistoplotpoints( x ( tk) ,y ( tk)), k =1 2 ,...,n ,correspondingto successivevaluesof t : t1
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16810.PLANARCURVES theoriginandthepositive x axiscountedcounterclockwise.Thus, thecurvehasthefollowinginterpretation.Asthepoint( x ( t ) ,y ( t )) rotatesabouttheorigin,thedistance R betweenitandtheorigin increaseslinearlywiththerotationangle.Suchamotionoccursalong anunwindingspiral.Intheinterval t [0 2 ],thespiralmakesone fullturnfromtheinitialpoint(0 0)totheterminalpoint(2 0). 65.1.ParametricCurvesandCurvesasPointSets.Ifacurveisde“ned asapointsetinthecoordinateplane,forexample,bytheCartesian equation F ( x,y )=0,thentherearemanyparametricequationsthat describeit.Forexample,thecircle(10.1)mayalsobedescribedbythe followingparametricequations: (10.3) x2+ y2=1= x =cos(3 ) ,y = Š sin(3 ) [0 2 ] Whatisthedierencebetween(10.2)and(10.3)?First,notethat, astheparameter t in(10.2)increases,thepoint(cos t, sin t )tracesout thecircle counterclockwise (theinitialpoint(1 0)movesupwardas y = sin t> 0for0
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65.PARAMETRICCURVES169 Agoodmechanicalanalogyisthemotionofaparticle.Aparametriccurvedescribestheactualmotion,thatis,howfastandinwhich directiontheparticlemovesalongitstrajectoryde“nedasapointset. Example 10.2 Supposeacurve C isdescribedbytheparametric equations x = x ( t ) y = y ( t ) if t [ a,b ] .Findtheparametricequations of C suchthatthecurveistracedoutbackward,thatis,fromthepoint ( x ( b ) ,y ( b )) to ( x ( a ) ,y ( a )) (theinitialandterminalpointsareswapped). Solution: Onehasto“ndanewparameter t = g ( ),suchthat g ( b )= a and g ( a )= b .When increasesfrom a to b ,theparameter t decreasesfrom b to a ,andthesought-afterparametricequationsare obtainedbythecomposition X ( )= x ( g ( ))and Y ( )=y ( g ( )).The simplestpossibilityistolookforalinearrelationbetween t and g ( )= c + d .Thecoecients c and d are“xedbytheconditions g ( a )= b or b = c + da and g ( b )= a or a = c + db .Therefore,by subtractingtheseequations, b Š a =( c + da ) Š ( c + db )= Š ( b Š a ) d or d = Š 1.Byaddingtheseequationswith d = Š 1, b + a =( c Š a )+( c Š b ) or c = a + b .Hence, t =( a +b ) Š ,sothattheparametricequations of C withreversedorientationare x = x ( a + b Š ) ,y = y ( a + b Š ) [ a,b ] Forexample,if C isthecircleorientedcounterclockwiseasin(10.2), thenthesamecircleorientedclockwiseisdescribedby x =cos(2 Š )=cos y =sin(2 Š )= Š sin [0 2 ]. 65.2.TheCycloid.Thecurvetracedbya“xedpointonthecircumferenceofacircleasthecirclerollsalongastraightlineiscalleda cycloid (seeFigure10.4).To“nditsparametricequations,suppose thatthecirclehasaradius R anditrollsalongthe xaxis .Letthe “xedpoint P onthecircumferencebeinitiallyattheoriginsothatthe centerofthecircleispositionedatthepoint(0 ,R )(onthe y axis).Let CP denotethestraightlinesegmentbetweenthecenterofthecircle C and P .Initially, CP isperpendiculartothe x axis.Asthecirclerolls, thesegment CP rotatesaboutthecenterofthecircle.Therefore,itis naturaltochoosetheangleofrotation asaparameter.Thecoordinatesof P arefunctionsof tobefound.Ifthecirclerollsadistance D sothatitscenterisat( D,R ),thenthearclength R ofthepart ofthecirclebetween P andthetouchpoint T hastobeequalto D thatis, D = R .Let Q beapointonthesegment CT suchthat PQ and CT areperpendicular.Considertheright-angledtriangle CPQ Itshypotenuse CP haslength | CP | = R ,andthelengthsofitscatheti are | CQ | = | CP | cos = R cos and | PQ | = | CP | sin = R sin .Let

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17010.PLANARCURVES Figure10.4. De“nitionofacycloid.Adiskofradius R isrollingalongthe x axis.Acurvetracedoutbya “xedpointonitsedgeiscalledacycloid. Figure10.5. Overallshapeofacycloid. ( x,y )becoordinatesofthepoint P .Theparametricequationsofthe cycloidare x = D Š| PQ | = R Š R sin = R ( Š sin ) y = R Š| CQ | = R Š R cos = R (1 Š cos ) Itlookslikeanupwardarcovertheinterval0 x 2 R ,withmaximalheight ymax=2 R ( = / 2),andthearcrepeatsitselfoverthe nextintervalofthelengthofcircumference2 R andsoon. Remark. In1696,theSwissmathematicianJohannBernoulli posedthe brachistochroneproblem :Findthecurvealongwhichaparticlewillslide(withoutfriction)intheshortesttime(underthein”uence ofgravity)fromapoint A toalowerpoint B notdirectlybeneath A Theparticlewilltaketheleasttimeslidingfrom A to B ifthecurveis apartofaninvertedarchofacycloid.

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65.PARAMETRICCURVES171 65.3.FamiliesofCurves.Dierentvaluesof R de“nedierentcycloids. Ingeneral,iftheparametricequationscontainanumericalparameter, thentheparametricequationsde“nea family ofcurves;eachfamily membercorrespondstoaparticularvalueofthenumericalparameter. Example 10.3 Investigatethefamilyofcurveswiththeparametric equations x = a cos t y = b sin(2 t ) t [0 2 ] ,where a and b are positivenumbers. Solution: Consider“rstthesimplestcase a = b =1.Thefunction x ( t )=cos t hasaperiodof2 ,and y ( t )=sin(2 t )hasaperiodof .Theinitialpointis( x (0) ,y (0))=(1 0).As t increases,thepoint movesupwardsothat x ( t )decreases(becomeslessthan1),while y ( t ) increases,reachingitsmaximumvalue1at t = / 4.Afterthat, y ( t ) beginstodecrease,while x ( t )continuestodecrease.At t = / 2,the pointarrivesattheoriginandpassesthroughitintothethirdquadrant sothat x ( t )and y( t )continuetodecrease.When t =3 / 4, y ( t )attains theminimumvalue Š 1andbeginstoincreasefor t> 3 / 4,while x ( t )=cos t isstilldecreasingtowarditsminimalvalue Š 1,whichis reachedat t = ,andthecurvecrossesthe x axismovingintothe secondquadrant.Inthesecondquadrant 1)orcompresses( a< 1)anygeometricalsethorizontallyinthe coordinateplane.Thetransformation y by doesthesamebutinthe verticaldirection.Sothefamilyofcurvesconsistsofcurvesofthe shapestretchedto“tintotherectangleboundedbythelines x = a and y = b 65.4.Exercises.In(1)…(9),sketchthecurvebyplottingitspoints.Includethearrow showingtheorientationofthecurve.Eliminatetheparameterto“nd aCartesianequationofthecurve.

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17210.PLANARCURVES (1) x =1+2 t,y =3 Š t (2) x = t,y =2 Š t (3) x = t2,y = t3(4) x =1+2 et,y =3 Š et(5) x =cosh t,y =sinh t (6) x =2sin t,y =3cos t (7) x =2 Š 3cos t,y = Š 1+sin t (8) x =cos t,y =sin(4 t )(9) x = t2sin t,y = t2cos t (10)Thecurves x = a sin( nt ), y = b cos t ,where n isapositiveinteger, arecalled Lissajous“gures .Investigatehowthesecurvesdependon a b ,and n (11)Consideradiskofradius R .Let P beapointonthediskata distance b fromitscenter.Findtheparametricequationsofthecurve tracedoutbythepoint P asthediskrollsalongastraightline.The curveiscalleda trochoid .Aretheequationswellde“nedif b>R Sketchthecurvefor bR (12)The swallowtailcatastrophecurves arede“nedbytheparametric equations x =2 ct Š 4 t3, y = Š ct2+3 t4.Sketchthesecurvesforafew valuesof c .Whatfeaturesdothecurveshaveincommon?Howdo theychangewhen c increases?66.CalculuswithParametricCurves66.1.TangentLinetoaParametricCurve.Consideraparametriccurve x = x ( t ), y = y ( t ),wherethefunctions x ( t )and y ( t )arecontinuouslydierentiableandthederivatives x( t )and y( t )donotvanish simultaneouslyforany t .Suchparametriccurvesarecalled smooth Theorem 10.1(TangentLinetoaSmoothCurve) Asmoothparametriccurve x = x ( t ) y = y ( t ) hasatangentlineatanypoint ( x0,y0) anditsequationis (10.4) x( t0)( y Š y0) Š y( t0)( x Š x0)=0 where ( x0,y0)=( x ( t0) ,y ( t0)) Proof. Takeapointofthecurve( x0,y0)=( x ( t0) ,y ( t0))correspondingtoaparticularvalue t = t0.Supposethat x( t0) =0.Then, bythecontinuityof x( t ),thereisaneighborhood I=( t0Š ,t0+ ) forsome > 0suchthat x( t ) =0forall t I;thatis,thederivativeiseitherpositiveornegativein I.Bytheinversefunction theorem(studiedinCalculusI),thereisaninversefunction t = f ( x ) thatisdierentiableinsomeopenintervalthatcontains x0.Substituting t = f ( x )intothesecondparametricequation y = y ( t ),one obtainsthatnearthepoint( x0,y0)thecurvecanberepresentedasa partofthegraph y = F ( x )suchthat y0= F ( x0).Thefunction F is

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66.CALCULUSWITHPARAMETRICCURVES173 dierentiableasthecompositionoftwodierentiablefunctions.The derivative F( x0)determinestheslopeofthetangenttothegraph,and theequationofthetangentlinereads (10.5) y = y0+ F( x0)( x Š x0) Byconstruction, y = F ( x )= y ( t )= F ( x ( t ))forall t I. Dierentiationofthisequationwithrespectto t bymeansofthechain ruleyields y( t )= F( x ( t )) x( t )= F( x ( t ))= y( t ) x( t ) = F( x0)= y( t0) x( t0) Substitutingthisequationinto(10.5),thelattercanbewrittenin theform(10.4).If x( t0)=0,then y( t0) =0bythede“nitionof asmoothcurvesothatthereisadierentiableinverse t = g ( y )and hence x = G ( y )= x ( g ( y )).Similarargumentsleadtotheconclusion thatthetangentlinetothegraph x = G ( y )hastheform(10.4).The detailsarelefttothereaderasanexercise. Theruleforcalculatingtheslopeofthetangentlinecanalsobe obtainedbymeansoftheconceptofthedierential.Recallthatthe dierentialsoftworelatedquantities y = F ( x )areproportional: dy = F( x ) dx .Ontheotherhand, x = x ( t ), y = y ( t )andtherefore dx = x( t ) dt and dy = y( t ) dt .Hence, F( x )= dy dx =dy dt dx dt= y( t ) x( t ) Thesemanipulationswithdierentialsarebasedonatacit assumption that,forasmoothcurve x = x ( t ), y = y ( t ),thereexistsa dierentiable function F suchthat y = F ( x ).Intheproofofthetangentlinetheorem,thishasbeenshowntobetrueasaconsequenceoftheinverse functiontheorem.Theuseofthedierentialsestablishesthefollowing helpfulrulestocalculatethederivatives: d dx =d dt dx dt= 1 x( t ) d dt and d dy =d dt dy dt= 1 y( t ) d dt .66.2.ConcavityofaParametricCurve.Theconcavityofagraph y = F ( x )isdeterminedbythesignofthesecondderivative F( x ).If F( x ) > 0,thegraphisconcavedownward,anditisconcaveupwardif

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17410.PLANARCURVES F( x ) < 0.If y ( t )and x ( t )aretwicedierentiable,thentheconcavity ofthecurvecanbedetermined: d2y d2x = d dx dy dx = 1 xd dt dy dx = y x x= yxŠ xy ( x)3. Example 10.4 Acurve C isde“nedbytheparametricequations x = t2, y = t3Š 3 t (i) Showthat C hastwotangentlinesatthepoint (3 0) (ii) Findthepointson C wherethetangentlineishorizontalorvertical. (iii) Determinewherethecurveisconcaveupwardordownward. Solution: (i)Notethat y ( t )= t ( t2Š 3)=0hasthreesolutions t =0 and t = 3.Butthecurvehasonlytwopointsofintersectionwith the x axis,(0 0)and(3 0),because x ( 3)=3;thatis,thecurveis self-intersectingatthepoint(3 0).Thisexplainswhythecurvemay havetwotangentlines.Onehas x( t )=2 t and y( t )=3 t2Š 3sothat x( 3)= 2 3and y( 3)=6.Sotheslopesofthetangentlines are( y/x)( 3)= 3,andtheequationsofthelinesread y = 3( x Š 3)and y = Š 3( x Š 3) (ii)Thetangentlinebecomeshorizontalwhen y( t )=3 t2Š 3=0 (seeEq.(10.4)).Thishappenswhen t = 1.Thus,thetangent lineishorizontalatthepoints(1 2).Thetangentlineisverticalif x( t )=2 t =0or t =0.Sothetangentlineisverticalattheorigin (0 0). (iii)Thesecondderivativeis d2y d2x = 1 xd dt dy dx = 1 2 t d dt 3 t2Š 3 2 t = 3 4 t d dt t Š 1 t = 3 4 t 1+ 1 t2 Thisequationshowsthatthecurveisconcavedownwardif t> 0(the secondderivativeispositive)andthecurveisconcaveupwardif t< 0 (thesecondderivativeisnegative). 66.3.CuspsofPlanarCurves.Consideracurvede“nedbytheCartesianequation x2Š y3=0.Thisequationcanbesolvedfor y y = x2 / 3, suchthat dy/dx =2 3xŠ 1 / 3.For x> 0,theslopeofthetangentline diverges, y( x ) as x 0+(as x approachesto0fromtheright). For x< 0,italsodiverges, y( x ) Š as x 0Š(as x approaches0 fromtheleft).Thetwobranchesofthecurve( x> 0and x< 0)are joinedat x =0andhavea common tangentline,whichisthevertical line x =0(the y axis)inthiscase,buttheslopesuersajumpdiscontinuity(from Š to ).Sothecurveisnotsmoothat x =0and

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66.CALCULUSWITHPARAMETRICCURVES175 Figure10.6. Plotof y = x2 / 3.Thecurvehasacuspat theorigin. exhibitsahornlikeshapenear x =0.Suchapointofaplanarcurve iscalleda cusp Aparametriccurve x = x ( t ), y = y ( t )mayhavecuspseventhough bothderivatives x( t )and y( t )arecontinuousforall t .Forexample, considertheparametriccurve x = t3, y = t2.Forallvaluesof t x2Š y3=0.Sothiscurvecoincideswiththatdiscussedaboveandhas acuspattheorigin( t =0).Thederivatives x( t )=3 t2and y( t )=2 t arecontinuouseverywhere,and,inparticular, x(0)= y(0)=0atthe cusppoint.Despitethecontinuityofthederivatives,theslopeofthe curveisnotde“nedsince dy/dx = y/xisanundeterminedform0 0. Acloserinvestigationshowsthattheslope y( t ) /x( t )=2 3tŠ 1suers ajumpdiscontinuity(from Š to+ as t changesfromnegativeto positive).Thede“nitionofasmoothparametriccurverequiresthat thederivatives x( t )and y( t )arecontinuousand donot vanishsimultaneouslyatany t Arationaleforthelatterconditionistoeliminate possiblecuspsthatmayoccuratpointswherebothderivativesvanish Furthermore,considerthecurve x = t2, y = t3.Theslope dy/dx = y/x=3 2t iscontinuouseverywhereand,inparticular,at t =0,where x(0)= y(0)=0.Nevertheless,thecurvehasacuspattheorigin.To seethis,letusinvestigatetheCartesianequationofthiscurve x3Š y2= 0,whichcanbesolvedfor x x = y2 / 3.Therefore,thederivative dx/dy =2 3yŠ 1 / 3exhibitsajumpdiscontinuityfrom Š to as y changesfromnegativetopositive.Thetwobranchesofthecurve( y> 0 and y< 0)haveacommontangentline(thehorizontalline y =0), butattheirjoiningpointacuspisformed.Notealsothattherate dx/dy = x/y=2 3tŠ 1suersafamiliarin“nitejumpdiscontinuity,thus indicatingacusp.Thisexampleshowsthatbothrates dy/dx = y/x

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17610.PLANARCURVES Figure10.7. Thecurve x = t2, y = t3.Thederivatives x(0)= y(0)=0vanishat t =0.Thecurvehasacusp atthepoint( x (0) ,y (0))=(0 0). and dx/dy = x/ymustbestudiedtodeterminewhetherthereisa cuspatthepointwhere y= x=0. Example 10.5 Findthetangentlinetotheastroidde“nedbythe parametricequation x = a cos3t y = a sin3t t [0 2 ] atthepoints t = / 4 .Determinethepointswherethetangentlineishorizontal andvertical.Isthecurvesmooth?Specifytheregionsofupwardand downwardconcavity.Usetheresultstosketchthecurve. Solution: Theslopeofthetangentlineatagenericpointis dy dx = 1 xd dt y = Š 1 3 a cos2t sin t 3 a sin2t cos t = Š tan t. Thevalue t = / 4correspondstothepoint x = a/ 23 / 2, y = a/ 23 / 2becausesin( / 4)=cos( / 4)=1 / 2,andtheslopeatthispointis Š 1.Sothetangentlineis y = a 2 2 Š x Š a 2 2 or y = a 2 Š x. Theslope dy/dt = Š tan t vanishesat t =0and t = sothetangent lineishorizontal( y =0)atthepoints( a, 0).However,thederivatives x( t )= Š 3 a cos2t sin t and y( t )=3 a sin2t cos t vanishsimultaneously

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66.CALCULUSWITHPARAMETRICCURVES177 at t =0and t = .Theinverseslope dx/dy =1 / ( dy/dx )= Š cot t exhibitsanin“nitejumpdiscontinuityat t =0and t = ,sothe curvehascuspsat( a, 0)andhenceitisnotsmoothatthesepoints. Theslope dy/dx isin“niteat t = / 2and t =3 / 2.Therefore,the curvehasaverticaltangentline( x =0)at(0 a ).However,theslope dy/dx = Š tan t hasanin“nitejumpdiscontinuityat t = / 2and t =3 / 2.Sothecurvehascuspsandisnotsmoothat(0 a ).Note alsothatbothderivatives xand yvanishatthesepoints.Thus,the curveconsistsoffoursmoothpieces,andthecurvehascuspsatthe joiningpointsofitssmoothpieces.Thesecondderivative d2y d2x = 1 xd dt dy dx = 1 3 a cos2t sin t (tan t )= 1 3 a sin t cos4t ispositiveifsin t> 0(or y> 0)andnegativeifsin t< 0(or y< 0).So thetwobranchesofthecurveabovethe x axisareconcavedownward, whilethetwobranchesbelowitareconcaveupward.Thecurveslook likeasquarewithvertices( a, 0),(0 a )whosesidesarebentinward towardtheorigin. 66.4.Exercises.In(1)and(2),“ndanequationofthetangentline(s)tothecurveat thegivenpoint.Sketchthecurveandthetangent(s). (1) x = t2+ t,y =4sin t, (0 0) (2) x =sin t +sin(2 t ) ,y =cos t +cos(2 t ) (1 Š 1) In(3)…(6),investigatetheconcavityofthecurve. (3) x = t3Š 12 t,y = t2Š 1(4) x =sin(2 t ) ,y =cos t (5) x = t2Š ln t,y = t2+ln t (6) x =3sin t3,y =2cos t3(7)Investigatetheslopeofthetrochoid x = R Š b sin y = R Š b cos intermsof .Findtheconditionontheparameters R and b suchthat thetrochoidhasverticaltangentlines. (8)Atwhatpointsonthecurve x =2 t3, y =1+4 t Š t2doesthe tangentlinehaveslope1? (9)Findequationsofthetangentstothecurve x =2 t3+1, y =3 t2+1 thatpassthroughthepoint(3 4). In(10)…(12).Investigatewhetherthecurvehascuspsornot.Ifitdoes, “ndtheirposition.Sketchthecurve. (10) x = t3,y = t3(11) x = t5,y = t2(12) x =( t2Š 1)3,y =( t3Š 1)2

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17810.PLANARCURVES 67.PolarCoordinates Apointonaplaneisdescribedbyanorderedpairofnumbers ( x0,y0)intherectangularcoordinatesystem.Thisdescriptionimplies ageometricalproceduretoobtainthepointastheintersectionoftwo mutuallyperpendicularlines x = x0and y = y0.Thesetofvertical andhorizontallinesformarectangulargridinaplane.Thereareother possibilitiestolabelpointsonaplanebyorderedpairofnumbers.Here thepolarcoordinatesystemisintroduced,whichismoreconvenientfor manypurposes. Fixapoint O onaplane.Let P beapointontheplane.A horizontalrayfrom O iscalledthe polaraxis ,andthepoint O iscalled the origin or pole .Let betheanglebetweenthepolaraxisandthe ray OP from O through P .Theangle iscountedcounterclockwise fromthepolaraxis.Thepositionofthepoint P ontheray OP is uniquelydeterminedbythedistance r = | OP | .Thus,anypoint P on aplaneisuniquelyassociatedwiththeorderedpair( r, ),and r are calledthe polarcoordinates of P .Thecoordinate r iscalledthe radial variable ,and iscalledthe polarangle Allpointsontheplanethathavethesamevalueoftheradialvariableformacircleofradius r centeredattheorigin(allpointsthathave thesamedistancefromtheorigin).Allpointsontheplanethathave thesamevalueofthepolarangleformaray(ahalf-lineboundedbythe Figure10.8. De“nitionofthepolarcoordinatesina plane. r isthedistance | OP | ,and istheanglecounted counterclockwisefromthehorizontalrayoutgoingfrom O totheright.Therectangularcoordinatesofapoint P arerelatedtothepolaronesas x = r cos y = r sin .

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67.POLARCOORDINATES179 origin).Soapoint P withpolarcoordinates( r, )istheintersection ofthecircleofradius r andtheraythatmakestheangle withthe polaraxis.Concentriccirclesandraysoriginatingfromthecenterof thecirclesformapolargridinaplane(seeFigure10.9). Torepresentallpointsofaplane,theradialvariablehastorange overtheinterval r [0 ),whilethepolarangletakesitsvaluesin theinterval[0 2 )becauseanyrayfromtheorigindoesnotchange afterrotationabouttheoriginthroughtheangle2 .Itisconvenient, though,tolet rangeoverthewholerealline.Positivevaluesof correspondtorotationanglescountedcounterclockwise,whilenegative valuesof areassociatedwithrotationanglescountedclockwise.All pairs( r, )witha“xedvalueof r andvaluesof dierentbyinteger multiplesof2 representthesamepointsoftheplane.Forexample, theorderedpairs( r, )=(1 Š )and(1 )correspondtothesame point.Indeed,bothpointsareonthecircleofunitradius.Theray = isobtainedfromthepolaraxisbycounterclockwiserotation ofthelatterthroughtheangle .Butthesamerayisobtainedby rotatingthepolaraxisthroughtheangle clockwise;thatis,therays = and = Š coincide. Figure10.9. Polargrid.Coordinatecurvesofthepolarcoordinates.Thecurvesofconstantvaluesof r are concentriccircles.Thecurvesofconstantvaluesof are raysoutgoingfromtheorigin.

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18010.PLANARCURVES Furthermore,themeaningoftheradialvariable r canbeextended tothecaseinwhich r isnegativebyagreeingthatthepairs( Š r, )and ( r, + ), r> 0,representthesamepoint.Geometrically,thepoints ( r, )lieonalinethroughtheoriginatthesamedistance | r | fromthe originbutontheoppositesidesoftheorigin.Withthisagreementon extendingthemeaningofthepolarcoordinates,eachpointonaplane mayberepresentedbycountablymanypairs: (10.6)( r, ) ( r, +2 n )or( Š r, +(2 n +1) ) where n isaninteger.67.1.RectangularandPolarCoordinates.Supposethatthepolaraxis issetsothatitcoincideswiththepositive x axisoftherectangular coordinatesystem.Everypointontheplaneiseitherdescribedby therectangularcoordinates( x,y )orthepolarcoordinates( r, ).Itis easyto“ndtherelationbetweenthepolarandrectangularcoordinates ofapoint P byexaminingtherectanglewiththediagonal OP .Its horizontalandverticalsideshavelengths x and y ,respectively.The lengthofthediagonalis r .Theanglebetweenthehorizontalsideand thediagonalis .Therefore,cos = x/r andsin = y/r ,or x = r cos ,y = r sin r2= x2+ y2, tan = y x Theserelationsallowustoconvertthepolarcoordinatesofapointto rectangularcoordinatesandviceversa. Example 10.6 Findtherectangularcoordinatesofapointwhose polarcoordinatesare (2 ,/ 6) .Findthepolarcoordinatesofapoint withrectangularcoordinates ( Š 1 1) Solution: For r =2and = / 6,onehas x =2cos( / 6)=2 3 / 2= 3and y =2sin( / 6)=2 / 2=1,so( x,y )=( 3 1).For x = Š 1and y =1,onehas r2=2or r = 2andtan = Š 1.Thepoint( Š 1 1) liesinthesecondquadrant,thatis, / 2 .Therefore, =3 / 4. Alternatively,onecantake =3 / 4 Š 2 = Š 5 / 4. 67.2.PolarGraphs.A polargraph isacurvede“nedbytheequation r = f ( )or,moregenerally, F ( r, )=0.Itconsistsofallpointsthat haveatleastonepolarrepresentation( r, )thatsatis“estheequation. Herepolarcoordinatesareunderstoodintheextendedsenseof(10.6) whentheyareallowedtotakeanyvalue. Thesimplestpolargraphisde“nedbyaconstantfunction r = a where a isreal.Since r representsthedistancefromtheorigin,the

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67.POLARCOORDINATES181 pairs( | a | )formacircleofradius | a | centeredattheorigin.Similarly, thegraph = b ,where b isreal,isthesetofallpoints( r,b ),where r rangesovertherealaxis,whichisthelinethroughtheoriginthat makesanangle b radianswiththepolaraxis.Noticethatthepoints ( r,b ), r> 0,and( r,b ), r< 0,lieintheoppositequadrantsrelativeto theoriginasthepairs( r,b )and( Š r,b + )representthesamepoint. Ingeneral,theshapeofapolargraphcanbedeterminedbyplotting points( f ( k) ,k), k =1 2 ,...,n ,forasetofsuccessivelyincreasing valuesof 1<2<
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18210.PLANARCURVES Figure10.10. Polarcurve r = .Itisaspiralbecause thedistancefromtheorigin r increaseswiththeangle asthepointrotatesabouttheoriginthroughtheangle y = f ( x )forsymmetric( f ( Š x )= f ( x ))orskew-symmetric( f ( Š x )= Š f ( x ))functions. (i) Ifapolarequationisunchangedwhen isreplacedby Š thecurveissymmetricaboutthepolaraxis.Notethatthe transformation( r, ) ( r, Š )meansthat( x,y ) ( x, Š y ), whichisthere”ectionaboutthe x axis(orthepolaraxis). (ii) Ifapolarequationisunchangedwhen( r, )isreplacedby ( Š r, )orby( r, + ),thecurveissymmetricabouttheorigin.Again,thesetransformationsareequivalentto( x,y ) ( Š x, Š y ),whichisthere”ectionabouttheorigin. (iii) Iftheequationisunchangedunderthetransformation( r, ) ( r, Š ),thenthecurveissymmetricabouttheverticalline = / 2.Intherectangularcoordinates,thistransformation is(x,y ) ( x, Š y ),whichisthere”ectionaboutthe y axis. Example 10.9 Describethe cardioid r =1+sin Solution: Theequationisunchangedunder Š sothecurve issymmetricabouttheverticalaxis(the y axis).Itissucientto investigatethecurveintheinterval [ Š / 2 ,/ 2](inthefourthand “rstquadrants).Consideraraythatrotatescounterclockwisefrom = Š / 2to = / 2(fromthenegative y axistothepositive y axis).When = Š / 2, r =0.As increasesfrom Š / 2to0(the fourthquadrant),thedistancefromtheorigin r =1+sin increases

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67.POLARCOORDINATES183 Figure10.11. Thecardioid r =1+sin monotonicallyfrom0to1( r =1onthepolaraxis).Intheinterval [0 ,/ 2](the“rstquadrant),thedistancefromtheorigin r continuesto increasemonotonicallyandreachesitsmaximalvalue2onthevertical axis. 67.4.TangenttoaPolarGraph.To“ndatangentlinetoapolargraph r = f ( ),thepolarangleisviewedasaparametersothattheparametricequationsofthegraphare x = r cos = f ( )cos ,y = r sin = f ( )sin Bytheproductruleforthederivative, dy dx =dy d dx d= f( )sin + f ( )cos f( )cos Š f ( )sin Inparticular,ifthecurvepassesthroughtheorigin, r =0,theequation fortheslopeattheoriginissimpli“ed dy dx =tan if dr d = f( ) =0 Notethatif f( )=0,thentheslopeisanundeterminedform0 0because x( )= y( )=0foranyvalueof suchthat f( )= f ( )=0. Thismeansthatthecurvemayhaveacuspattheoriginandhenceis notsmooth. Example 10.10 Findtheslopeofthecardioid r =1+sin in termsof .Investigatethebehaviorofthecardioidneartheorigin.

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18410.PLANARCURVES Solution: Here f ( )=1+sin and f( )=cos .Thisleadstothe slope dy dx = cos sin +(1+sin )cos cos2 Š (1+sin )sin = cos (1+2sin ) (1+sin )(1 Š 2sin ) wheretheidentitycos2 =1 Š sin2 hasbeenusedtotransformthedenominator.Thecardioidpassesthroughtheoriginas passesthrough Š / 2.Theslope dy/dx isundeterminedbecausethenumeratorand denominatoroftheratiovanishat = Š / 2(bothderivatives dx/d and dy/d vanish).Theleftandrightlimitshavetobeinvestigatedto seeiftheslopehasajumpdiscontinuitythusindicatingacusp.The numeratorvanishesbecauseofthefactorcos ,whilethedenominator vanishesbecauseofthefactor(1+sin ).Hence, lim ( Š / 2)dy dx = Š 1 3 lim ( Š / 2)cos 1+sin = Š 1 3 lim ( Š / 2)Š sin cos = wherelHospitalsrulehasbeenusedtoresolvetheundeterminedform0 0andthepropertythattan as ( Š / 2)hasbeeninvoked to“ndthelimit.Thecardioidhasaverticaltangentlineattheorigin. Theslopehasanin“nitejumpdiscontinuity,meaningthatthecardioid hasacuspattheorigin(seeFigure10.11). 67.5.Exercises.In(1)…(3),convertthepolargraphequationtoaCartesianequation andsketchthecurve. (1) r =4sin (2) r =tan sec (3) r =2sin Š 4cos In(4)…(12),sketchthecurvewiththegivenpolarequation. (4) r = 0(5) r =ln 1(6) r2Š 3 r +2=0 (7) r =4cos(6 )(8) r2=9sin(2 )(9) r =1+2cos(2 ) (10) r =2+sin(3 )(11) r =1+2sin(3 )(12) r2 =1 (13)Sketchthecurve( x2+ y2)2=4 x2y2. Hint :Usepolarcoordinates. (14)Investigatethedependenceoftheshapeofthecurve r =cos( n ) astheinteger n increases.Whathappensif n isnotaninteger? (15)Showthatthecurve r =1+ a sin hasaninnerloopwhen | a | > 1 and“ndtherangeof thatcorrespondstotheinnerloop. (16)Forwhatvaluesof a isthecurve r =1+ a sin smooth? In(17)and(18),“ndtheslopeofthetangentlinetothegivencurve atthepointspeci“edbythevalueof andgiveanequationofthe

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68.PARAMETRICCURVES:THEARCLENGTHANDSURFACEAREA185 tangentline. (17) r =2sin = / 3(18) r =1 Š 2cos = / 6 (19)Showthatthecurves r = a sin and r = a cos intersectatright angles.68.ParametricCurves:TheArcLengthandSurfaceArea68.1.ArcLengthofaSmoothCurve.Let C beasmoothcurvede“ned bytheparametricequations x = x ( t ), y = y ( t ),where t [ a,b ]. Supposethat C istraversedexactlyonceas t increasesfrom a to b andconsiderapartitionoftheinterval[ a,b ]suchthat t0= a and tk= t0+ k t k =0 1 2 ,...,n ,aretheendpointsofthepartition intervalsofwidth t =( b Š a ) /n .Thenthepoints Pkwithcoordinates ( x ( tk) ,y ( tk))lieonthecurvesothat P0and Pnaretheinitialand terminalpoints,respectively.Thecurve C canbeapproximatedbya polygonalpathwithvertices Pk.Byde“nition,thelength L of C is thelimitofthelengthsoftheseapproximatingpolygonsas n : (10.7) L =limn nk =1| Pk Š 1Pk| providedthelimitexists,and,inthiscase,thecurveiscalled measurable Figure10.12. Thearclengthofasmoothparametric curveisapproximatedbythelengthof n straightline segmentsconnectingpointsonthecurve.Thearclength isde“nedin(10.7)asthelimit n .

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18610.PLANARCURVES Bythemeanvaluetheorem,whenappliedtothefunctions x ( t )and y ( t )ontheinterval[ tk Š 1,tk],therearenumbers t kand t kin( tk Š 1,tk) suchthat xk= x ( tk) Š x ( tk Š 1)= x( t k) t, yk= y ( tk) Š y ( tk Š 1)= y( t k) t. Therefore, | Pk Š 1Pk| = ( xk)2+( yk)2= ( x( t k))2+( y( t k))2 t. Thesumin(10.7)resemblesaRiemannsumforthefunction F ( t )= ( x( t ))2+( y( t ))2.ItisnotexactlyaRiemannsumbecause t k = t kingeneral.However,if x( t )and y( t )arecontinuous,itcanbeshown thatthelimit(10.7)isthesameasif t kand t kwereequal,namely, L istheintegralof F ( t )over[ a,b ]. Theorem 10.2(ArcLengthofaCurve) Ifacurve C isdescribed bytheparametricequations x = x ( t ) y = y ( t ) t [ a,b ] ,where x( t ) and y( t ) arecontinuouson [ a,b ] and C istraversedexactlyonceas t increasesfrom a to b ,thenthelengthof C is L = b a dx dt 2+ dy dt 2dt. If C isagraph y = f ( x ),then x = t y = f ( t ),and dx = dt ,and thelengthisgivenbythefamiliarexpression L = b a 1+ dy dx 2dt. Itisconvenienttointroducethearclengthofanin“nitesimalsegment ofacurve(the dierentialofthearclength ) ds = ( dx )2+( dy )2= L = Cds = b ads dt dt. Thesymbol Cmeansthesummationoverin“nitesimalsegmentsofthe curve S (theintegralalongacurve C )andexpressesasimplefactthat thetotallengthisthesumofthelengthsofits(in“nitesimal)pieces.68.2.IndependenceofParameterization.Byitsveryde“nition,thearc lengthis independent oftheparameterizationofthecurve.Ifacurve C isde“nedasapointset,then any parametricequationscanbeused toevaluatethearclength.Let C betracedoutonlyonceby x = x ( t ), y = y ( t ),where t [ a,b ],andby x = X ( ), y = Y ( ),where [ ]. Asnoted,thereisarelationbetweentheparameters t and = g ( t ), suchthat g ( t )increasesfrom to as t increasesfrom a to b ,that

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68.PARAMETRICCURVES:THEARCLENGTHANDSURFACEAREA187 is,d dt= g( t ) 0,suchthat x ( t )= X ( g ( t ))and y ( t )= Y ( g ( t )). Therefore,theintegrals(10.7)correspondingtodierentparametric equationsofthesamecurvearerelatedbyachangeoftheintegration variable: L = b a dx dt 2+ dy dt 2dt = b a dx d d dt 2+ dy d d dt 2dt = b a dx d 2+ dy d 2d dt dt = dx d 2+ dy d 2d. Thus, thearclengthisindependentofthecurveparameterizationand canbecomputedinanysuitableparameterizationofthecurve Acircleofradius R isdescribedbytheparametricequations x = R cos t y = R sin t t [0 2 ].Then dx = Š R sin tdt and dy = R cos tdt .Hence, ds2=( R sin tdt )2+( R cos tdt )2= R2(sin2t +cos2t ) dt2= R2dt2,or ds = Rdt ,and L = Cds = 2 0Rdt = R 2 0dt =2 R. Example 10.11 Findthelengthofonearchofthecycloid x = R ( Š sin ) y = R (1 Š cos ) Solution: Accordingtothedescriptionofthecycloid,onearchcorrespondstotheinterval [0 2 ].Thearclengthdierential ds is foundasfollows: dx = R (1 Š cos ) d,dy = R sin d, ds2= dx2+ dy2=[(1 Š cos )2+sin2 ] R2d2=[1 Š 2cos +cos2 +sin2 ] R2d2=(2 Š 2cos ) R2d2. ds = 2(1 Š cos ) Rd. Toevaluatetheintegralof 2(1 Š cos ),thedouble-angleidentityis invoked,sin2( / 2)=(1 Š cos ) / 2.Since0 / 2 when [0 2 ], thesinusisnonnegative,sin( / 2) 0,intheintegrationinterval,and hence,aftertakingthesquareroot( u2= | u | ),theabsolutevaluecan beomitted.Thus, L = R 2 0 4sin2( / 2) d =2 R 2 0sin( / 2) d =2 R [ Š 2cos( / 2)] 2 0=8 R.

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18810.PLANARCURVES 68.3.AreaofaPlanarRegion.Theareaunderthecurve y = f ( x ) andabovetheinterval x [ a,b ]isgivenby A = b af ( x ) dx ,where f ( x ) 0.Supposethatthecurveisalsodescribedbyparametric equations x = x ( t ), y = y ( t ),sothatthefunction x ( t )isone-to-one. Then,bychangingtheintegrationvariable, dx = x( t ) dt and A = b aydx = y ( t ) x( t ) dt. Thenewintegrationlimitsarefoundasusual.When x = a t iseither or ,andwhen x = b t istheremainingvalue. Example 10.12 Findtheareaunderonearchofthecycloid x = R ( Š sin ) y = R (1 Š cos ) Solution: When [0 2 ], x [0 2 R ]foronearchofthecycloid, and y ( ) 0.Usingthedierential dx foundinthepreviousexample, A = 2 R 0ydx = R22 0(1 Š cos )2d = R22 0(1 Š 2cos +cos2 ) d = R22 0[1+1 2(1+cos(2 ))] d = R22 0(1+1 2) d =3 R2, where 2 0cos d =0and 2 0cos(2 ) d =0bythe2 periodicityof thecosinefunction. 68.4.SurfaceAreaofAxiallySymmetricSurfaces.Anaxiallysymmetric surfaceisasurfacesymmetricrelativetorotationsaboutaline.Sucha lineiscalledthe symmetryaxis .Forexample,acylinderissymmetric relativetorotationsaboutitsaxis,asphereissymmetricrelativeto rotationsaboutitsdiameter,andsoon.Anaxiallysymmetricsurface issweptbyaplanarcurvewhenthelatterisrotatedaboutaline.A cylinderofradius R andheight h isobtainedbyrevolvingastraightline segmentoflength h aboutalineparalleltothesegmentatadistance R .Asphereofradius R isobtainedbyrevolvingacircleofradius R aboutitsdiameter. Let ds bethearclengthofanin“nitesimalsegmentofasmooth curve C positionedatapoint( x,y ).Ifthedistancebetweenthepoint ( x,y )andthesymmetryaxisis R ( x,y ),thenthearea dA ofthepart ofthesurfacesweptbythecurvesegmentwhenthelatterisrotated aboutthesymmetryaxisistheareaofacylinderofradius R ( x,y )and

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68.PARAMETRICCURVES:THEARCLENGTHANDSURFACEAREA189 height ds : dA =2 R ( x,y ) ds. Thetotalsurfaceareaisthesumofareasofallsuchpartsofthesurface (10.8) A =2 CR ( x,y ) ds =2 b aR ( x ( t ) ,y ( t )) ds dt dt, where x = x ( t ), y = y ( t ), a t b areparametricequationsof C Hereitisagainassumedthatthepoint( x ( t ) ,y ( t ))tracesoutthecurve C onlyonceas t increasesfrom a to b .Inparticular,ifthesymmetry axiscoincideswiththe x axis,then R ( x,y )= | y | (thedistanceofthe Figure10.13. Asurfaceisobtainedbyrotationofa smoothcurveaboutaverticalline.If ds isthearclength ofanin“nitesimalsegmentofthecurveatapoint P and R isthedistanceofthepoint P fromtherotation axis,thenthesurfaceareasweptbythecurvesegment is dA =2 Rds (thesurfaceareaofacylinderofradius R andheight ds ).

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19010.PLANARCURVES point( x,y )tothe x axis)and A =2 C| y | ds =2 b a| y ( t ) | dx dt 2+ dy dt 2dt. Example 10.13 Findtheareaofthesurfaceobtainedbyrevolving onearchofthecycloid x = R ( Š sin ) y = R (1 Š cos ) aboutthe x axis. Solution: Thedierentialofthearclengthofthecycloidhasbeen computedinExample10.11.Since y ( t ) 0here,theabsolutevalue maybeomittedand A =2 Cyds =2 2 0R (1 Š cos ) 2(1 Š cos ) Rd =8 R22 0sin3( / 2) d =16 R2 0sin3udu =16 R2 0(1 Š cos2u )sin udu =16 R21 Š 1(1 Š z2) dz =16 R2( z Š z3/ 3) 1 Š 1= 64 R2 3 wherethedouble-angleidentityhasbeenusedagain,sin2( / 2)=(1 Š cos ) / 2,andthentwosuccessivechangesoftheintegrationvariable havebeendonetoevaluatetheintegral, u = / 2 [0 ]and z = cos u [ Š 1 1]. 68.5.Exercises.In(1)…(3),“ndthearclengthofthecurve. (1) x =2+3 t2, y =1 Š 2 t3betweenthepoints(2 1)and(5 Š 1). (2) x =3sin t Š sin(3 t ), y =3cos t Š cos(3 t ),0 t (3) x = t/ (1+ t ), y =ln(1+ t )betweenthepoints(0 0)and(2 / 3 ln3). (4)Findtheareaoftheregionenclosedbythecurve x = a cos3t y = a sin3t (theastroid). (5)Findtheareaoftheregionenclosedbythecurve x = a cos t y = b sin t (anellipse). (6)Findtheareaenclosedbythecurve x = t2Š 2 t y = t andthe y axis. In(7)…(10),“ndtheareaofasurfacegeneratedbyrotatingthegiven curveaboutthespeci“edaxis.Sketchthesurface. (7) x = a cos3t y = a sin3t (aboutthe x axis). (8) x2+ y2= a2(aboutthe y axis). (9) x = t3, y = t2,0 t 1(aboutthe x axis).

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69.AREASANDARCLENGTHSINPOLARCOORDINATES191 (10) x = etŠ t y =4 et/ 2,0 t 1(aboutthe y axis). (11)Let V bethevolumeasolidboundedbyanaxiallysymmetric surface.Showthat V = C[ R ( x,y )]2ds ,where C isthecurvewhose revolutionaboutthesymmetryaxisgivestheboundarysurfaceand R ( x,y )isde“nedin(10.8).Findthevolumeofthesolidboundedby thesurfacedescribedinExample10.13.69.AreasandArcLengthsinPolarCoordinates69.1.AreaofaPlanarRegion.Theorem 10.3(AreaofaPlanarRegioninPolarCoordinates) Letaplanarregion D beboundedbytworaysfromtheorigin = a = b andapolargraph r = f ( ) ,where f ( ) 0 ,thatis, D = { ( r, ) | 0 r f ( ) ,a b } Thentheareaof D is A = 1 2 b a[ f ( )]2d. Proof. Considerapartitionoftheinterval[ a,b ]bypoints k= a + k k =0 1 ,...,n ,where =( b Š a ) /n .Let mkand Mkbetheminimumandmaximumvaluesof f on[ k Š 1,k].Recallthata continuousfunction f alwaysattainsitsmaximumandminimumvalues onaclosedinterval.Thearea Akoftheplanarregionboundedby therays = k Š 1, = kandthepolargraph r = f ( )isnotlessthan theareaofthedisksectorwithradius r = mkandangle andisnot greaterthantheareaofthedisksectorwithradius r = Mkandangle .Theareaofasectorofadiskwithradius R andangle radians is A =1 2R2 .Therefore, 1 2 m2 k Ak 1 2 M2 k andthetotalarea A oftheplanarregioninquestionsatis“estheinequality AL n= 1 2nk =1m2 k A 1 2nk =1M2 k = AU n,A =nk =1 Ak, whichistrueforany n .Let F ( )=1 2[ f ( )]2.Thefunction F iscontinuouson[ a,b ].Then1 2m2 kand1 2M2 karetheminimumandmaximum valuesof F onthepartitioninterval[ k Š 1,k].Thisshowsthatthe lowerandupperbounds, AL nand AU n,are loweranduppersums for thefunction F on[ a,b ].Bythede“nitionofthede“niteintegraland

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19210.PLANARCURVES integrabilityofacontinuousfunction(seeCalculusI),theupperand lowersumsconvergetotheintegralof F over[ a,b ], AL n b aFd and AU n b aFd as n .Theconclusionofthetheorem, A = b aFd followsfromthesqueezeprinciple. Example 10.14 Findtheareaenclosedbyoneloopofthefour-leaf rose r =cos(2 ) Solution: Notethat r =1when =0,whichisthemaximalvalueof r .Thefunctioncos(2 )hastworoots = / 4thatarethenearest to =0.Hence,oneloopcorrespondstotheinterval [ Š / 4 ,/ 4]. Theareais A = 1 2 / 4 Š / 4cos2(2 ) d = 1 4 / 4 Š / 4[1+cos(4 )] d = 1 4 [ +1 4sin(4 )] / 4 Š / 4= 8 Let D beaplanarregionthatliesbetweentwopolargraphs r = f ( )and r = g ( )suchthat f ( ) g ( ) 0if [ a,b ]and0 < b Š a 2 ;thatis, D isthesetofpointswhosepolarcoordinates satisfytheinequalities: D = { ( r, ) | 0 g ( ) r f ( ) ,a b } Thentheareaof D isgivenby A = 1 2 b a[ f ( )]2d Š 1 2 b a[ g ( )]2d = 1 2 b a [ f ( )]2Š [ g ( )]2 d. Example 10.15 Findtheareaofaregion D boundedbythecardioid r =1+sin andthecircle r =3 / 2 thatliesabovethepolaraxis (inthe“rstandsecondquadrants). Solution: Thepolargraphs r =1+sin = f ( )and r =3 / 2= g ( ) areintersectingwhen f ( )= g ( )or1+sin =3 / 2orsin =1 / 2. Sincetheregion D liesinthe“rsttwoquadrants,thatis,0 thevaluesof forthepointsofintersectionhavetobechosenas = / 6= a and = Š / 6= b .Therefore, D = { ( r, ) | 3 / 2 r 1+sin ,/ 6 5 / 6} ,

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69.AREASANDARCLENGTHSINPOLARCOORDINATES193 andhencetheareaof D is A = 1 2 b a[(1+sin )2Š9 4] d = 1 2 b a[ Š5 4+2sin +sin2 ] d = 1 2 b a[ Š5 4+2sin +1 2(1 Š cos(2 ))] d = 1 2 b a[ Š3 4+2sin Š1 2cos(2 )] d =1 2[ Š3 4 Š 2cos Š1 4sin(2 )] 5 / 6 / 6= 9 3 Š 2 8 Remark. When“ndingpointsofintersectionoftwopolargraphs, r = f ( )and r = g ( ),bysolvingtheequation f ( )= g ( ),one hastokeepinmindthatasinglepointhasmanyrepresentationsas describedin(10.6).Sosomeofthepairs( f ( ) ),where ranges oversolutionsoftheequation f ( )= g ( ),maycorrespondtothe samepoint.Toselectdistinctpoints,allpairs( f ( ) )satisfyingthe intersectionconditioncanbetransformedbymeansof(10.6)sothat r [0 )and [0 2 ).Inthisrangeofpolarcoordinates,thereis aone-to-onecorrespondencebetweenpointsonaplaneandpairs( r, ) withjustoneexceptionwhen r =0;allthepairs(0 )correspondto theoriginofthepolarcoordinatesystem.69.2.ArcLength.Supposethatacurve C istraversedbythepoint ( r, )=( f ( ) )onlyonceas increasesfrom a to b .Choosing asaparameter,thecurveisdescribedbytheparametricequations x = r cos y = r sin ,where r = f ( ).To“ndthearclengthof C onehasto“ndtherelationbetweenthearclengthdierential ds and d .Onehas dx = dr d cos Š r sin d,dy = dr d sin + r cos d Therefore, ds2= dx2+ dy2= dr d cos Š r sin 2+ dr d sin + r cos 2 d2= dr d 2(cos2 +sin2 )+ r2(cos2 +sin2 ) d2= dr d 2+ r2 d2.

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19410.PLANARCURVES Thearclengthofthecurve C is L = Cds = b ads d d = b a r2+ dr d 2d. where r = f ( )and b>a Example 10.16 Findthelengthofthecardioid r =1+sin Solution: Onehas r2+ dr d 2=(1+sin )2+(cos )2=2(1+sin ) wherethetrigonometricidentitysin2 +cos2 =1hasbeenused.The cardioidistraversedonceif [ Š ].Therefore,thelengthis L = 2 Š 1+sin d =2 2 / 2 Š / 2 1+sin d, sincethecardioidissymmetricabouttheverticalline(the y axis). Thisintegralcanbeevaluatedbythesubstitution u =1+sin [0 2] sothat du =cos d ,wherecos = 1 Š sin2 = 1 Š ( u Š 1)2= u (2 Š u ).Hence, L =2 2 2 0 u u (2 Š u ) du =2 2 2 0du 2 Š u = Š 4 2 2 Š u 2 0=8 69.3.SurfaceArea.Ifasurfaceisobtainedbyrotatingapolargraph r = f ( )aboutaline,the(10.8)canbeusedto“ndtheareaofthe surfacewherethedistance R ( x,y )andthearclengthdierential ds havetobeexpressedinthepolarcoordinateswith r = f ( ). Example 10.17 Findtheareaofthesurfaceobtainedbyrotating thecardioid r =1+sin aboutitssymmetryaxis. Solution: Thesymmetryaxisofthecardioidisthe y axis.Sothe distancefromthe y axistoapoint( x,y )is R ( x,y )= | x | .Thesurface canbeobtainedbyrotatingthepartofthecardioidthatliesinthe fourthand“rstquadrants,thatis, x 0or [ Š / 2 ,/ 2].Since x = r cos ,thesurfaceareais A =2 C| x | ds =2 / 2 Š / 2r cos ds d d =2 / 2 Š / 2r cos r2+ dr d 2d.

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69.AREASANDARCLENGTHSINPOLARCOORDINATES195 Thederivative ds/d hasbeencalculatedinthepreviousexample. Therefore, A =2 2 / 2 Š / 2(1+sin )3 / 2cos d =2 2 1 Š 1(1+ u )3 / 2du =2 2 (1+ u )5 / 2 5 / 2 1 Š 1= 32 5 wherethesubstitution u =sin hasbeenmadetoevaluatethe integral. 69.4.Exercises.In(1)…(4),sketchthecurveand“ndtheareathatitencloses. (1) r =4cos(2 )(2) r = a (1+cos )(3) r =2 Š cos(2 ) (4) r2=4cos(2 ) In(5)…(7),sketchthecurveand“ndtheareaofoneloopofthecurve. (5) .r =9sin(3 )(6) r =1+2sin (innerloop)(7) r =2cos Š sec In(8)and(9),“ndtheareaoftheregionthatliesinsidethe“rstcurve andoutsidethesecondcurve.Sketchthecurves. (8) r =2sin ,r =1(9) r =3cos ,r =1+cos In(10)…(13),“ndtheareaoftheregionboundedbythecurves.Sketch theregion. (10) r =2 ,r = [0 2 ](11) r2=sin(2 ) ,r2=cos(2 ) (12) r =2 a sin ,r =2 b cos ,a,b> 0 (13) r =3+2cos ,r =3+2sin (14)Findtheareainsidethelargerloopandoutsidethesmallerloop ofthelima con r =1 / 2 Š cos In(15)…(17),sketchthecurveand“nditslength. (15) r =2 a sin (16) r = [0 2 ](17) r = a +cos ,a 1 In(18)…(20),“ndtheareaofthesurfaceobtainedbyrotatingthecurve aboutthespeci“edaxis.Sketchthesurface. (18) r = a> 0aboutalinethroughtheorigin. (19) r =2 a cos a> 0,(i)aboutthe y axisand(ii)aboutthe x axis. (20) r2=cos(2 )aboutthepolaraxis. (21) = a and0 r R ,where0 0aboutthe polaraxis.

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19610.PLANARCURVES 70.ConicSections Considertwointersectinglinesinspace, L1and L2.Asurfaceswept bytheline L2whenitisrotatedabouttheline L1isacirculardouble cone .Theline L1isthesymmetryaxisofthecone.Thepointofintersectionofthelinesiscalledthe vertex ofacone.Anyplanethatdoes notpassthroughthevertexintersectstheconealongacurve.ItappearsthatallsuchcurvesfallintothreetypesasshowninFigure10.14. Ifthecurveofintersectionisaloop,thenitisanellipse.Iftheplane isparalleltotheline L2,thenthecurveisaparabola.Iftheplane isparalleltotheaxisofthecone,thenthecurveisahyperbola.The curvesofintersectionofaplaneandaconearecalled conicsections or conics .Theyhaveapuregeometricaldescription,whichwillbe presentedhere. Remark. Atrajectoryofanymassiveobjectinthesolarsystem (e.g.,comet,asteroid,planet)isaconicsection„thatis,aparabola, hyperbola,orellipse.ThisfactfollowsfromNewtonsLawofGravity andwillbeprovedinCalculusIII.70.1.Parabolas.A parabola isthesetofpointsinaplanethatare equidistantfroma“xedpoint F (calledthe focus )anda“xedline Figure10.14. Conicsectionsarecurvesthatareintersectionsofaconewithvariousplanes.Theshapeof aconicsectiondependsontheorientationoftheplane relativetotheconesymmetryaxis.

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70.CONICSECTIONS197 Figure10.15. Left:Geometricaldescriptionofa parabolaasasetofpoints P inaplanethatareequidistantfroma“xedpoint F ,calledthe focus ,anda“xed linecalledthe directrix (ahorizontallineinthe“gure). Right:Acircularparaboloidisthesurfaceobtainedby rotatingaparabolaaboutthelinethroughitsfocusand perpendiculartoitsdirectrix. (calledthe directrix ).Let P beapointinaplane.Considertheline through P thatisperpendiculartothedirectrixandlet Q bethepoint oftheirintersection.Then P liesonaparabolaif | FP | = | QP | .This conditionisusedtoderivetheequationofaparabola. Aparticularlysimpleequationofaparabolaisobtainedifthecoordinatesystemissetsothatthe y axiscoincideswiththelinethrough thefocusandperpendiculartothedirectrix.Theorigin O ischosen sothat F =(0 ,p )andhencetheparabolacontainstheorigin O ,while thedirectrixistheline y = Š p paralleltothe x axis(theoriginis atdistance | p | from F andfromthedirectrix).If P =( x,y ),then | FP | = x2+( y Š p )2,thepoint Q hasthecoordinates( x, Š p ),and | PQ | = ( y + p )2. Anequationoftheparabolawithfocus (0 ,p ) and directrix y = Š p is | FP |2= | PQ |2= x2+( y Š p )2=( y + p )2 x2=4 py. Inthe16thcentury,Galileoshowedthatthepathofaprojectilethatis shotintotheairatanangletothegroundisaparabola.Thesurface obtainedbyrotatingaparabolaaboutitssymmetryaxisiscalleda paraboloid .Ifasourceoflightisplacedatthefocusofaparaboloid mirror,then,afterthere”ection,thelightformsabeamparalleltothe symmetryaxis.Thisfactisusedtodesign”ashlights,headlights,and

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19810.PLANARCURVES soon.Conversely,abeamoflightparalleltothesymmetryaxisofa paraboloidmirrorwillbefocusedtothefocuspointafterthere”ection, whichisusedtodesignre”ectingtelescopes.70.2.Ellipses.An ellipse isthesetofpointsinaplane,thesumof whosedistancesfromtwo“xedpoints F1and F2isaconstant.The “xedpointsarecalled foci (pluralof focus ).Let P beapointona plane.Then P belongstoanellipseif | PF1| + | PF2| =2 a ,where a> 0 isaconstant(thefactor2ischosenforconveniencetobeseenlater). Evidently, | F1F2| < 2 a ;otherwise,noellipseexists. Aparticularlysimpleequationofanellipseisobtainedwhenthe coordinatesystemissetsothatthefocilieonthe x axisandhave thecoordinates F1=( Š c, 0)and F2=( c, 0),where | c | 0.Theellipseintersectsthe x axis at( a, 0)andthe y axisat(0 b )(calledthe vertices ofanellipse). Thelinesegmentjoiningthepoints( a, 0)iscalledthe majoraxis Ifthefociofanellipsearelocatedonthe y axis,then x and y are swappedinthisequation,andthemajoraxisliesonthe y axis.This showsthattherestriction a b canbedroppedintheellipseequation. Inparticular,anellipsebecomesacircleofradius a if a = b OneofKeplerslawsisthattheorbitsoftheplanetsinthesolar systemareellipseswiththeSunatonefocus.

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70.CONICSECTIONS199 Figure10.16. Left:Anellipseisthesetofpointsina plane,thesumofwhosedistancesfromtwo“xedpoints F1and F2(thefoci)isaconstant.Right:Acircular ellipsoidisthesurfaceobtainedbyrotatinganellipse aboutthelinethroughitsfoci.70.3.Hyperbolas.A hyperbola isthesetofallpointsinaplane,the dierenceofwhosedistancesfromtwo“xedpoints F1and F2(thefoci) isaconstant.Foranypoint P onahyperbola, | PF1|Š| PF2| = 2 a (asthedierenceofthedistancescanbenegative).Letthefocibeat ( c, 0).Followingthesameprocedureusedtoderiveanequationofan ellipse, anequationofahyperbolawithfoci ( c, 0) isfoundtobe x2 a2Š y2 b2=1 where c2= a2+ b2.Thedetailsarelefttothereaderasanexercise. Thisequationshowsthat x2/a2 1forany y ,thatis, x a or x Š a .Ahyperbolathereforehastwo branches .Thebranchin x Š a intersectsthe x axisat x = Š a ,whilethebranchin x a doessoat x = a .Thepoints( a, 0)arecalled vertices .Furthermore, intheasymptoticregion | x | ,ahyperbolahasslantasymptotes y = ( b/a ) x .Indeed, y = b x2 a2Š 1= b | x | a 1 Š a2 x2 b | x | a 1 Š a2 2 x2 b | x | a as | x | .Herethelinearization 1+ u 1+ u/ 2hasbeenusedto obtaintheasymptoticbehaviorforsmall u = Š a2/x2 0.

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20010.PLANARCURVES Figure10.17. Left:Ahyperbolaisthesetofallpoints inaplane,thedierenceofwhosedistancesfromtwo “xedpoints F1and F2(thefoci)isaconstant.Top right:Acircularhyperboloidofonesheetisthesurface obtainedbyrotatingahyperbolaaboutthelinethrough themidpointofthesegment F1F2andperpendicularto it(theverticallineintheleftpanel).Bottomright:A circularhyperboloidoftwosheetsisthesurfaceobtained byrotatingahyperbolaaboutthelinethroughitsfoci (thehorizontallineintheleftpanel). Ifthefociofahyperbolaareonthe y axis,then,byreversingthe rolesof x and y ,itfollowsthat thehyperbola y2 a2Š x2 b2=1 hasfoci (0 c ) ,where c2= a2+ b2,vertices (0 a ) ,andslantasymptotes y = ( a/b ) x .70.4.ShiftedConics.Consideracurvede“nedbyaquadraticCartesianequation Ay2+ Bx2+ y + x + =0 .

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70.CONICSECTIONS201 Supposethat A =0and B =0.Bycompletingthesquares,this equationcanbetransformedtothestandardform A y Š 2 A 2+ B x Š 2 B 2= 2 4 A + 2 4 B Š = d or ( y Š y0)2 A/d + ( x Š x0)2 B/d =1 where x0= / (2 B )and y0= / (2 A ),provided d =0.Depending onthesignsof A/d and B/d ,thisequationdescribeseitheranellipse orahyperbolaasiftheoriginwasmovedtothepoint( x0,y0).If A/d and B/d arebothnegative,thentheequationhasnosolution. Ifeither A or B vanishes,butnotboth,thenthequadraticCartesian equationdescribesaparabola(thedetailsarelefttothereaderasan exercise).If A = B =0,thetheequationdescribesastraightline. If d =0,solutionsoftheequationformasetoftwostraightlines, y Š y0= Š ( B/A )( x Š x0),throughthepoint( x0,y0),provided AB< 0.WhensolutionsoftheCartesianequationformahyperbola ( d =0, AB< 0),theselinesareitsslantasymptotes.70.5.ConicSectionsinPolarCoordinates.Thefollowingtheoremoers auniformdescriptionofconicsections. Theorem 10.4(ConicSections) Let F bea“xedpoint(calledthe focus )and L bea“xedline(calledthe directrix )inaplane.Let e bea “xedpositivenumber(calledthe eccentricity ).Thesetofpoints P in theplanewhosetheratioofthedistancefrom F tothedistancefrom L istheconstant e isaconicsection.Theconicis (1)anellipseif e< 1 (2)aparabolaif e =1 (3)ahyperbolaif e> 1 e = | PF | | PL | Proof. Setthecoordinatesystemsothat F isattheoriginand thedirectrixisparalleltothe y axisand d unitstotheright.Thus, thedirectrixhastheequation x = d> 0andisperpendiculartothe polaraxis.Ifthepoint P haspolarcoordinates( r, )andrectangular coordinates( x,y ),then | PF | = r = x2+ y2and | PL | = d Š x = d Š r cos .Thecondition | PF | = e | PL | yieldstheequation r = e ( d Š x ). Bysquaringit,oneinfersa quadratic Cartesianequation x2+ y2= e2( d Š x )2 (1 Š e2) x2+ y2+2 e2dx Š e2d =0 whichhasbeeninvestigatedintheprecedingsection.If e =1,then theequationdescribestheshiftedparabola y2= Š 2 d ( x Š 1 / 2).When

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20210.PLANARCURVES e =1,bycompletingthesquares,thisequationisbroughttothe standardform x + e2d 1 Š e22+ y2 1 Š e2= e2d2 (1 Š e2)2. If e< 1,thenallthecoecientsarepositive,andtheequationdescribes ashiftedellipse ( x Š x0)2 a2+ y2 b2=1 ,a2= b2 1 Š e2,b2= e2d2 1 Š e2,x0= e2d e2Š 1 = Š c, where c isthedistancefromtheorigintothefocioftheellipse, c2= a2Š b2.Theeccentricityisthen e = c/a .Similarly,if e> 1,thenthe coecientshaveoppositesigns,andtheequationdescribesashifted hyperbola ( x Š x0)2 a2Š y2 b2=1 ,e = c a ,c2= a2+ b2. Inthebeginningoftheproof,thepolarequationforconicsections wasgivenas r = e ( d Š r cos ).Ifthedirectrixischosentobetothe leftofthefocusas x = Š d ,thencos isreplacedby Š cos inthepolar equation.Ifthedirectrixischosentobeparalleltothepolaraxisas y = d ,thentheconicsectionsare r = e ( d y )= e ( d r sin ).These equationscanbesolvedfor r toobtainconicsectionsaspolargraphs. Corollary 10.5(ConicsinPolarCoordinates) Apolarequation oftheform r = ed 1 e cos or r = ed 1 e sin representsaconicsectionof eccentricity e .Theconicsectionisan ellipseif e< 1 ,aparabolaif e =1 ,andahyperbolaif e> 1 .70.6.Exercises.In(1)…(9),classifytheconicsection.Findthevertices,foci(orfocus), directrix,andasymptotes(ifthecurveisahyperbola).Sketchthe curve. (1) y2=16 x (2) x2= Š 4 y (3) y +12 x Š 2 x2=18 (4) x2+4 y2=16(5)9 x2Š 18 x +4 y2=27 (6) x2+3 y2+2 x Š 12 y +10=0(7)4 x2Š 9 y2=36 (8) y2Š 2 y =4 x2+3(9) y2Š 4 x2+2 y +16 x +3=0

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70.CONICSECTIONS203 In(10)Along-rangeradionavigationsystemusestworadiostations, locatedatpoints A and B alongthecoastline,thattransmitsimultaneoussignalstoashiplocatedatpoint P inthesea.Theonboard computerconvertsthetimedierenceinreceivingthesesignalsintoa distancedierence | PA |Š| PB | .Thislocatestheshipononebranch ofahyperbola.Supposethatstation B islocated D milesfromstation A .Ashipreceivesthesignalfrom B microseconds( s)beforeit receivesthesignalfrom A .Thesignaltravelswiththespeedoflight, c =980ft / s.Howfarothecoastlineistheship?Ifthecoordinate systemissetsothattheline AB coincideswiththe x axisand A isat theorigin,“ndthecoordinatesoftheshipasfunctionsof In(11)…(13),classifytheconicsection.Findtheeccentricity,anequationofthedirectrix,andsketchtheconic. (11) r = 8 4+sin (12) r = 10 2 Š 5cos (13) r = 1 3+3cos (14)Showthattheconicsections r = a/ (1 Š cos )and r = b/ (1+cos ) intersectatrightangles. (15)TheorbitofHalleysComet,lastseenin1986andduetoreturn in2062,isanellipsewitheccentricity0 97andonefocusattheSun. Thelengthofitsmajoraxisis36.18AU.Anastronomicalunit(AU) isthemeandistancebetweentheEarthandtheSun,about93million miles.FindapolarequationfortheorbitofHalleysComet.Whatis themaximalandminimaldistancefromthecomettotheSun?



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ConceptsinCalculusIIUNIVERSITYPRESSOFFLORIDAFloridaA&MUniversity,Tallahassee FloridaAtlanticUniversity,BocaRaton FloridaGulfCoastUniversity,Ft.Myers FloridaInternationalUniversity,Miami FloridaStateUniversity,Tallahassee NewCollegeofFlorida,Sarasota UniversityofCentralFlorida,Orlando UniversityofFlorida,Gainesville UniversityofNorthFlorida,Jacksonville UniversityofSouthFlorida,Tampa UniversityofWestFlorida,Pensacola OrangeGroveTexts Plus

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ConceptsinCalculusII Mikl osB onaandSergeiShabanov UniversityofFloridaDepartmentof MathematicsU niversity P ressof F lorida Gainesville  Tallahassee  Tampa  BocaRaton Pensacola  Orlando  Miami  Jacksonville  Ft.Myers  Sarasota

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Copyright2012bytheUniversityofFloridaBoardofTrusteesonbehalfoftheUniversityof FloridaDepartmentofMathematics ThisworkislicensedunderamodiedCreativeCommonsAttribution-Noncommercial-No DerivativeWorks3.0UnportedLicense.Toviewacopyofthislicense,visithttp:// creativecommons.org/licenses/by-nc-nd/3.0/.Youarefreetoelectronicallycopy,distribute,and transmitthisworkifyouattributeauthorship. However,allprintingrightsarereservedbythe UniversityPressofFlorida(http://www.upf.com).PleasecontactUPFforinformationabout howtoobtaincopiesoftheworkforprintdistribution .Youmustattributetheworkinthe mannerspeciedbytheauthororlicensor(butnotinanywaythatsuggeststhattheyendorse youoryouruseofthework).Foranyreuseordistribution,youmustmakecleartoothersthe licensetermsofthiswork.Anyoftheaboveconditionscanbewaivedifyougetpermissionfrom theUniversityPressofFlorida.Nothinginthislicenseimpairsorrestrictstheauthor'smoral rights. ISBN978-1-61610-161-9 OrangeGroveTexts Plus isanimprintoftheUniversityPressofFlorida,whichisthescholarly publishingagencyfortheStateUniversitySystemofFlorida,comprisingFloridaA&M University,FloridaAtlanticUniversity,FloridaGulfCoastUniversity,FloridaInternational University,FloridaStateUniversity,NewCollegeofFlorida,UniversityofCentralFlorida, UniversityofFlorida,UniversityofNorthFlorida,UniversityofSouthFlorida,andUniversityof WestFlorida. UniversityPressofFlorida 15Northwest15thStreet Gainesville,FL32611-2079 http://www.upf.com

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Contents Chapter6.ApplicationsofIntegration1 36 .TheAreaBetweenCurves1 37 .Volumes8 38 .CylindricalShells17 39 .WorkandHydrostaticForce22 40 .AverageValueofaFunction28 Chapter7.MethodsofIntegration33 41 .IntegrationbyParts33 42 .TrigonometricIntegrals36 43 .TrigonometricSubstitution41 44 .IntegratingRationalFunctions45 45 .StrategyofIntegration51 46 .IntegrationUsingTablesandSoftwarePackages54 47 .ApproximateIntegration59 48 .ImproperIntegrals67 Chapter8.SequencesandSeries77 49 .In“niteSequences77 50 .SpecialSequences85 51 .Series92 52 .SeriesofNonnegativeTerms99 53 .ComparisonTests106 54 .AlternatingSeries111 55 .RatioandRootTests117 56 .Rearrangements126 57 .PowerSeries133 58 .RepresentationofFunctionsasPowerSeries139 59 .TaylorSeries147 Chapter9.FurtherApplicationsofIntegration157 60 .ArcLength157 Chapterandsectionnumberingcontinuesfromthepreviousvolumeintheseries, ConceptsinCalculusI .

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viCONTENTS 61 .SurfaceArea162 62 .ApplicationstoPhysicsandEngineering168 63 .ApplicationstoEconomicsandtheLifeSciences176 64 .Probability182 Chapter10.PlanarCurves193 65 .ParametricCurves193 66 .CalculuswithParametricCurves201 67 .PolarCoordinates207 68 .ParametricCurves:TheArcLengthandSurfaceArea214 69 .AreasandArcLengthsinPolarCoordinates221 70 .ConicSections225

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CHAPTER6 ApplicationsofIntegration 36.TheAreaBetweenCurves36.1.TheBasicProblem.Inthepreviouschapter,welearnedthatif thefunction f satis“es f ( x ) 0forallrealnumbers x intheinterval [ a,b ],thentheareaofthedomainwhosebordersarethegraphof f thehorizontalaxis,andtheverticallines x = a and x = b isequalto b af ( x ) dx .Ifthereisnodangerofconfusionastowhat a and b are, thenthisfactissometimesinformallyexpressedbythesentencethe integralof f isequaltotheareaofthedomainthatis under thegraph of f .Ž Whatcanwesayabouttheareaofthedomain betweentwocurves ? Thereareseveralwaystoaskthisquestion.Theeasiestversion,discussedbythefollowingtheorem,diersfromtheprevioussituation onlyinthatthehorizontallineisreplacedbyanotherfunction g Theorem 6.1 Let f and g betwofunctionssuchthat,forallreal numbers x [ a,b ] ,theinequality f ( x ) g ( x ) holds.Thenthedomain whosebordersarethegraphof f ,thegraphof g ,andtheverticallines x = a and x = b hasarea A = b a( f ( x ) Š g ( x )) dx. SeeFigure6.1foranillustrationofthecontentofTheorem6.1. Thereaderisinvitedtoexplainwhythistheoremisadirectconsequenceofthefactthatwerecalledinthe“rstparagraphofthissection. Thereaderisalsoinvitedtoexplainwhythetheoremholdsevenif f and g takenegativevalues. Example 6.1 Computethearea A ( D ) ofthedomain D whose bordersarethegraphofthefunction f ( x )= x3+1 ,thefunction g ( x )= x2+2 ,andtheverticallines x =2 and x =3 .SeeFigure6.2foran illustrationofthisspeci“cexample. Solution: InordertoseethatTheorem6.1isapplicable,wemust “rstshowthat,forall x [2 3],theinequality f ( x ) g ( x )holds.1

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26.APPLICATIONSOFINTEGRATION Figure6.1. Areaenclosedby f ( x )and g ( x )between x = a and x = b Figure6.2. Areaenclosedbythegraphsof f ( x )= x3+ 1and g ( x )= x2+2between x =2and x =3. Thisisnotdicult,sinceweonlyneedtoshowthatif x [2 3],then f ( x ) g ( x ),thatis, x3+1 x2+2 x3Š x2 1 x2( x Š 1) 1 andthisisclearlytruesince x 2,so x2 4,and x Š 1 1,forcing x2( x Š 1) 4.

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36.THEAREABETWEENCURVES3 Therefore,Theorem6.1applies,andwehave A ( D )= 3 2( x3+1) Š ( x2+2) dx = 3 2( x3Š x2Š 1) dx = x4 4 Š x3 3 Š x 3 2=8 11 12 36.2.IntersectingCurves.Sometimes,thepoints a and b aredeterminedbythecurvesthemselves,andnotgiveninadvance.Inthat case,wehavetocomputethembeforewecanapplyTheorem6.1. Example 6.2 Findthearea A ( D ) ofthedomain D whoseborders arethegraphsofthefunctions f ( x )= x2+3 x +5 and g ( x )=2 x2+ 7 x +8 .SeeFigure6.3foranillustration. Solution: Letus“ndthepointsinwhichthegraphsof f and g intersect.Inthesepoints,wehave x2+3 x +5=2 x2+7 x +8 0= x2+4 x +3 0=( x +3)( x +1) Thatis,thetwocurvesintersectintwopoints,andthesepointshave horizontalcoordinates a = Š 3and b = Š 1.Furthermore,if x Figure6.3. Areaenclosedbythegraphsof f ( x )and g ( x )between x = Š 3and x = Š 1.

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46.APPLICATIONSOFINTEGRATION [ Š 3 Š 1],thatis,if x isbetweenthosetwointersectionpoints,then f ( x ) Š g ( x )=( x2+3 x +5) Š (2 x2+7 x +8) = Š ( x2+4 x +3) = Š ( x +3)( x +1) 0 since x +3 0and x +1 0.Therefore,if x [ Š 3 Š 1],then f ( x ) g ( x ),andTheorem6.1applies.Sowehave A ( D )= Š 1 Š 3( f ( x ) Š g ( x )) dx = Š ( x2+4 x +3) dx = Š x3 3 Š 2 x2Š 3 x Š 1 Š 3=1 1 3 Thesituationbecomesslightlymorecomplicatedif f g doesnot holdthroughouttheentireinterval[ a,b ].Forinstance,itcouldhappen that f ( x ) g ( x )atthebeginningoftheinterval[ a,b ],andthen,froma givenpointon, g ( x ) f ( x ).Inthatcase,wesplit[ a,b ]upintosmaller intervalssothatoneachofthesesmallerintervals,either f ( x ) g ( x ) or g ( x ) f ( x )holds.ThenwecanapplyTheorem6.1toeachof theseintervals.Asonsomeoftheseintervals f ( x ) g ( x )holds,while onsomeothers g ( x ) Š f ( x )holds,theapplicationofTheorem6.1will sometimesinvolvethecomputationof ( f ( x ) Š g ( x )) dx andsometimes ( g ( x ) Š f ( x )) dx.Thefollowingtheoremformalizesthisidea. Theorem 6.2 Let f and g betwofunctions.Thentheareaofthe domainwhosebordersarethegraphof f ,thegraphof g ,thevertical line x = a andtheverticalline x = b isequalto b a| f ( x ) Š g ( x ) | dx. NotethatTheorem6.1isaspecialcaseofTheorem6.2,namely, thespecialcasewhen f ( x ) Š g ( x )= | f ( x ) Š g ( x ) | forall x [ a,b ]. Example 6.3 Let f ( x )= x3+3 x2+2 x andlet g ( x )= x3+ x2. Computethearea A ( D ) ofthedomainwhosebordersarethegraphsof f and g andtheverticallines x = Š 2 and x =1 .SeeFigure6.4for anillustration.

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36.THEAREABETWEENCURVES5 Figure6.4. Graphsof f ( x )and g ( x )on[ Š 2 1]. Solution: InordertouseTheorem6.2,we“rstneedtocompute | f ( x ) Š g ( x ) | .Wehave f ( x ) Š g ( x )=( x3+3 x2+2 x ) Š ( x3+ x2) 2 x2+2 x =0 2 x ( x +1)=0 Thatis,thereareonlytwopointswherethesetwocurvesintersect, namely,at x = Š 1and x =0.If x Š 1orif x 0,then f ( x ) Š g ( x )= 2 x ( x +1) > 0,so | f ( x ) Š g ( x ) | = f ( x ) Š g ( x )=2 x2+2 x .If Š 1
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66.APPLICATIONSOFINTEGRATION Figure6.5. Graphof | f ( x ) Š g ( x ) | on[ Š 2 1]. 36.3.CurvesFailingtheVerticalLineTest.Sometimeswewanttocomputetheareabetweentwocurvesthatdonotpasstheverticalline test;thatis,theycontaintwoormorepointsonthesameverticalline. Suchcurvesare not graphsoffunctionsofthevariable x .Iftheypass the horizontallinetest ,thatis,iftheydonotcontaintwopointson thesamehorizontalline,thentheycanbeviewedasfunctionsof y Wecanthenchangetherolesof x and y inTheorems6.1and6.2and proceedasintheearlierexamplesofthissection. Example 6.4 Computethearea A ( D ) ofthedomainbetweenthe verticalline x =4 andthecurvegivenbytheequation y2= x .See Figure6.6foranillustration. Solution: Neithercurvesatis“estheverticallinetest,butbothsatisfy thehorizontallinetest.Therefore,weset f ( y )=4and g ( y )= y2.Itis Figure6.6. Graphof y2= x and x =4.

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36.THEAREABETWEENCURVES7 clearthatthetwocurvesintersectatthepointsgivenby y = Š 2and y =2.Betweenthesetwocurves,thevalueof f ( y )islarger.Therefore, Theorem6.1applies(withtherolesof x and y reversed).Sowehave A ( D )= 2 Š 2( f ( x ) Š g ( x )) dx = 2 Š 24 Š y2dy = 4 y Š y3 3 2 Š 2=16 Š 16 3 =10 2 3 Notethatthegeometricmeaningofreversingtherolesof x and y issimplyre”ectingallcurvesthroughthe x = y line.Thatre”ection doesnotchangetheareaofanydomain,soonecanexpectanalogous methodsofcomputingareasbeforeandafterthatre”ection.36.4.Exercises.(1)Findtheareaofthedomainwhosebordersaretheverticalline x =0,theverticalline x =2,andthegraphsofthefunctions f ( x )= x2+3and g ( x )=sin x (2)Findtheareaofthedomainwhosebordersaretheverticalline x =2,theverticalline x =4,andthegraphsofthefunctions f ( x )= x2+3and g ( x )= x (3)Findtheareaofthedomainbetweenthehorizontalaxisand thegraphofthefunction f ( x )= x2+6 x (4a)Findtheareaofthedomainbetweenthehorizontalaxisand thegraphofthefunction f ( x )=18 Š 2 x2. (4b)Findtheareaofthedomainbetweenthehorizontalaxisand thegraphofthefunction f ( x )=16 Š x4. (5)Findtheareaofthedomainwhosebordersaretheverticalline x =1,theverticalline x =3,andthegraphsofthefunctions f ( x )= x3and g ( x )= eŠ x. (6)Findtheareaofthedomainwhosebordersarethevertical line x = Š 1,theverticalline x =1,andthegraphsofthe functions f ( x )= x3and g ( x )= Š 2.

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86.APPLICATIONSOFINTEGRATION (7)Findtheareaofthedomainbetweenthegraphsofthefunctions f ( x )= exand g ( x )= x +1andtheverticallines x =0 and x =1.Howdoweknowthatbetweenthosetwovertical lines,thegraphof f isabovethegraphof g ? (8)Findtheareaofthedomainbetweenthegraphsofthefunctions f ( x )= x2+2and g ( x )=4 x Š 1. (9)Findtheareaofthedomainbetweenthegraphsofthefunctions f ( x )= x3Š 3 x2and g ( x )= x2. (10)Findtheareaofthedomainbetweenthegraphsofthefunctions f ( x )= x2and g ( x )= x1 / 5. (11)Findtheareaofthedomainbetweenthegraphsofthefunctions f ( x )= x3and g ( x )= x4. (12)Findtheareaofthedomainbetweenthegraphsofthefunctions f ( x )= x and g ( x )= x (13)Computetheareabetweenthecurvesgivenbytheequations x =5 y and x = y2+6. (14)Computetheareabetweenthecurves y = x Š 2and y = ( x Š 2)3. (15)Computetheareabetweenthe y axisandthecurve x = y2Š 7 y +6. (16)Computetheareabetweenthe y axisandthecurve x = y3Š 11 y2+6 y Š 6. (17)Computetheareabetweenthecurves y = x2and y =32 Š x2. (18)Computetheareabetweenthe three curves f ( x )= x g ( x )= Š x ,and h ( x )=4. (19)Computetheareabetweenthecurves y = | x | and y = x2Š 12. (20)Computetheareabetweenthecurves y = | x | and y =20 Š x2. 37.Volumes37.1.ExtendingtheDenitionofVolumes.Ifasolid S canbebuiltup usingunitcubes,thenwecansimplysaythatthevolume V ( S )of S is thenumberofunitcubesusedtobuild S .However,ifthebordersof S arenotplanes,thenthismethodwillhavetobemodi“ed.Aballor aconeisanexampleofthis. Sowewouldliketode“nethenotionofvolumesothatitisapplicabletoalargeclassofsolids,notjusttothosesolidsthatarebordered byplanes.Thisde“nitionshouldagreewithourintuition.Itshould alsobeinaccordancewiththefactthatwecan approximate allsolids withacollectionofverysmallcubes;therefore, V ( S )mustbecloseto thenumberofunitcubesusedintheapproximation.

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37.VOLUMES9 Withthesegoalsinmind,werecallthatwealreadyde“nedthe area ofadomainintheplanewhosebordersarethegraphsofcontinuous functions.Buildingonthatde“nition,wesaythatthe volume ofa prism isitsbaseareatimesitsheight.Moreformally,let S beasolid whosebaseandcoverareidenticalcopiesoftheplate P ,locatedat distance h fromeachother,ontwoparallelplanesthatareatdistance h fromeachother.Thenwede“nethevolumeof S tobe V ( S )= A ( P ) h, where A ( P )istheareaof P .SeeFigure6.7foranillustration.In particular,thevolumeofacylinderwhosebaseisacircleofradius r andwhoseheightis h is r2h Nowlet S beanysolidlocatedbetweentheplanesgivenbythe equations x = a and x = b .Inordertode“neandcomputethevolume V ( S )of S ,wecut S into n partsbytheplanes x = xifor i =0 1 ,...,n where a = x0
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106.APPLICATIONSOFINTEGRATION Figure6.8. Thevolumeoftheball B approximatedbycylinders. If n islarge,thentheunionoftheprisms Piapproximates S well, so V ( S )shouldbede“nedinawaythatassuresthat V ( S )iscloseto (6.1)ni =1V ( Pi)=ni =1A ( Ti) x. As n goestoin“nity,theRiemannsumontheright-handsideof (6.1)hasalimit.Wede“ne thatlimit tobethevolume V ( S )of S ,so V ( S )=limn ni =1A ( Ti) x. Ontheotherhand,bythede“nitionofthede“niteintegral,wehave limn ni =1A ( Ti) x = b aA ( t ) dt, where A ( t )istheareaoftheintersectionof S andtheplane x = t Therefore,thisintegralisequaltothevolume V ( S )ofthesolid S Thatprovesthefollowingtheorem,whichwillbeourmaintoolinthis section. Theorem 6.3 Let S beasolidlocatedbetweentheplanes x = a and x = b ,andlet A ( t ) betheareaoftheintersectionof S andthe plane x = t .Thenthevolume V ( S ) of S satis“estheequation V ( S )= b aA ( t ) dt. Example 6.5 Computethevolumeoftheball B whosecenteris attheoriginandwhoseradiusis r Solution: Foranygiven t [ Š r,r ],theintersectionoftheplane x = t andtheball B isacircle.Let Ctdenotethiscircle.Bythe

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37.VOLUMES11 Pythagoreantheorem, Cthasradius r2Š t2,andtherefore,thearea of Ctis A ( t )=( r2Š t2) .SeeFigure6.8foranillustration. NowwecanuseTheorem6.3tocomputethevolumeof B .Weget V ( B )= r Š r r2Š t2 dt = r2t Š 1 3 t3 r Š r=2 r3 Š 2 3 r3 = 4 3 r3. Thereisnothingmagicalaboutthe x axisasfarasTheorem6.3is concerned.Theargumentthatyieldedthattheoremcanberepeated forthe y axisinsteadofthe x axis,yieldingthefollowingtheorem. Theorem 6.4 Let S beasolidlocatedbetweentheplanes y = a and y = b ,andlet B ( t ) betheareaoftheintersectionof S andthe plane y = t .Thenthevolume V ( S ) of S satis“estheequation V ( S )= b aB ( t ) dt. Example 6.6 Let S betherightcircularconewhosesymmetry axisisthe y axis,whoseapexisat y = h ,andwhosebaseisacircle intheplane y =0 withitscenterattheoriginandwithradius r .Find thevolumeof S Solution: Thecone S isbetweentheplanes y =0and y = h ,and B ( t )ofTheorem6.4iseasiertocomputethan A ( t )ofTheorem6.3,so weusetheformer. Theintersectionoftheplane y = t and S isacircle.Theradius rtofthiscircle,bysimilartriangles,satis“es rt r = h Š t h ,

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126.APPLICATIONSOFINTEGRATION showingthat rt= r ( h Š t ) /h .Therefore, B ( t )= r2( h Š t )2/h2,and Theorem6.4implies V ( S )= h 0( r2( h Š t )2/h2) dt = r2 h2h 0( h2Š 2 ht + t2) dt = r2 h2 h2t Š ht2+ t3 3 h 0= r2 h2 h3 3 = 1 3 hr2. SeeFigure6.9foranillustration. 37.2.AnnularRings.Intheexamplesthatwehavesolvedsofar,the computationof A ( t )or B ( t ),thatis,thecomputationoftheareaof theintersectionbetweenasolidandahorizontalorverticalplane,was notdicult.Thatcomputationcouldbedonedirectly. Therearesituationsinwhichthedomainswhoseareasweneedto computearenot convex ;thatis,visuallyspeaking,thereisaholein them.Thishappensparticularlyoftenwhen S isobtainedbyrotating adomain D aroundaline. Figure6.9. Rightcircularcone.

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37.VOLUMES13 Example 6.7 Let D bethedomainbetweenthetwocurves y = x2= f ( x ) and y =2 x = g ( x ) andlet S bethesolidobtainedby rotating D abouttheline x = Š 3 .Findthevolumeof S Solution: Thetwocurvesintersectatthepoints(0 0)and(2 4).The intersectionofthehorizontalplane y = t with S hastheformofan annularring ,whichissometimesinformallycalleda washer .Thisis simplyasmallercirclecutoutothemiddleofalargercircle,sothat thetwocirclesareconcentric.Ifthelargercirclehasradius r1andthe smallercirclehasradius r2,thentheannularringhasarea ( r2 1Š r2 2). Thisgeneralrecipeenablesustocompute B ( t )intheexampleat hand.Thepointsin D satisfy x [0 2]and y [0 4].As0 x 2, theinequality x2 2 x holds.So,for t [0 4],thepoint Pi=( t/ 2 ,t ) onthegraphof g ( x )= y =2 x isclosertothe y axisthanthepoint Po=( t,t )onthegraphof y = x2= f ( x ).(Ittakesalargervalue of x togetthesamevalue t = y by f thanittakestogetthesame valueby g .)Sotheoutercircleoftheannuluswillbegivenbythe rotatedimageofthecurveof f (theparabola),andtheinnercircleof theannuluswillbegivenbytherotatedimageofthecurveof g (the straightline). Inparticular,for“xed t ,theradiiareobtainedasthedistanceof Po(resp. Pi)fromtheaxisofrotation,thatis,theline x = Š 3.For theinnerradius,thisyields r2= Š 3 Š t 2 = t +6 2 whilefortheouterradius,thisyields r1= Š 3 Š t = t +3 Therefore,theareaoftheannularringthatistheintersectionof S andtheplane y = t isgivenby B ( t )= ( r2 1Š r2 2)= t +3 2Š t +6 2 2= 6 t0 5Š t2 4 Š 2 t NowwecanapplyTheorem6.4tocompute V ( S ).Weobtain V ( S )= 4 0B ( t ) dt = 4 06 t0 5Š t2 4 Š 2 tdt = 4 t1 5Š t3 12 Š t2 4 0 77 01

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146.APPLICATIONSOFINTEGRATION 37.3.SpecialCasesofTheorems6.3and6.4.Notethatthesolidswe discussedsofarinthissectioncouldbeobtainedbyrotatingthegraph ofafunctionaroundanaxis.Suchvolumesarecalled volumesofrevolution .Indeed,theballofExample6.5canbeobtainedbyrotating thegraphofthefunction f ( x )= r2Š x2(asemicircle)aboutthe x axis.TheconeofExample6.6canbeobtainedbyrotatingthegraph ofthefunction f ( x )= Š xh r+ r = y (astraightline)aboutthe y axis. Forsuchsolids,theareas A ( t )and B ( t )appearinginTheorems6.3 and6.4areeasytocompute,sincetheintersectionsappearinginthose theoremswillbe circles or annularrings .If S isasolidobtainedby rotatingacurveaboutthe x axis,thentheintersectionoftheplane x = t and S isacircleofradius f ( x ),andhence A ( t )= f ( x )2 .If S is asolidobtainedbyrotatingthecurveofthefunction g ( y )= x about the y axis,thentheintersectionof S andtheplane y = t isacircle ofradius g ( t ),andso B ( t )= g ( t )2 .Thisyieldsthefollowingspecial versionsofTheorems6.3and6.4. Theorem 6.5 Let S beasolidbetweentheplanes x = a and x = b obtainedbyrotatingthegraphofthefunction f ( x )= y aboutthe x axis.Thenwehave V ( S )= b af ( t )2dt. Theorem 6.6 Let S beasolidbetweentheplanes y = a and y = b obtainedbyrotatingthegraphofthefunction g ( y )= x aboutthe y axis. Thenwehave V ( S )= b ag ( t )2dt. Theexercisesattheendofthissectionwillprovidefurtherexamples fortheusesofthesetheorems. Ifthedomaintoberotateddoesnotincludethe entire areabetween thecurveandandcoordinateaxis(forinstance,becauseitisadomain between twocurves),thenwegetannularrings,whichwediscussedin thelastsection.37.4.ASolidNotObtainedbyRevolution.Whilevolumesofrevolution areaveryfrequentapplicationofTheorems6.3and6.4,theyarenot theonlyapplicationsofthosetheorems. Example 6.8 Let S beapyramidwhosebaseisasquareofside length a andwhoseheightis h .Computethevolumeof S Solution: The“rststepistoplace S inacoordinatesystemsothat Theorem6.3canbeapplied.Letusplacetheaxisof S onthe x axisof

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37.VOLUMES15 thecoordinatesystem,sothatthecenterofthebaseof S isattheorigin andthecuspof S isat x = h .Thisdoesnotcompletelydeterminethe positionof S ,because S couldstillrotatearoundthe x axis.However, suchrotationswillnotchangethevalueof A ( t )forany t [0 ,h ],and sotheyareinsigni“cantforthecomputationof V ( S ). Nownotethat,forany t [0 ,h ],theintersectionof S andtheplane x = t isasquareofsidelength a ( h Š t ) /h (bysimilartriangles).See Figure6.10foranillustration.So A ( t )= a2( h Š t )2/h2,andTheorem 6.3implies V ( S )= h 0a2( h Š t )2 h2dt = a2 h2 h2t Š ht2+ t3 3 h 0= a2h 3 37.5.TheBigPicture.Notethatthetheoremspresentedinthissectiononintegralscon“rmourintuitionthatifacertainfunctionmeasuresaquantity,then,undertheappropriateconditions,theintegral ofthatfunctionmeasuresaquantitythatissomehowinaspacethat isonedimensionhigher.Forinstance,wesawearlierthatif f ( x )measuredtheheightofacurveatagivenhorizontalcoordinate x ,then, Figure6.10. Pyramid.

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166.APPLICATIONSOFINTEGRATION undertheappropriateconditions, f ( x ) dx measuredthe area under thecurve.Sotakingintegralsmeantmovingfromonedimensionto two.Inthissection,thefunctions A ( t )and B ( t )measuredareas ofdomainsinagivenplane,while A ( t ) dt and B ( t ) dt measured volumes .Sotakingintegralsmeantmovingfromtwodimensionsto three.37.6.Exercises.(1)Let T bethetrianglewhosebordersarethecoordinateaxes andtheline y +4 x =4.Rotate T aboutthe y axisand computethevolumeoftheobtainedsolid. (2)Computethevolumeofthesolidbetweentheplanes x = Š 1 and x =1obtainedbyrotatingthegraphofthefunction f ( x )= x2= y aboutthe x axis. (3)Computethevolumeofthesolidbetweentheplanes x = Š 2 and x =2obtainedbyrotatingthegraphofthefunction f ( x )= x4= y aboutthe x axis. (4)Computethevolumeofthesolidbetweentheplanes x = Š 1 and x =1obtainedbyrotatingthegraphofthefunction f ( x )= x3+2 x = y aboutthe x axis. (5)Computethevolumeofthesolidbetweentheplanes x = Š 3 and x =3obtainedbyrotatingthegraphofthefunction f ( x )=2 x2+5 x = y aboutthe x axis. (6)Computethevolumeofthesolidbetweentheplanes x =0 and x = obtainedbyrotatingthegraphofthefunction f ( x )=sin x = y aboutthe x axis. (7)Computethevolumeofthesolidbetweentheplanes x =0 and x = / 4obtainedbyrotatingthegraphofthefunction f ( x )=tan x = y aboutthe x axis. (8)Computethevolumeofthesolidobtainedbyrotatingthe domainbetweenthecurves y = x and y = x abouttheline y = Š 2. (9)Computethevolumeofthesolidobtainedbyrotatingthe domainbetweenthecurves y = x4and y = x abouttheline x = Š 1. (10)Computethevolumeofthesolidobtainedbyrotatingabout the x axisthedomainbetweenthe x axisandthecurve y = e2 x, betweentheplanes x =2and x =4. (11)Computethevolumeofthesolidobtainedbyrotatingabout the x axisthedomainbetweenthe x axisandthecurve y =3 x ,betweentheplanes x =0and x =6.

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38.CYLINDRICALSHELLS17 (12)Computethevolumeofa regulartetrahedron ofsidelength z Aregulartetrahedronisasolidthathasfourfaces,eachof whichisaregulartriangle. (13)Let P beapyramidwhosebaseisatriangleofarea A and whoseheightis h .Findandproveaformulaforthevolumeof P (14)Let P beapyramidwhosebaseisarectangleofarea A and whoseheightis h .Findandproveaformulaforthevolumeof P (15)Computethevolumeofthesolidbetweentheplanes x =0and x =1obtainedbyrotatingthecurve f ( x )= ex= y aboutthe y axis. (16)Computethevolumeofthesolidbetweentheplanes x =0 and x =1obtainedbyrotatingthecurveofthefunction f ( x )= x (1 Š x )aboutthe x axis. (17)Computethevolumeofthesolidobtainedbyrotatingthe domainbetweenthecurves y = x3and y = x4aboutthe y axis. (18)Let D bethedomainthatliesabovethe x axisandonthe rightofthe y axisandisborderedbythecurve y =9 Š x2. Rotate D abouttheline x = Š 1.Computethevolumeofthe obtainedsolid. (19)Let D bede“nedasintheprecedingexercise.Rotate D about theline y = Š 1.Computethevolumeoftheobtainedsolid. (20)UseTheorem6.6todeducetheformulaforthevolumeofa sphereofradius r 38.CylindricalShells38.1.AnAlternativeMethodtoComputeVolumes.Inprinciple,Theorems6.3and6.4aresimplemethodstocomputethevolumesofsolids. Inpractice,however,theareas A ( t )and B ( t )thatappearinthesetheoremsmaybediculttoexplicitlyevaluate.Onesituationinwhich theseareasareoftendiculttocomputeiswhenthesolidinquestion isobtainedbyrotatingadomainaroundsomeline;thatis,itisasolid ofrevolution. Asanexample,letustrytocomputethevolumeofthesolid S obtainedbyrotatingaboutthe y axisthedomainborderedbythelines x =0, x =3,and y =0andthegraphofthefunction f ( x )=3 x3Š x4. IfwetrytosolvethisproblemusingTheorems6.3or6.4,weruninto diculties,because A ( t )and B ( t )willnotbeeasytocompute.For instance,ifwewantedtouseTheorem6.4,then,inordertocompute

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186.APPLICATIONSOFINTEGRATION Figure6.11. Singlecylindricalshell. B ( t ),wewouldneedtodescribetheintersectionof S andtheplane y = t .Forthis,wewouldhaveto“ndthe x coordinatesofthepoints ofthatintersection;thatis,wewouldneedto“ndallrealnumbers x [0 3]forwhich y =3 x3Š x4= t .Thisisafourth-degreeequation for x ,whichisverydicultandcumbersometosolve.Ifwewanted touseTheorem6.3,then,inordertocompute A ( t ),wewouldneed todescribetheintersectionoftheplane x = t and S ,whichisnot straightforwardtodo. Insituationslikethis,thatis,whentheapplicationofTheorems 6.3and6.4leadstotechnicaldiculties,itoftenhelpstouseanother methodcalledthemethodof cylindricalshells .Acylindricalshell C issimplyacylinder C1ofwhichasmallercylinder C2isremoved,so that C1and C2havethesamesymmetryaxis.SeeFigure6.11foran illustration.If C2isjustalittlebitsmallerthan C1,then C lookslike ashell,explainingthename cylindricalshell If C1and C2bothhaveheight h and Cihasradius ri,thenthe volumeof C canbecomputedas V ( C )= V ( C1) Š V ( C2) = hr2 1 Š hr2 2 = h ( r2 1Š r2 2) .

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38.CYLINDRICALSHELLS19 Notethatthelastformof V ( C )canberearrangedas (6.2) V ( C )= h ( r1Š r2) 2 ( r1+ r2) 2 Thiswayofwriting V ( C )mightseemcontrivedat“rstsight.However,ithasthefollowingmotivation.Notethat r1Š r2isthewidthŽ of C ,while h isitsheight.Finally,ifwe”atten C outintheplane, itwillbecomeabrickwithsidelengths h r1Š r2,and,2 r1+ r2 2,since thelengthofthemissingsideisequaltothecircumferenceofacircle whoseradiusistheaverageoftheradiiof C1and C2. Inotherwords,(6.2)saysthatthevolumeofacylindricalshellis equaltotheproductofitsheight,width,andlengthŽ(ifthelatteris interpretedproperly). Wearenowinapositiontousecylindricalshellstocomputevolumes.Let S beasolidthatisobtainedbyrotatingthedomain D whichliesbelowthecurveof f ( x )= y from x = a to x = b ,aboutthe y axis.Inordertoestimate V ( S ),wecut[ a,b ]into n intervalsofequal lengthusingpoints a = x0
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206.APPLICATIONSOFINTEGRATION Figure6.12. (a)Thecurveof y =3 x3Š x4and(b)the solidobtainedbyitsrotation. Theorem 6.7 Thevolume V ( S ) ofthesolidobtainedbyrotating thedomain D whosebordersarethecurveof f ( x )= y ,thelines x = a and x = b ,andthehorizontalaxis y =0 aboutthe y axisisequalto V ( S )= b a2 xf ( x ) dx. Example 6.9 Computethevolumeofthesolid S obtainedbyrotatingaboutthe y axisthedomainborderedbythelines x =0 x =3 and y =0 andthegraphofthefunction f ( x )=3 x3Š x4. Solution: ByTheorem6.7,wehave V ( S )=2 3 0x (3 x3Š x4) dx =2 3 0(3 x4Š x5) dx =2 3 x5 5 Š x6 6 3 0=2 729 5 Š 729 6 =152 677 Theaxisaroundwhichwerotateadomaindoesnothavetobe acoordinateaxisinorderforthemethodofcylindricalshellstobe applicable.Wecanapplythemethodaslongaswecandecomposethe solidinquestionintocylindricalshellswhoseheightandradiuswecan compute.

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38.CYLINDRICALSHELLS21 Figure6.13. (a)Thecurveof y = x2/ 2and(b)the solidobtainedbyitsrotation. Example 6.10 Let S bethesolidobtainedbyrotatingthedomain whosebordersarethehorizontalaxis,theverticallines x =0 and x =2 ,andthegraphofthefunction f ( x )=2 x Š x2aboutthevertical line x =3 .Compute V ( S ) Solution: Wecandecompose S intocylindricalshellswhosecenter isontheverticalline x =3.Theshellcontainingthepoint x of thehorizontalaxiswillhaveheight f ( x )=2 x Š x2andradius3 Š x Therefore,wehave V ( S )= 2 02 (2 x Š x2)(3 Š x ) dx =2 2 0 x3Š 5 x2+6 x dx =2 x4 4 Š 5 x3 3 +3 x2 2 0=2 8 3 16 755 38.2.Exercises.In(1)…(8),usethemethodpresentedinthissectiontocomputethe volumeofthesolidobtainedbyrotatingthedomainbetweenthegiven curvesaboutthe y axis.

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226.APPLICATIONSOFINTEGRATION (1) y = f ( x )= x3, x =1, y =0. (2) y = f ( x )=1 x2, x =2, x =3, y =0. (3) y = f ( x )= x g ( x )= Š x x =2. (4) y = f ( x )=3 x g ( x )= x3, y =0, y =1. (5) y = f ( x )= x4, x =0, y =1. (6) y = f ( x )=1 Š x2, x + y =1. (7) y = f ( x )= eŠ x, x =0, y =0, y =1. (8) y = f ( x )= e2 x, x =0, x =1, y = e2. In(9)…(16),usethebestavailablemethodtocomputethevolumeof thesolidobtainedbyrotatingthedomainbetweenthegivencurves aboutthegivenaxis. (9) y = f ( x )= x y = g ( x )= Š x y =2,about x = Š 1. (10) y = f ( x )= x y = g ( x )= Š x y =2,about x =3. (11) y = f ( x )= x y = g ( x )= Š x y =2,aboutthe x axis. (12) y = f ( x )=6 Š x y = g ( x )= x y =6,aboutthe y axis. (13) y = f ( x )=6 Šx y = g ( x )= x y =6,abouttheline y =6. (14) y = f ( x )=6 Š x y = g ( x )= x y =6,abouttheline y =7. (15) y = f ( x )= Š ( x2+4 x Š 21), y =0,aboutthe y axis. (16) y = f ( x )= x +(1 /x ), y =4,about x =1. (17)Describeasolidwhosevolumeisgivenby 1 02 x cos xdx (18)Describeasolidwhosevolumeisgivenby2 1 0 4 Š x x2+1dx (19)Useacomputersoftwarepackagetocomputethevolumeof thesolidobtainedbyrotatingtheregionbetweenthecurves y = eŠ xand y = eŠ x2aboutthe y axis. (20)Useacomputersoftwarepackagetocomputethevolumeof thesolidobtainedbyrotatingtheregionbetweenthecurves y =sin x and y =sin( x2)andthelines x =0and x =1about the y axis. 39.WorkandHydrostaticForce39.1.WorkMovingaPointlikeObject.Inphysics,theword work has amorespeci“cmeaningthanineverydaylife.Workinphysicsmeans thata force isextendedtomoveanobjectacertaindistance. Theforce F movinganobjectiscomputedbytheformula F = m d2s dt2, whichiscalled Newtonssecondlaw .Here m isthemassoftheobject, while a =d2s dt2isits acceleration .SoNewtonssecondlawsaysthatthe

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39.WORKANDHYDROSTATICFORCE23 massofanobjectisindirectproportiontotheforceneededtomove itatconstantacceleration. Ifaconstantforce F isexertedwhileanobjectmovesdistance d thentheworkdonebythatconstantforceiscomputedbytheformula W = Fd. Notethatinthemetricsystem,distanceismeasuredinmeters(m), timeismeasuredinseconds(s),andthereforeaccelerationismeasured inm / s2.Massismeasuredinkilograms(kg),soforceismeasuredin kg m s2,whicharecalled newtons (N),and,“nally,workismeasuredin N m,whicharealsocalled joules (J).Onejouleistheworkthatis donewhenaforceequalto1newtonmovesanobjectadistanceof1 meter. Example 6.11 Howmuchworkisneededtoliftachildof20kgto aheightof0.5meters?Usethefactthatgravitationcausesdownward accelerationof g =9 8m / s2. Solution: Inordertoliftthechild,oneneedstoovercomethedownwardaccelerationcausedbygravity.Thismeansthatanupwardforce of m g =20kg 9 8 m s2=198N hastobeexertedacrossadistanceof d =0 5meters.Thisyields W = Fd =198N 0 5m=99J Sotheworkneededis99J. Iftheforceexertedisnotconstantacrosstheentiredistance,but thedistancecanbesplitupintoafewpartssothattheforceisconstant oneachpart,thenwecancomputetheworkdonebytheforceoneach partjustasinthepreviousexample,andthenwecanaddtheobtained amountstogetthetotalamountofworkdoneacrosstheentiredistance. Iftheforceexertedchangesaccordingtoacontinuousfunction f ( d ), thenwecanapproximatetheworkdoneusingtheideaoftheprevious paragraphandthenuseintegrationtocomputethetotalworkdoneby theforceasfollows. Let a and b berealnumbersandletusassumethatanobjectis movingfrom a to b ,andtheforcemovingtheobjectatagivenpoint x isequalto f ( x ),where f isacontinuousfunction.Inordertocompute theworkdoneacrosstheentiredistance,letussplittheinterval[ a,b ] into n equalintervals,usingpoints a = x0
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246.APPLICATIONSOFINTEGRATION andset x =( a Š b ) /n .Thentheworkdonebytheforceontheinterval Ii=[ xi Š 1,xi]isabout f ( x i) x ,where x iissomesamplepointin Ii. Indeed, f iscontinuous,soif n islargeandtherefore Iiisshort,then f doesnotchangemuchonthatinterval,sotheshapeofthedomain underthecurveof f andabove Iiisroughlyarectangle.Thismeans thatthetotalworkdonebytheforceon[ a,b ]iscloseto (6.4)ni =1f ( x i) x. As n getslarger,thisapproximationgetsbetter,andsowede“nethe totalworkdonebytheforceacrosstheinterval[ a,b ]asthe limit ofthe sumin(6.4)as n goestoin“nity.Ontheotherhand,thatsumisa Riemannsum,soitslimit,as n ,isthede“niteintegral b af ( x ) dx Inotherwords,wehaveprovedthefollowingtheorem. Theorem 6.8 Let a and b berealnumbers.Ifanobjectismoved from a to b byaforcethatisequalto f ( x ) atpoint x ,where f is continuouson [ a,b ] ,thenthetotalworkdonebytheforceon [ a,b ] is W = b af ( x ) dx. Example 6.12 Theforceneededtoextendagivenspring x centimetersoveritsnaturallengthisgivenbythefunction f ( x )=70 x Howmuchworkisneededtoextendthespring10cmoveritsnatural length? Solution: ByTheorem6.8,wehave W = 0 1 070 xdx =35 x2 0 1 0=0 35J So12.25Jofworkisneededtostretchthespring10cmoveritsnatural length. Wepointoutthatthelawofphysicsthatsaysthattheforceneeded toextendaspringby x unitsoveritsnaturallengthisequalto kx is called Hookeslaw ,and k iscalledthe springconstant .39.2.HydrostaticForce.Letussaythatwewanttopumpwaterout ofatankthathastheshapeofthesouthernhalfofaballofradius 1(m).Howmuchworkisneededtodothat?Thisquestionismore

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39.WORKANDHYDROSTATICFORCE25 i i * Figure6.14. Thetankinacoordinatesystem,andits layeratheight z complexthanthepreviousonesincedeeperlayersofthehemisphere aresmaller,andwaterinthoselayershastotravelfartherinorderto reachthetopoftank. Therefore,wewillcutthetankupintosmalllayersandestimate theamountofworkneededtopumpouteachlayerofwater. Let x =0denotethebottomofthetankandlet x =1denote thecenterofthetopcircleofthetank.Cutthetankinto i horizontal layersbyplanesthatareatheights 0= x0
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266.APPLICATIONSOFINTEGRATION As n grows,theexpressiondisplayedin(6.5)approximatesthe neededworkbetterandbetter.Wede“nethetotalworkneeded(to pumpallthewateroutofthetank)tobethelimitofthesumshown in(6.5)as n goestoin“nity.AsthatsumisaRiemannsum,itslimit, as n goestoin“nity,isthede“niteintegral W = g 1 0(2 x Š x2)(1 Š x ) dx. Astheintegrandisapolynomial,thisintegralisveryeasytocompute. Wegetthat W = g 1 0( x3Š 3 x2+2 x ) dx = g x4 4 Š x3+ x2 1 0= g/ 4 =7696 675J Soittakesalmost7700Jofworktopumpoutallthewaterfromthe tank.39.3.Exercises.(1)Howmuchworkisdonewhenabookofmass2kgislifted1.5 metersfromitsoriginallocation? (2)Howmuchworkisdonewhenasuitcaseofmass15kgislifted 0.5metersfromitsoriginallocation? (3)Asuitcaseofmass20kgislifted0.6metersandputonachair. Someobjectsareremoved,andthenthesuitcase,whichnow hasamassof12kg,isliftedanadditional0.4meters.How muchworkwasdoneduringthetotalascentofthesuitcase? (4)Ifittakes10Jofworktoliftanobject2meters,howmuch workdoesittaketoliftthatobjectanadditional3meters? (5)Anelevatorandthepassengersenteringitonthe“rst”oor haveatotalmassof2000kg.Thedistancebetweentwoadjacent”oorsservedbythiselevatoris6meters.Howmuchwork isdoneiftheelevator“rststopsonthethird”oor,wheretwo additionalpeopleofmass75kgeachenter,andthentheelevatorstopsonthesixth”oor? (6)Ifittakes20Jofworktostretchaspring20cmoveritsnatural length,howmuchworkdoesittaketostretchthatspringan additional 5cm?

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39.WORKANDHYDROSTATICFORCE27 (7)Ifittakes30Jofworktostretchaspring10cmoveritsnatural length,howmuchworkdoesittaketostretchthatspringan additional 2cm? (8)Letusassumethat8Jofworkisneededtostretchaspring from20cmto21cm,and10Jofadditionalworkisneeded tostretchthatsamespringfrom21cmto22cm.Whatisthe naturallengthofthespring? (9)Aropeofmass5kgandlength30mislyingontheground. Howmuchworkisneededtoliftoneendofthatropetoa heightof30meters? (10)Wehave n copiesofabook,withamassof2kgeach.How muchworkisneededtoarrangethesecopiesinatowerifeach copyis0.08metersthick? (11)Howmuchworkisneededtopumpoutallthewaterfroma tankoftheshapeofacubeofsidelength1?(Thelengthis measuredinmeters.) (12)Howmuchworkisneededtopumpoutallthewaterfroma tankoftheshapeofaninvertedconeofheight10whosetop circlehasradius2? (13)Howmuchworkisneededtopumpoutallthewaterfroma tankoftheshapeofaninvertedconeofheight25whosetop circlehasradius5? (14)Considerthetankofthepreviousexercise.Howmuchwork isneededifwewanttopumpout half thewaterfromthis tank?Youcanuseasoftwarepackagetosolvethehigh-degree polynomialequationencounteredwhilesolvingthisexercise. (15)Howmuchworkisneededtopumpoutallthewaterfroma tankoftheshapeofacylinderofheight20whosebasecircle hasradius30? (16)Howmuchworkisneededtopumpoutallthewaterfroma tankoftheshapeofaninvertedpyramidofheight15whose topplateisasquareofsidelength10? (17)Considerthetankofthepreviousexercise.Howmuchwork isneededifwewanttopumpout half thewaterfromthis tank?Youcanuseasoftwarepackagetosolvethehigh-degree polynomialequationencounteredwhilesolvingthisexercise. (18)Howmuchworkisneededtopumpoutallthewaterfroma tankoftheshapeofaregulartetrahedronofsidelength1? (19)Considerthetankofthepreviousexercise.Howmuchwork isneededifwewanttopumpout seven-eighths ofthewater fromthistank?

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286.APPLICATIONSOFINTEGRATION (20)ConsiderthetankdiscussedinSection39.2.Howmuchwork isneededifwewanttopumpout half thewaterfromthis tank?Youcanuseasoftwarepackagetosolvethehigh-degree polynomialequationencounteredwhilesolvingthisexercise. 40.AverageValueofaFunction Theconceptof average isasimpleoneaslongaswetaketheaverage ofa “nitenumber ofvalues,suchastheaveragepriceofahouseina givenneighborhoodortheaveragedailyhightemperatureinagiven cityinagivenmonth.If a1,a2,...,anarerealnumbers,then (6.6) A =( a1+ a2+ + an) /n istheiraverage.However,whatcanwesayabouttheaveragevalue ofafunctionoveragiveninterval[ a,b ]?Wewillclearlyneedanew de“nitionforthatsincetherearein“nitelymanyrealnumbersin[ a,b ], sosummingallofthemandthendividingtheirsumbythenumberof summandsisnotanoption. Hereisanintuitivewayofextendingthede“nitionofaverageto thevaluestakenbyafunctionoveraninterval.Itfollowsfrom(6.6) that A istheonlyrealnumberwiththepropertythatifwereplace each of a1,a2,...,anby A ,thenthesum( a1+ a2+ + an)doesnot change.Thisobservationsuggeststhefollowingde“nition. Definition 6.1 Let f beafunctionsuchthat b af ( x ) dx exists. Thentheaveragevalueof f ontheinterval [ a,b ] istherealnumber c = b af ( x ) dx b Š a Indeed, c istheonlyrealnumberwiththepropertythatifwereplace f bytheconstantfunction f ( x )= c ,thentheintegral b af ( x ) dx does notchange. Amoresystematicapproachisthefollowing.Aswesawwhenwe “rstlearnedaboutintegrals, b af ( x ) dx canbeapproximatedinthe followingway.Split[ a,b ]into n equalintervalsandchooseapoint xiinthe i thsuchinterval.Takearectangleofheight f ( xi)overthe i th interval.Theaveragevalueofthe n valuesof f takenatthepoints xiis,ofcourse, An= f ( x1)+ f ( x2)+ + f ( xn) n .

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40.AVERAGEVALUEOFAFUNCTION29 Figure6.15. Theaveragevalueofsin x on[0 ]. Ontheotherhand,thetotalareaofthe n rectangleswehavejust de“nedis Rn= b Š a n ( f ( x1)+ f ( x2)+ + f ( xn)) Comparingthelasttwodisplayedequations,weseethat (6.7) An= Rn b Š a If n goestoin“nity,thenthe n rectangleswillapproximatethedomain underthegraphof f ,andsotheright-handsideof(6.7)willconverge tob af ( x ) dx b Š a,whiletheleft-handsidewillapproximatetheaveragevalue of f on[ a,b ]. Example 6.13 Whatistheaveragevalue A of sin x ontheinterval [0 ] ? Solution: Wehave A = 0sin xdx = Š cos x 0 = 1 Š ( Š 1) = 2 SeeFigure6.15foranillustration.

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306.APPLICATIONSOFINTEGRATION Itisworthpointingoutthatacontinuousfunction f willactually takeitsaveragevalueoneachinterval.Thisisthecontentofthe followingtheorem. Theorem 6.9 [MeanValueTheorem]Let f beacontinuousfunctionon [ a,b ] andlet c betheaveragevalueof f on [ a,b ] .Thenthere existsarealnumber x [ a,b ] suchthat f ( x )= c Proof. Itsucestoshowthatif m istheminimumof f on[ a,b ] and M isthemaximumof f on[ a,b ],then m c M ,andourclaim followsfromtheintermediatevaluetheorem. Weknowthat m ( b Š a ) b af ( x ) dx M ( b Š a ) forobviousgeometricreasons.Nowdivideallthreetermsby b Š a to get m c M Example 6.14 Thereisarealnumber x [0 ] suchthat sin x = 2 / Solution: ThisfollowsfromthepreviousexampleandTheorem6.9. 40.1.Exercises.(1)Findtheaveragevalueof xnon[0 1]. (2)Comparetheresultsofthepreviousexercisefor n =1, n =2, and n =3.Explainthepatternthatyoudetect. (3)Findtheaveragevalueoftan x on[0 ,/ 4]. (4)Findtheaveragevalueofx x2+1on[0 2]. (5)Findtheaveragevalueof exon[0 1]. (6)Findtheaveragevalueof ex ex+1on[0 3]. (7)Findtheaveragevalueof1 / (1 Š x2)on[2 3]. (8)Findtheaveragevalueof x cos( x2)on[0 ]. (9)Findtheaveragevalueofcos x on[ Š / 2 ,/ 2].Trytodeduce youranswerfromtheresultofExample6.13withoutadditionalcomputation. (10)Whichislarger,theaveragevalueofsin x ortheaveragevalue ofcos x ,ifbothaveragesaretakenontheinterval[0 ,/ 2]? (11)Whichislarger,theaveragevalueofsin x on[0 ]orthe averagevalueofsin x on[ Š 14 17 ]?Canyou“ndananswer thatdoesnotinvolvecomputation?

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40.AVERAGEVALUEOFAFUNCTION31 (12)Provethatif f ( x ) g ( x )forall x [ a,b ],and f and g are continuousfunctionson[ a,b ],thentheaveragevalueof f on [ a,b ]isatleastaslargeastheaveragevalueof g on[ a,b ]. (13)Whichislarger,theaveragevalueof f ( x )= x2on[2 3]orthe averagevalueof g ( x )= xŠ 2on[1 / 3 1 / 2]? (14)Abicyclistcoveredacertaindistanceatanaveragevelocityof 25milesperhour.Provethattherewasamomentatwhichthe instantaneousvelocity ofthebicyclistwas25milesperhour. (15)Startingfromastandstill,acarneeded10secondstoreacha travelingvelocityof20meterspersecond.Provethatthereis amomentatwhichtheinstantaneousaccelerationofthecar is2m / s2. (16)Letusassumethat 4 0f ( x ) dx =10.Provethatthereisa number x [0 4]suchthat f (4)=2 5. (17)Let f beanoddfunction.Whatistherelationbetweentheaveragevalueof f on[ a,b ]andtheaveragevalueof f on[ Š a, Š b ]? (18)Let f beanoddfunction.Computetheaveragevalueof f for anyinterval[ Š R,R ]. (19)Let f beanevenfunction.Whatistherelationbetweenthe averagevalueof f on[ Š R,R ]andtheaveragevalueof f on [0 ,R ]? (20)Let a bea“xedrealnumberandlet f beacontinuousfunction. De“ne g ( x )astheaveragevalueof f ontheinterval[ a,x ]. Compute g ( x )for f ( x )=3, f ( x )=2 x +7,and f ( x )= x2+ 2 x +1.

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CHAPTER7 MethodsofIntegration 41.IntegrationbyParts41.1.MethodofIntegrationbyParts.Let u and v betwodierentiable functionsofthevariable x .Weusedthesimpleproductrule (7.1)( uv )= uv + uvtocomputethe derivative oftheproductofthesetwofunctions.Is thereasimilarruleforcomputingthe integral oftheproductoftwo functions?Ingeneral,theanswerisno.Thereisnorulethatprovides theintegraloftheproductoftwofunctionsthatwouldworkinevery case.However,therearemanycasesinwhicharelativelysimpleway ofreversingŽtheproductruleofdierentiationwillgiveustheanswer wearetryingtoobtain. Indeed,integratingbothsidesoftheproductrule(7.1)ofdierentiation(7.1)withrespectto x ,wegettheidentity u ( x ) v ( x )= ( u( x ) v ( x )) dx + ( u ( x ) v( x )) dx or,afterrearrangement, (7.2) ( u( x ) v ( x )) dx = u ( x ) v ( x ) Š ( u ( x ) v( x )) dx. Formula(7.2)isveryusefulifwewanttocomputetheintegralof theproductoftwofunctions,oneofwhichcanplaytheroleof uand theotheroneofwhichcanplaytheroleof v .Ifwecancompute u and ( uv),thenformula(7.2)enablesustocompute ( uv )aswell.If wecannotcarryoutoneorbothofthesecomputations,thenformula (7.2)willnothelp. Example 7.1 Compute xexdx Solution: Weset u( x )= exand v ( x )= x .Thenformula(7.2)iseasy toapply,since v ( x )= x and v( x )=1.Therefore,(7.2)impliesthat xexdx = ex x Š ex 1 dx = ex x Š ex+ C = ex( x Š 1)+ C. 33

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347.METHODSOFINTEGRATION Thereaderisencouragedtoverifythattheobtainedsolutionis correctbycomputingthederivativeof ex( x Š 1)andcheckingthatit isindeedequalto ex x Atthispoint,thereadermaybeaskinghowweknewthatwe neededtoset u( x )= exand v ( x )= x ,andnottheotherwayaround. Theansweristhattheotherdistributionofroles,thatis, u( x )= x and v ( x )= exwouldnothavehelped.Indeed,ifwehadchosen uand v inthatway,wewouldhaveneededtocompute ( uv) dx = ( exx2) / 2 dx .Thatwouldhavebeenmorecomplexthantheoriginal problem.Weshouldalwayschoose uand v sothat ( uv) dx iseasy tocompute.Thatusuallymeansselecting v sothatitbecomessimpler whendierentiated,andtoselect usothat udoesnotgetmuch morecomplexwhenintegrated(oratleastoneofthesetwodesirable outcomesoccur). Example 7.2 Compute x cos xdx Solution: Weset u( x )=cos x and v ( x )= x ,whichmeansthat u ( x )=sin x and v( x )=1.Soformula(7.2)implies x cos xdx = x sin x Š sin xdx = x sin x +cos x + C. Thetechniqueofintegrationwehavejustexplainediscalled integrationbyparts .41.2.AdvancedExamples.Sometimestheintegranddoesnotseemto beaproduct,butitcanbetransformedintoone.Thefollowingisa classicexample. Example 7.3 Compute ln xdx Solution: Thecrucialobservationisthatwritingln x =1 ln x helps. Let u( x )=1and v ( x )=ln x .Then u ( x )= x and v( x )=1 /x ,so, crucially, u ( x ) v( x )=1.Therefore,formula(7.2)yields ln xdx = 1 ln xdx = x ln x Š 1 dx = x ln x Š x + C. Sometimesintegrationbypartsleadstoanequationorasystemof equationsthatneedstobesolvedinordertogetthesolutiontoour problem. Example 7.4 Compute excos x .

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41.INTEGRATIONBYPARTS35 Solution: Weset u( x )= exand v ( x )=cos x .Then u ( x )= exand v( x )= Š sin x ,andformula(7.2)yields (7.3) excos xdx = excos x + exsin xdx. Sowecouldsolveourproblemifwecouldcomputetheintegral exsin xdx .Wecandothatbyapplyingthetechniqueofintegrationbyparts again,setting u( x )= exand v ( x )=sin x .Weobtain (7.4) exsin xdx = exsin x Š excos xdx. Finally,notethat(7.3)and(7.4)isasystemofequationswithunknowns excos xdx and excos xdx .Wecansolvethissystem,for instance,byaddingthesetwoequationsandnotingthat exsin x cancels.Wegettheequation excos xdx = ex(cos x +sin x ) Š excos xdx or excos xdx = ex 2 (cos x +sin x )+ C. Notethatsubstitutingtheobtainedexpressionfor excos xdx into (7.4),wegetaformulafor exsin xdx ,namely, exsin xdx = ex 2 (sin x Š cos x )+ C.41.3.DeniteIntegrals.Ifweevaluatebothsidesofformula(7.2)from a to b andweapplythefundamentaltheoremofcalculus,wegetthe identity (7.5) b a( uv ) dx =( uv ) b aŠ b a( uv) dx. Example 7.5 Evaluate 2 1ln xdx Solution: AswesawinExample7.3,wecanset u ( x )= x and v ( x )= ln x .Then u( x )=1, v( x )=1 /x ,andformula(7.5)yields 2 1ln x =( x ln x ) 2 1Š 2 11 dx =2ln2 Š 1

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367.METHODSOFINTEGRATION 41.4.Exercises.(1)Compute x sin xdx (2)Compute x2exdx (3)Compute x2cos xdx (4)Compute x ln xdx (5)Compute x2ln x (6)Compute e2 xsin xdx (7)Compute xeŠ xdx (8)Compute ( x +cos x )2dx (9)Compute x ln xdx (10)Compute x tanŠ 1( x ) dx (11)Compute x sinŠ 1( x ) dx (12)Compute (ln x )2dx (13)Evaluate 2 1x cos2 xdx (14)Evaluate 1 0x2sin xdx (15)Evaluate 1 0xexdx (16)Evaluate 1 0xe2 xdx (17)Evaluate 3 0x3exdx (18)Evaluate 3 2x ln xdx (19)Evaluate 2 1(ln x )2dx (20)Useintegrationbypartstocompute (cos x )2dx 42.TrigonometricIntegrals42.1.Powersofsinandcos.Inthissection,weconsiderfunctionsofthe form f ( x )=sinmx cosnx anddiscusstechniquesfortheirintegration. Itseemsnaturalto“rstconsiderthecaseswhen m or n is0,thatis, when f isjustapowerofsinorcos.Eventhesespecialcaseswillbreak upintofurthersubcases.Theeasiestsubcaseiswhentheexponents are even numbers.Inthatcase,wecanusethetrigonometricidentities (7.6)cos2 x =2cos2x Š 1 and (7.7)sin2 x =2sin x cos x toeliminatehighpowersoftrigonometricfunctionsintheintegrand. Example 7.6 Compute cos4xdx .

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42.TRIGONOMETRICINTEGRALS37 Solution: Using(7.6),wegetthatcos2x =1+cos2 x 2,andso cos4x = 1+cos2 x 2 2= 1 4 + cos2 x 2 + cos22 x 4 Applying(7.6)again,with2 x replacing x ,wegetthatcos22 x =1+cos4 x 2, sothepreviousdisplayedequationturnsinto cos4x = 3 8 + cos2 x 2 + cos4 x 8 Havingeliminatedthepowersofcos,theintegrationiseasytocarry outasfollows: cos4xdx = 3 8 + cos2 x 2 + cos4 x 8 dx = 3 x 8 + sin2 x 4 + sin4 x 32 + C. Thecomputationismorecomplexiftheintegrandisan odd power ofsinorcos.Inthatcase,weseparateonefactorandconverttherest intothe other trigonometricfunction,usingtherulecos2x +sin2x =1. Example 7.7 Compute sin3xdx Solution: Wehave sin3x =sin x sin2x =sin x (1 Š cos2x )=sin x Š sin x cos2x. Theadvantageofthisformisthatitmakesintegrationbysubstitutioneasy.Indeed,set u =cos x ,then du/dx = Š sin x ,andso Š sin x cos2xdx = u2du = u3 3 + C = cos3x 3 + C. Comparingthetwodisplayedequationsofthissolutionandnotingthat sin xdx = Š cos x ,weget sin3x = Š cos x + cos3x 3 + C. Themethodsshownabovecanbeusedtocomputetheintegralof products ofpowersofsin x andcos x .Inotherwords,themethodallows ustocompute cosmx sinnxdx asshownbelow. Example 7.8 Compute cos2x sin3xdx .

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387.METHODSOFINTEGRATION Solution: JustasinExample7.7,weseparateonesin x factor.This accomplishestwothings.Itallowsustoconverttheremainingeven numberofsin x factorstocos x factors,anditallowsustointegrateby substitution. Indeed, cos2x sin3xdx = cos2x sin2x sin xdx = cos2x (1 Š cos2x )sin x = cos2x sin x Š cos4x sin x = Š u2du + u4du = Š u3 3 + u5 5 + C = Š cos3x 3 + cos5x 5 + C, whereweusedthesubstitution u =cos x Wecanalwaysproceedthiswayifatleastoneof m and n inthe integrandcosmx sinnx isodd.Indeed,inthatcase,afterseparating onefactorfromthatoddpower,anevenpowerremains,andthatcan beconvertedtothe other trigonometricfunctionusingtheidentity sin2x +cos2x =1.Ifboth m and n areeven,thenwecanusethat identityrightaway. Example 7.9 Compute cos4x sin2xdx Solution: Wehave cos4x sin2xdx = cos4x (1 Š cos2x ) dx = cos4x Š cos6xdx. Nownotethatwecomputed cos4xdx inExample7.6.Youareasked tocompute cos6x inExercise42.4.4.Thedierenceofthesetwo resultsthenprovidesthesolutionofthepresentexample. 42.2.Powersoftanandsec.Whenintegratingaproductoftheform tanmx secnx ,wewillusetheidentitysec2x =tan2x +1andthedifferentiationrules(tan x )=sec2x and(sec x )=sec x tan x Therearetwoeasycases,namely,when m isodd(and n isatleast 1)andwhen n 2iseven.

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42.TRIGONOMETRICINTEGRALS39 Inthe“rstcase,thatis,when m isoddand n 1,weseparateone factoroftan x sec x andexpresstheremainingfactorsintermsofsec x bytheidentity Š 1+sec2x =tan2x .Thenwesubstitute u =sec x whichleadstodu dx=tan x sec x Example 7.10 Compute tan3x sec xdx Solution: Followingthestrategyexplainedabove,wehave tan3x sec xdx = tan2x tan x sec xdx = ( Š 1+sec2x )tan x sec xdx = ( Š 1+ u2) du = Š u + u3 3 + C = Š sec x + sec3x 3 + C. Inthesecondcase,thatis,when n 2iseven,weseparateone factorofsec2x ,expresstheremainingfactorsintermsoftan x using theidentitysec2x =1+tan2x ,andsubstitute u =tan x ,whichleads todu dx=sec2x Example 7.11 Compute sec4x dx. Solution: Wehave sec4xdx = sec2x sec2xdx =(1+tan2x )sec2xdx =(1+ u2) du = u + u3 3 + C =tan x + tan3x 3 + C. Ifwearenotinthesetwoeasycases,thenthereisnorecipethat willalwayswork.Wethenneedtohaveaseparatestrategyforeach problem.WewillshowexamplesofthatinExercises42.4.13,42.4.14, 42.4.15,and42.4.16.

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407.METHODSOFINTEGRATION 42.3.SomeOtherTrigonometricIntegrals.Ifourgoalistocomputeintegralsoftheform cos mx sin nxdx cos mx cos nxdx ,and sin mx sin nxdx ,thenwecanoftenmakeuseofthefollowingidentities: (7.8)sin a cos b = 1 2 sin( a Š b )+ 1 2 sin( a + b ) (7.9)cos a cos b = 1 2 cos( a Š b )+ 1 2 cos( a + b ) (7.10)sin a sin b = 1 2 cos( a Š b ) Š 1 2 cos( a + b ) Example 7.12 Compute cos3 x cos5 xdx Solution: Using(7.9)with a =3 x and b =5 x andnotingthat cos Š 2 x =cos2 x ,wegetthatcos3 x cos5 x =1 2cos2 x +1 2cos8 x ,and so cos3 x cos5 xdx = 1 2 cos2 x + 1 2 cos8 x dx = 1 4 sin2 x + 1 16 sin8 x + C. 42.4.Exercises.(1)Compute cos3xdx (2)Compute sin5xdx (3)Compute cos5xdx (4)Compute cos6xdx (5)Compute sin4xdx (6)Compute sin3x cos2xdx (7)Compute sin x cos3xdx (8)Compute sin3x cos3xdx (9)Compute sin2x cos2xdx (10)Compute sin4x cos4xdx (11)Compute tan2x sec4xdx (12)Compute tan3x sec5xdx (13)Compute tan5xdx byseparatingonefactoroftan2x inthe integrandandexpressingitintermsofsec2x (14)Compute tan3xdx usingintegrationbyparts. (15)Compute sec3xdx usingintegrationbyparts,with u( x )= sec2x and v ( x )=sec x .

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43.TRIGONOMETRICSUBSTITUTION41 (16)Compute sec5xdx .( Hint: RecalltheresultofExample 7.11.) (17)Compute sin x cos xdx (18)Compute sin2 x cos4 xdx (19)Compute sin3 x sin7 xdx (20)Compute cos2 x cos4 xdx 43.TrigonometricSubstitution43.1.ReversingtheTechniqueofSubstitutions.Letusassumethatwe wanttocomputetheareaofacirclebyviewingone-fourthofthatcircle asthedomainunderacurve.Let r betheradiusofthecircle,andlet usplacethecenterofthecircleattheorigin.Thenthenortheastern quarterofthecircle,showninFigure7.1,isjustthedomainunderthe graphofthefunction f ( x )= r2Š x2,where x rangesfrom0to r .In otherwords,weneedtocomputetheintegral (7.11) r 0 r2Š x2dx. InChapter5,wepresentedthetechniqueofintegrationbysubstitution.Thistechniqueworkedinsituationswhenthebestwaytocomputeanintegralwastode“neasimplefunctionof x ,suchas y ( x )= x2, andthencontinuetheintegrationintermsofthatnewvariable y Inordertocomputetheintegralin(7.11),weusethe reverse of thestrategymentionedinthepreviousparagraph.Wede“neanother variable y sothat x isasimplefunction f of y .Itisimportantto de“ne f and y sothat f isone-to-one,sincethatassuresthat f ( y )= x isequivalentto fŠ 1( x )= y Incomputingtheintegralin(7.11),wecanset x = r sin y .Then dx/dy = r cos y ,andthelimitsofintegrationare y =0and y = / 2. Figure7.1. Thenortheasternquadrantoftheunitcircle.

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427.METHODSOFINTEGRATION Thisyields r 0 r2Š x2dx = / 2 0 r2Š r2sin2yr cos ydy = r2/ 2 0 1 Š sin2ydy = r2/ 2 0cos2ydy = r2y +sin2 y 2 / 2 0= r2 4 Notethatwecouldwritecos y for 1 Š sin2y ,since0 y / 2, andinthatinterval,cos y isnonnegative. Theprecedingcomputationresultedin“ndinga de“nite integral. Wepointoutthatbyconvertingtheinde“niteintegral r2y +sin2 y 2 backtoafunctionof x usingtherule x = r sin y ,we getthecorresponding inde“nite integral,thatis r2Š x2dx = r2 2 sinŠ 1 x r + 1 2 x r2Š x2+ C. Theresultthatwearegoingtocomputeinthenextexamplewillbe usefulinthenextsection,whenwewilllearnatechniquetointegrate rationalfunctions. Example 7.13 Computetheintegral 1 (1+ x2)2dx Solution: Weusethesubstitution x =tan y .Then y =tanŠ 1( x ),and so dy/dx =1 / (1+ x2),andhence dy = dx/ (1+ x2).Thisyields 1 (1+ x2)2dx = 1 1+ x2dy = 1 1+tan2y dy = cos2ydy = y 2 + sin2 y 4 + C

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43.TRIGONOMETRICSUBSTITUTION43 = y 2 + sin y cos y 2 + C = 1 2 tanŠ 1( x )+ 1 2 x x2+1 + C. Thelaststepisjusti“edsince (7.12) x x2+1 = tan y 1+tan2y =tan y cos2y =sin y cos y. Figure7.2illustratesthistrigonometricargument. Example 7.14 Computetheintegral 1 x2Š 1. Solution: Thedenominatorremindsusofthetrigonometricidentity tan2y =sec2y Š 1,andso,if y [0 ,/ 2),thentan y = sec2y Š 1. Therefore,weusethesubstitution x =sec y .Then dx/dy =tan y sec y Hence,wehave 1 x2Š 1 = 1 sec2y Š 1 dx = 1 tan y dx = tan y sec y tan y dy = sec ydy =ln | sec y +tan y | + C =ln | x + x2Š 1 | + C. Figure7.3illustratesthistrigonometricargument. Figure7.2. Someexpressionsfrom(7.12).

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447.METHODSOFINTEGRATION Figure7.3. Someexpressionsoccuringinthesolution ofExample7.14.43.2.SummaryoftheMostFrequentlyUsedTrigonometricSubstitutions.Thethreeexamplesthatwehaveseensofarinthissectionshowthe threemostfrequentlyusedreversesubstitutions.Thatis, (i)Tocompute r2Š x2dx ,usethereversesubstitution x = r sin y (ii)Tocomputeintegralsinvolving( r2+ x2)underarootsignor inthedenominatorofafraction,usethereversesubstitution x = r tan y (iii)Tocompute x2Š r2dx ,usethereversesubstitution x = r sec y Finally,awordofcaution.Theavailabilityofthemethodofreverse substitutiondoesnotmeanthatthismethodis always thebestoneto computeanintegralthatcontainsasquarerootsign.Someofthe followingexercisescanbesolvedbyanothermethodfaster(and,no, wearenotrevealingwhichones).43.3.Exercises.(1)Usethemethodpresentedinthissectiontocomputethearea ofanellipsedeterminedbytheequation x2 a2+ y2 b2=1 (2)Compute 1 Š 4 x2dx (3)Compute x3 x2+16dx (4)Compute x3 25 Š x2dx (5)Compute 4 Š 36 x2dx (6)Compute 1+ x2dx (7)Compute 1 4+ x2dx (8)Compute x x Š 5dx (9)Compute 1 x2 x2Š 4dx .

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44.INTEGRATINGRATIONALFUNCTIONS45 (10)Compute x x2Š 4 dx (11)Compute x2Š 2 xdx (12)Compute x2 9+ x2dx (13)Compute 1 4 x2Š 25dx (14)Compute 1 x2Š 4 x +13dx (15)Compute 1 ( x2+4 x +5)2dx (16)Compute 2 x Š x2dx (17)Compute x2 6 x Š x2dx (18)Compute e2 xŠ 16 dx (19)Compute ex 4 Š e2 xdx (20)Findtheaveragevalueof1 x2+9ontheinterval[0 3]. 44.IntegratingRationalFunctions44.1.Introduction.Recallthata rationalfunction istheratiooftwo polynomials,suchas R ( x )= P ( x ) Q ( x ) = 3 x +5 2 x2+4 x +9 Integratingrationalfunctionsisrelativelysimple,becausemostofthese functionscanbeobtainedassumsofevensimplerfunctions.Ifthe degreeof P ( x )isatleastaslargeasthedegreeof Q ( x ),thenwecan divide P ( x )by Q ( x ),gettingapolynomialasaquotient,andpossibly aremainder.Thatis,ifthedegreeof P isatleastaslargeasthe degreeof Q ,thenthereexistpolynomials P1( x )and P2( x )suchthat thedegreeof P2( x )is less thanthedegreeof Q ( x )and R ( x )= P ( x ) Q ( x ) = P1Q(x )+ P2( x ) Q ( x ) = P1( x )+ P2( x ) Q ( x ) As P1( x )isapolynomial,itiseasytointegrate.Therefore,thedicultyofintegrating R ( x )liesinintegratingR2( x ) Q ( x ),whichisarational functionwhose denominator isofhigherdegreethanitsnumerator. Forthisreason,intherestofthissection,wefocusonintegrating rationalfunctionswiththatproperty,thatis,whenthedegreeofthe denominatorishigherthanthedegreeofthenumerator. Example 7.15 Let R ( x )=x3+2 x +1 x2Š x +1.Thendividing P ( x ) by Q ( x ) usinglongdivision,weget P ( x )=( x +1)( x2Š x +1)+2 x,

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467.METHODSOFINTEGRATION so R ( x )= P ( x ) Q ( x ) = x +1+ 2 x x2Š x +1 andintegrating Q ( x ) boilsdowntointegrating2 x x2Š x +1.44.2.BreakingUptheDenominator.Inordertodecidehowtobreak uparationalfunction R ( x )intothesumofsimplerterms,weanalyze thedenominator Q ( x )of R ( x ).Atheoremincomplexanalysis,sometimescalledthefundamentaltheoremofalgebra,impliesthatif q ( x ) isapolynomialwhosecoecientsarerealnumbers,then q ( x )canbe writtenasaproductofpolynomialsthatareofdegree1or2. Thisdecomposition,orfactorization,of Q ( x )willdeterminethe wayinwhichwebreakupourrationalfunctionintothesumofsimpler terms.Thereareseveralcasestodistinguish,basedonthefactorization of Q ( x ).44.2.1.DistinctLinearFactors.Theeasiestcaseiswhen Q ( x )factors intotheproductofpolynomialsofdegree1,andeachoftheseterms occursonlyonce. Example 7.16 Compute 1 x2+3 x +2dx Solution: Notethat x2+3 x +2=( x +1)( x +2).Usingthatobservation, wearelookingforrealnumbers A and B suchthat (7.13) 1 x2+3 x +2 = A x +1 + B x +2 asfunctions,thatis,suchthat(7.13)holdsfor all realnumbers x Multiplyingbothsidesby x2+3 x +2,weget (7.14)1= A ( x +2)+ B ( x +1) If(7.14)holdsfor all realnumbers x ,itmustholdfor x = Š 1and x = Š 2aswell.However,if x = Š 1,then(7.14)reducesto1= A andif x = Š 2,then(7.14)reducesto1= Š B .Soweconcludethat A =1and B = Š 1arethenumberswewantedto“nd.Itisnoweasy tocomputetherequestedintegralasfollows: 1 x2+3 x +2 dx = 1 x +1 dx Š 1 x +2 dx =ln( x +1) Š ln( x +2)+ C. Theabovemethodcanalwaysbeappliedif Q ( x )factorsintoa productoflinearpolynomials,eachofwhichoccursonlyonce.In

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44.INTEGRATINGRATIONALFUNCTIONS47 particular,if Q ( x )decomposesas a ( x Š a1)( x Š a2) ( x Š ak),then wecandecompose R ( x )intoasumoftheform A1 x Š a1+ A2 x Š a2+ + Ak x Š ak. Afterdeterminingthenumbers Ai,wecanintegrateeachoftheabove k summands.44.2.2.RepeatedLinearFactors.Thenextcaseiswhen Q ( x )factors intolinearterms,butsomeofthesetermsoccurmorethanonce. Example 7.17 Compute 2 x +7 ( x +1)2( x Š 1)dx Solution: Justasinthepreviouscase,wedecomposetheintegrand intoasumofsimplerfractions.Wearelookingforrealnumbers A B and C suchthat 2 x +7 ( x +1)2( x Š 1) = A x +1 + B ( x +1)2+ C x Š 1 Multiplyingbothsidesbythedenominatoroftheleft-handside,we get 2 x +7= A ( x +1)( x Š 1)+ B ( x Š 1)+ C ( x +1)2. Substituting x =1inthelastdisplayedequationyields9=4 C ,so C =2 25.Substituting x = Š 1yields5= Š 2 B ,so B = Š 2 5.Finally, thecoecientof x2ontheleft-handsideis0,whileontheright-hand side,itis A + C .So A + C =0,yielding A = Š 2 25. Nowweareinapositiontocomputetherequestedintegral. 2 x +7 ( x +1)2( x Š 1) dx = Š 2 25 x +1 dx + Š 2 5 ( x +1)2dx + 2 25 x Š 1 dx = Š 2 25 x +1 dx + Š 2 5 ( x +1)2dx + 2 25 x Š 1 dx = Š 2 25ln( x +1)+ 2 5 x +1 +2 25ln( x Š 1) Ingeneral,ifaterm( x + a )koccursin Q ( x ),thenthepartialfraction decompositionof R ( x )willcontainonetermwithdenominator( x + a )iforeach i { 1 2 ,...,k } .Forinstance,if Q ( x )=( x +2)3( x +5)2( x Š 10), then R ( x )willhaveapartialfractiondecompositionoftheform A1 x +2 + A2 ( x +2)2+ A3 ( x +2)3+ A4 x +5 + A5 ( x +5)2+ A6 x Š 10 .

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487.METHODSOFINTEGRATION 44.2.3.DistinctQuadraticFactors.Thethirdcaseiswhenthefactorizationof Q ( x )containssomequadraticfactorsthatareirreducible (i.e.,theyarenottheproductoftwolinearpolynomialswithrealcoecients),butnoneoftheseirreduciblequadraticfactorsoccursmore thanonce.Inthatcase,afterobtainingthepartialfractiondecompositionof R ( x ),wemayhavetoresorttotheformulas 1 x2+1 dx =tanŠ 1x + C and 1 x2+ a2dx = 1 a tanŠ 1 x a + C. Example 7.18 Computetheintegral 4 x +2 x3+ x2+ x +1dx Solution: Itiseasytonoticethatsetting x = Š 1turnsthedenominatorto0;hence,thedenominatorisdivisibleby x +1.Dividingthe denominatorby x +1,weget x2+1,sothedenominatorfactorsas ( x +1)( x2+1).Thefactor x2+1isirreducible(itisnotdivisibleby x Š b foranyrealnumber b ,sincenorealnumber b satis“estheequation b2+1=0). Therefore,wearelookingforrealnumbers A B ,and C suchthat (7.15) 4 x +2 x3+ x2+ x +1 = A x +1 + B x2+1 + Cx x2+1 Thereaderisinvitedtoverifythatthethirdsummandoftherighthandsideisnecessary;thatis,ifthesummandCx x2+1isremoved,then nopairofrealnumbers( A,B )willsatisfy(7.15). Inorderto“ndthecorrectvaluesof A B ,and C ,multiplyboth sidesof(7.15)by( x +1) ( x2+1)andrearrange,toget 4 x +2=( A + C ) x2+( B + C ) x + A + B. Thecoecientof x2is0ontheleft-handside,soithastobe0onthe right-handside.Therefore, A + C =0.Similarly,thecoecientof x is 4ontheleft-handside,soithastobe4ontheright-handside,forcing B + C =4.Similarly,theconstanttermsofthetwosideshavetobe equal,and,consequently, A + B =2.Solvingthissystemofequations, weget A = Š 1, B =3,and C =1.Therefore, 4 x +2 x3+ x2+ x +1 dx = Š 1 x +1 dx +3 1 x2+1 dx + x x2+1 dx = Š ln( x +1)+3tanŠ 1x + 1 2 ln( x2+1)

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44.INTEGRATINGRATIONALFUNCTIONS49 Ingeneral,if x2+ ax + b isaquadraticfactorin Q ( x ),thenthepartial fractiondecompositionwillcontainasummandoftheformE x2+ ax + band asummandoftheformFx x2+ ax + b.Again,thelatterisnecessary,since arationalfractionoftheformE + Fx x2+ ax + bwillnotequaloneoftheformE x2+ ax + bforanychoiceof E if F =0.44.2.4.RepeatedQuadraticFactors.Finally,itcanhappenthatthe factorization Q ( x )containsirreduciblequadraticfactors,someofwhich occurmorethanonce. Example 7.19 Computetheintegral x3+2 x2+3 x +7 x4+2 x2+1dx Solution: Itiseasytoseethatthedenominatorfactorsas( x2+1)2. Hence,wearelookingforrealnumbers A B C ,and D suchthat x3+2 x2+3 x +7 x4+2 x2+1 = A x2+1 + Bx x2+1 + C ( x2+1)2+ Dx ( x2+1)2. Multiplyingbothsidesby x4+2 x2+1andrearranging,weget x3+2 x2+3 x +7= Bx3+ Ax2+( B + D ) x +( A + C ) Foreach k ,thecoecientsof xkmustbethesameonbothsides. Hence, A =2and B =1,so C =5and D =2. Nowwecancomputetherequestedintegralusingthepreceding partialfractiondecompositionasfollows: x3+2 x2+3 x +7 x4+2 x2+1 dx = 2 x2+1 + x x2+1 + 5 ( x2+1)2+ 2 x ( x2+1)2 dx =2 tanŠ 1x + 1 2 ln( x2+1)+ 5 x 2( x2+1) + 5 2 tanŠ 1x Š 1 x2+1 = 1 2 ln( x2+1)+4 5tanŠ 1x + 1 2 5 x Š 2 x2+1 Hereweusedtheformulafor 1 ( x2+1)2thatwecomputedinthelast section,inExample7.13. Bynow,thereadermustknowwhatthegeneralversionofthe techniqueoftheprecedingexampleis.Ifthefactorizationof Q ( x ) contains( x2+ ax + b )k,then,foreachinteger i suchthat1 i k thepartialfractiondecompositionof R ( x )willcontainasummandof theformEi ( x2+ ax + b )iandasummandoftheformFix ( x2+ ax + b )i.

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507.METHODSOFINTEGRATION 44.3.RationalizingSubstitutions.Therearesituationswhenafunction thatis not arationalfunctioncanbeturnedintoonebyanappropriate substitution,andthenitcanbeintegratedbythemethodspresented inthissection.Themostfrequentscenarioinwhichthishappensis whentheintegrandcontainsroots,butifthoserootsarereplacedby anothervariable,wegetarationalfunctioninthatothervariable. Example 7.20 Compute x x +1dx Solution: Weusethesubstitution x = y .Then dy/dx =1 2 x=1 2 y. Thisleadsto x x +1 dx = y y +1 2 ydy = 2 y2 y +1 dy = 2( y Š 1)+ 2 y +1 dy = y2Š 2 y +2ln( y +1) = x Š 2 x +2ln( x +1) Notethatthecomputationwouldhavebeenverysimilarifthe integrandcontainedsomeotherrootof x .Indeed,iftheintegrand containedr x insteadof x ,thenwewouldhavesubstituted y =r x andthatwouldhaveturnedtheintegrandintoarationalfunctionof y .Indeed, y = x1 /rimplies dy dx = x(1 /r ) Š 1 r dy dx = y1 Š r r andtherefore dx = ryr Š 1dy. Inotherwords, dx isaequalto dy timesa polynomialfunction of y so,indeed,theintegrandwillbearationalfunctionof y .44.4.Exercises.(1)Compute 5 x2+5 x +4dx (2)Compute x +3 x2Š 6 x +5dx (3)Compute x2+4 x +1 x3Š 6 x2+11 x Š 6dx (4)Compute 3 x3+ x2Š x Š 1dx (5)Compute 2 x +7 x3Š x2Š x +1dx .

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45.STRATEGYOFINTEGRATION51 (6)Compute 1 x4+5 x2+4dx (7)Compute 1 x4+4 x2+4dx (8)Compute 1 1 Š x4dx (9)Compute 3 x +7 x4Š 1dx (10)Compute x2+4 x3Š xdx (11)Compute x +3 x3+1dx (12)Compute 1 x3Š 1dx (13)Compute x +1 x Š 3dx (14)Compute 3 x 3 x +2dx (15)Compute 1 1+ xdx (16)Compute ex e2 x+1dx (17)Compute ex e2 xŠ 1dx (18)Compute ex e2 xŠ 4 ex+3dx (19)Compute sin x cos x +cos2xdx (20)Compute 1 10sin x +8dx .( Hint: Usethesubstitution u = tan( x/ 2)andobservethatthisimpliesthatsin x =2 u 1+ u2.) Notethatthesubstitutiondescribedaboveiscalledthe Weierstrasssubstitution ,namedaftertheGermanmathematician KarlWeierstrass.Itisapowerfultoolforcomputingtheintegralsoffunctionsofthetype f g ,where f isatrigonometric functionand g isarationalfunction. 45.StrategyofIntegration Wepresentedvariousintegrationtechniquesinthischapterand insomeprecedingchapters.Themostgeneraloneswereintegration bypartsandintegrationbysubstitution.Themostfrequentlystudied specialcaseswererelatedtotrigonometricfunctionsandtheirinverses. Reversesubstitutioncameupinsomespecialcases.Wealsodiscussed theintegrationofrationalfunctions,usingthetechniqueofpartial fractions. Inshort,wehavelearnedadecentnumberofmethods.Forthis veryreason,itissometimesnotobviouswhichmethodweshoulduse whentryingtointegrateafunction.Whilethereisnogeneralrule,in thissectionwewillprovideafewguidelines. Solet f beafunctionthatisnotequaltooneofthefunctions whoseintegralweeitherknowohandorhaveadeterministicmethod

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527.METHODSOFINTEGRATION tocompute.Thatis, f isnotapolynomial, f isnotarationalfunction, f isnotthefunction f ( x )= axor f ( x )=logax forsomepositivereal number a ,and f isnotoneofthebasictrigonometricfunctionslike sin x ortan x .Letusalsoassumethatsimplealgebrawillnothelp, thatis,that f cannotbetransformedintooneoftheseelementary functionsbysimplealgebraictransformations.Thenhowdowedecide whichmethodtouse?45.1.Substitution.Themethodthatneedstheleastamountofwork, whenitisavailable,isasimplesubstitution,soitisreasonableto trytousethatmethod“rst.Thereisaparticularlygoodchancefor thisapproachtoworkwhen f isthecompositionoftwofunctions,one ofwhichhasaconstantderivative,orwhen f isoftheform f ( x )= h( g ( x )) g( x ),sincethen f ( x )=dh dx( x ),andso f ( x ) dx = h ( x ).Inthe languageofsubstitutions,thismeansthatsubstituting y = g ( x )will work,since fdx = h( g ( x )) g( x ) dx = h( y ) dy dx dx = h( y ) dy. Inotherwords,theintegralofthecompositefunction f isturnedinto somethingsimpler,theintegralofthefunction h Example 7.21 Let f ( x )=sin2 x .Then,usingthesubstitution y =2 x ,wegetthat sin2 xdx =1 2 sin ydy = Š1 2cos y + C = Š1 2cos 2 x + C Example 7.22 Let f ( x )=3 x x2+1.Thenweset y = x2+1 ,so dy/dx =2 x .Thisleadsto 3 x x2+1 dx = 3 x y dx = 3 2 1 y dy = 3 2 ln y + C = 3 2 ln( x2+1)+ C. Thereadershouldcomputetheintegral sinnx cos xdx atthis point.45.2.IntegrationbyParts.If f istheproductoftwofunctions,but substitutiondoesnotseemtohelp,thenintegrationbypartsisthe logicalnextstep.Thistechniqueisparticularlyusefulwhenoneof thetwofunctionswhoseproductis f ismadesigni“cantlysimplerby dierentiation.

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45.STRATEGYOFINTEGRATION53 Example 7.23 Computetheintegral xeŠ xdx Solution: Consideringtheintegrand,wenoticethatsubstitutionis unlikelytohelp,since x and eŠ xarenotcloselyrelated.Ontheother hand,theintegrandisaproduct,andoneoftheterms, x ,ismade simplerbydierentiation.Therefore,wechoosethetechniqueofintegrationbyparts,with x = u and v= eŠ x.Then u=1,while v = Š eŠ x,andweget xeŠ xdx = Š xeŠ xŠ eŠ xdx = Š xeŠ x+ eŠ x+ C =(1 Š x ) eŠ x+ C. 45.3.Radicals.Ifthetwomostgeneralmethods(substitutionandintegrationbyparts)arenothelpful,thenitisquitepossiblethatthereis arootsignintheintegrand.Inthatcase,therearetwospeci“cmethodsthatwecantry,reversesubstitutionandrationalizingsubstitution. Theeasiestwaytoknowwhentouseeachofthesetwomethodsisto remembertherelativelyfewcasesinwhichreversesubstitutionworks directly.Aswehaveseen,thesearetheintegralsinvolving 1 r2+ x2dx r2Š x2dx ,and x2Š r2dx .Iftheintegranddoesnotcontain anyofthesefunctions,thenitmaybesimplertousearationalizing substitution,suchasincomputingtheintegral x x +4dx .Theexercisesattheendofthissectionwillaskthereadertodecidewhich methodtouseforafewspeci“cexamples.45.4.IfEverythingElseFails.Ifnoneofourmethodswork,thenitmay bethatanunexpectedtransformationoftheintegrandmayhelp,at leastwithrelatingtheintegraltoonethatisnotquiteaschallenging tocompute. Example 7.24 Compute sin4x tan2xcos22 xdx Solution: Weusethetrigonometricidentitytan x =sin x/ cos x ,to rewritetheintegrand.Weget sin4x tan2x cos22 xdx = sin2x cos2x cos22 xdx = 1 4 sin22 x cos22 xdx = 1 16 sin24 xdx,

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547.METHODSOFINTEGRATION whichiseasytointegratewiththemethodwelearnedforpowersof trigonometricfunctions. Itisimportanttopointoutthatsometimesthereisindeed nosolution ;thatis,thereexistelementaryfunctions f suchthatthereisno elementaryfunction F ( x )satisfying F( x )= f ( x ).Examplesofsuch functions f include ex2, ex/x ,and1 ln x.45.5.Exercises.Inalloftheexercisesbelow,computetheintegral. (1) x2eŠ 5 xdx (2) x x2Š 1dx (3) x +1 x3+1dx (4) esin xcos xdx (5) ex exŠ 1dx (6) 1 x2+4dx (7) 1 x +4dx (8) 1 x ln xdx (9) 1 x ln x ln(ln x )dx (10) 1 1 Š xdx (11) 1+sin x cos xdx (12) e3 x ex+1dx (13) ( x +cos x )2dx (14) exsin3 xdx (15) x2ln xdx (16) 1 Š x2 xdx (17) 1 Š 4 x Š 4 x2dx (18) x 5 Š xdx (19) tan4x sec2xdx (20) 1 1+sin xdx 46.IntegrationUsingTablesandSoftwarePackages46.1.TablesofIntegrals.Tablesofintegralscanbefoundinmany calculustextbooksandontheInternet.Thewebsitewww.integraltable.comisagoodexample,andwewilluseitasareferenceinthis subsection.(Whenwesaythetableofintegrals,Žwemeanthattable.)

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46.INTEGRATIONUSINGTABLESANDSOFTWAREPACKAGES55 Nomatterhowextensiveatableofintegralsis,itcannotcontain all integrals.Itisthereforeimportanttoknowhowtousethesetables tocomputeintegralsthatarenotcontainedinthetablesinthesame form. Theeasiestcaseiswhentheintegraltobecomputedisa special case ofamoregeneralintegralthatisinthetable. Example 7.25 Usethetableofintegralstocompute ln( x2+9) dx Solution: Lookingatthetableofintegrals,we“ndthattheintegrand isaspecialcaseofintegral(45),with a =3.Usingtheformulagiven inthetableforthegeneralcasewith a =3,wegettheresult ln( x2+9) dx = x ln( x2+9)+6tanŠ 1 x 3 Š 2 x + C. Sometimes,wehavetoresorttointegrationbysubstitutiontobe abletousethetableofintegrals. Example 7.26 Usethetableofintegralstocompute 9 Š 4 x2dx Solution: Takingalookatthetableofintegrals,we“ndthatformula (30)providesaformulafor a2Š x2dx .Inordertobeabletouse thatformula,weset y =2 x ,whichimplies dy/dx =2.Therefore, 9 Š 4 x2dx = 1 2 9 Š y2dy. Nowwecanapplyformula(30)fromthetableofintegrals,toget 1 2 9 Š y2dy = y 4 9 Š y2+ 9 2 tanŠ 1 y 9 Š y2 + C = x 2 9 Š 4 x2+ 9 2 tanŠ 1 2 x 9 Š 4 x2 + C. Sometimesweneedtocarryoutsomealgebraicmanipulationbefore wecanusethetechniqueofsubstitutionsinconnectionwithusingthe tableofintegrals. Example 7.27 Usethetableofintegralstocompute x2+2 x +1 x2+2 x +10dx Solution: Thereisnointegralinthetableofintegralsthatwouldimmediatelystandoutasonethatisverysimilartothisone.Thecrucial observationisthatthesubstitution y = x +1signi“cantlysimpli“esour integrand.Thatsubstitutionleadstotheintegral y2 y2+32dy ,which,

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567.METHODSOFINTEGRATION inturn,canbedirectlyfoundinthetableofintegralsasitem(36). Substituting x backintotheobtainedformula,weget x2+2 x +1 x2+2 x +10 dx = x2+2 x +10 2 Š 9 2 ln x +1+ x2+2 x +10 + C. 46.2.SoftwarePackages.ComputersoftwarepackagessuchasMaple andMathematicaareveryusefultoolsofintegration.Thesepackages willcomputethede“niteorinde“niteintegralsofalargeclassoffunctions,thentheywillpresenttheresultsinaformthatisusually,but notalways,intheformtheuserexpected.Inthissection,weshowa fewexamplesoftheseunexpectedresultsandexplainhowtointerpret them. Tostartwithaverybasicexample,type int(x^2+3x,x); intoMaple.Wegettheanswer1 3x3+3 2x2.Thisisthecorrectanswer havingconstantterm0 .Experimentingwithotherfunctions,wenote thatMaplealwaysanswersinthisway,thatis,withouttheconstant C attheend.Itisimportantnottoforgetthisifwewillbeusingthe obtainedfunctioninsomefurthercomputation. Mapledoesnotalwaysprovidethe simplestform fortheintegral thatitcomputes.Forinstance,ifweaskMapletocomputetheindefiniteintegral x ( x2+3)6dx ,wegettheoutput (7.16) 1 14 x14+ 3 2 x12+ 27 2 x10+ 135 2 x8+ 405 2 x6+ 729 2 x4+ 729 2 x2. However,itisveryeasytocompute x ( x2+3)6dx byhand,using thesubstitution u = x2.Thatsubstitutionleadstothesamesolution,butinamuchsimplerform,namely,1 14( x2+3)7.Ifwewant toverifythatthisresultindeedagreeswiththeonegivenbyMaple anddisplayedin(7.16),wecanaskMapleto expand theexpression H =1 14( x2+3)7usingthe expand command.Weseethattheexpandedexpressionindeedagreeswiththeonegivenin(7.16),upto theconstanttermsattheend. Thereareothercommandslike expand that,areusefulifwewant totransformtheoutputofanintegrationsoftwarepackage.Thecommands rationalize and simplify areexamplesofthese.

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46.INTEGRATIONUSINGTABLESANDSOFTWAREPACKAGES57 Thereisoftenmorethanonewaytoexpressintegralsinvolving hyperbolicfunctions.Astrikingexampleisthefollowing.Ifweask Mapletocompute 1 1 Š x2dx bytyping int(1/(1-x^2),x); MaplereturnstheanswertanhŠ 1x .Thisisverysurprising,sinceitis notdiculttointegratetheintegrandasarationalfunction,withno hyperbolicfunctionsinvolved.Indeed, 1 1 Š x2= 0 5 x +1 Š 0 5 x Š 1 andso (7.17) 1 1 Š x2dx = 1 2 ln( x +1) Š 1 2 ln( x Š 1)=ln x +1 x Š 1 TheresultobtainedbyMapleactually agrees withtheresultgiven in(7.17),evenifthatisnotobvious.Indeed,if y =tanhŠ 1x ,then, byde“nition, x =( eyŠ eŠ y) / ( ey+ eŠ y).Solvingthisequationfor y isnotcompletelytrivial,butattheend,ityields y =ln x +1 x Š 1. ( Hint: Multiplyboththenumeratorandthedenominatorby eytoget x =( e2 yŠ 1) / ( e2 y+1).) Finally,whencomputingde“niteintegrals,Maplesometimesanswersbyusingtheacronymsofsomerarefunctions.Forinstance,if wewanttocompute 1 0sin( x2) dx bytyping int(exp(sin(x^2)),x=0..1). thenwegettheanswer (1/2)*FresnelS(sqrt(2)/sqrt(Pi))*sqrt(2)*sqrt(Pi). Here FresnelS referstotheFresnelsineintegral,aconceptbeyond thescopeofthisbook.Ifwesimplywanttoknowanumericalvalue for 1 0ex2dx ,wecantype evalf(int(sin(x^2),x=0..1)) instead.Mapleoutputs0.4596976941astheanswer.46.3.Exercises.(1)Usethetableofintegralstocompute x ln(3 x +5) dx (2)Usethetableofintegralstocompute x cos 2Š x dx (3)Usethetableofintegralstocompute x +7 x+4 x +5dx (4)Usethetableofintegralstocompute x2 x6+16dx (5)Usethetableofintegralstocompute 8 x3 (4 x2+9)dx (6)Usethetableofintegralstocompute 8 x2+3 dx .

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587.METHODSOFINTEGRATION (7)Thetableofintegralsprovidesaformulafor 1 ax2+ bx + cdx Describethevaluesof a b ,and c forwhichthatformulawill not workandcomputetheintegralinthoseexceptionalcases. (8)Thetableofintegralsprovidesaformulafor ln( ax2+ bx + c ) dx. Describethevaluesof a b ,and c forwhichthatformulawill not workandcomputetheintegralinthoseexceptionalcases. (9)Useyourfavoritesoftwarepackagetocompute xexdx x2exdx ,and x3exdx .Doyouseeapattern?Trytoguess what x4exdx is,thenverifyyourguessbyusingyoursoftwarepackageagain. (10)Useyourfavoritesoftwarepackagetocompute ln xdx x ln xdx ,and x2ln xdx .Doyouseeapattern?Tryto guesswhat x3ln xdx is,thenverifyyourguessbyusingyour softwarepackageagain. (11)Useyourfavoritesoftwarepackagetocompute ln xdx (ln x )2dx ,and (ln x )3dx .Doyouseeapattern?Tryto guesswhat (ln x )4dx is,thenverifyyourguessbyusingyour softwarepackageagain. (12)Let f ( x )=1 1+ x2andlet g ( x )= 1 1+ x2.Clearly, f ( x )= g ( x ) forall x forwhichthesefunctionsarede“ned.Computethe integralofbothfunctionswithMaple.Iftheresultsseem dierent,showthattheyareinfactequal. (13)Compute x ( x2+1)8dx “rstbyMaple,thenbysubstitution. Whichanswerissimpler? (14)Compute x x2+4 x +29 dx usingasoftwarepackage.If theresultappearstobedierentfromwhatyou“ndinthe tableofintegrals,explainwhythetworesultsareequivalent. (15)Compute tan2x sec4xdx .IftheresultappearstobedierentfromwhatyougetusingthemethodsofSection42,explain whythetworesultsareequivalent. (16)Compute sin2x cos4xdx .Iftheresultappearstobedierent fromwhatyougetusingthemethodsofSection42,explain whythetworesultsareequivalent. (17)Useasoftwarepackagetocompute 1 0sinnxdx for n =3, n =4,and n =5.(Thecaseof n =2wasdiscussedinthe text.)Explainthetrendyoudetect. (18)Useasoftwarepackagetocompute 1 0cosnxdx for n =2, n =3, n =4,and n =5.Explainthetrendyoudetect.

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47.APPROXIMATEINTEGRATION59 (19)Useasoftwarepackagetocomputethevalueoftheexpression1 106Š 106eŠ x2/ 2dx (20)Useasoftwarepackagetocomputethevalueoftheexpression1 1060 x4 x6+1dx 47.ApproximateIntegration Sometimesitisnotpossibleto“ndtheexactvalueofade“nite integral b af ( x ) dx .Itcouldhappenthatwecannot“ndtheantiderivativeof f ( x )orthattheantiderivativeof f ( x )isnotanelementary function.Oritcouldhappenthat f itselfisnotgivenbyaformula, butinstead, f isgivenbyitsgraph,whichisplottedbyacomputer program.Inthiscase,weresorttomethodsof approximateintegration .47.1.BasicApproximationMethods.Thekeyobservationbehindthe approximationmethodsisthefactthatif f ( x ) 0for x [ a,b ],then b af ( x ) dx isequaltotheareabelowthegraphofthefunction f on thatinterval.Moreprecisely, b af ( x ) dx isequaltotheareaofthe domainborderedbythehorizontalaxis,theverticallines x = a and x = b ,andthegraphof f Inordertoestimatetheareaofthisdomain D ,wecut D intosmall verticalstrips.Todoso,wechooserealnumbers a = x0
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607.METHODSOFINTEGRATION (ii)Theright-endpointapproximationshows b af ( x ) dx ni =1( xiŠ xi Š 1) f ( xi) (iii)Themidpointapproximationshows b af ( x ) dx ni =1( xiŠ xi Š 1) f ( xi) Astheaboveformulassuggest,itwillbeparticularlyeasytowork withtheseformulasifthepoints x1,x2,...,xn Š 1splittheinterval[ a,b ] into equal parts,sinceinthatcase xiŠ xi Š 1=( b Š a ) /n forall i Example 7.28 Findtheapproximatevalueof 1 0ex2dx Solution: Letususetheleft-endpointmethodwith n =4and x1, x2, and x3splitting[0 1]intofourequalparts.Thatmeansthat x1=1 / 4, x2=1 / 2,and x3=3 / 4.Then(7.18)implies 1 0ex2dx 1 4 e0+ e1 / 16+ e1 / 4+ e9 / 16 1 2759 since xiŠ xi Š 1=1 / 4forall i Figure7.4. Left-endpointmethod.

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47.APPROXIMATEINTEGRATION61 Figure7.5. Right-endpointmethod. Ifweusetheright-endpointmethod,withthesamesetofpoints xi,weget 1 0ex2dx 1 4 e1 / 16+ e1 / 4+ e9 / 16+ e 1 7055 Itisnotsurprisingthatthesecondmethodyieldsthelargerresult, sincetheintegrand ex2isan increasingfunction ,so f ( xi) >f ( xi Š 1). Furthermore,foreachpoint x [ xi Š 1,xi],wehave f ( xi Š 1) f ( x ) f ( xi).Sotheleft-endpointmethod underestimates theareaofeach strip Si,whiletheright-endpointmethod overestimates it.Therefore, theareaof D „andhencethecorrectvalueof 1 0ex2dx „isbetween thetwovaluesof1.2759and1.7055computedabove. Replacingthevalueof n =4bysomelargernumberwillresultin amorepreciseapproximation(andmorework).Usingthemidpoint methodwillresultinanapproximation A thatisclosertotheactual valueoftheintegral 1 0ex2dx ,butitisnotcompletelyobviousfrom whichside A approximates 1 0ex2dx ,thatis,whether A< 1 0ex2dx or A> 1 0ex2dx .47.2.MoreAdvancedApproximationMethods. 47.2.1.TrapezoidMethod.Ifthedierencebetween f ( xi Š 1)and f ( xi) islarge,thenestimatingtheareaof Sibyusingrectanglescouldlead

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627.METHODSOFINTEGRATION Figure7.6. Trapezoidmethod. tolargeerrors.Amorere“nedapproachistoestimatetheareaof Sibycomputingtheareaofthe trapezoid whoseverticesarethepoints ( xi Š 1, 0),( xi, 0), f ( xi),and f ( xi Š 1).Weknowthattheareaofthis trapezoidistheaveragelengthofitsparallelsidestimesthedistance ofthoseparallelsidesfromeachother,thatis,( f ( xi)+ f ( xi Š 1))( xiŠ xi Š 1) 2. Summingoverallpossiblevaluesof i ,wegetanestimateforallthe areaof D ,thatis,for b af ( x ) dx .Indeed,weobtaintheformula b af ( x ) dx =ni =1( f ( xi)+ f ( xi Š 1))( xiŠ xi Š 1) 2 Inparticular,ifthe xiarechosensothattheysplittheinterval[ a,b ] into n equalparts,thenthelastdisplayedequationsimpli“esto b af ( x ) dx = b Š a 2 nni =1( f ( xi)+ f ( xi Š 1)) = b Š a 2 n ( f ( a )+2 f ( x1)+2 f ( x2)+ +2 f ( xn Š 1)+ f ( b )) Notethat f ( x0)= f ( a )and f ( xn)= f ( b )occuronlyonceinthe suminthelastlinesince a and b areeachpartofonlyoneofthe intervals[ xi Š 1,xi]. Example 7.29 Usethetrapezoidmethodwith n =4 to“ndthe approximatevalueof 1 0ex2dx .

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47.APPROXIMATEINTEGRATION63 Figure7.7. Trapezoidmethod. Solution: Wewillhave x1=1 / 4, x2=1 / 2,and x3=3 / 4,justasin Example7.28.Thisyields 1 0ex2dx = 1 8 e0+2 e1 / 16+2 e1 / 4+2 e9 / 16+ e =1 4907 Thealertreadermayhavenoticedthattheresultweobtainedis preciselytheaverageoftheleft-endpointandright-endpointapproximationsweobtainedforthesameintegralintheprevioussection. (Thismeansthatitisabetterapproximationthanatleastoneofthe twoearlierones.)Thisisnotanaccident,andinExercise47.4.17, thereaderwillbeaskedtoprovethat,undercertainconditions,this phenomenonwillalwaysoccur.47.2.2.Simpson’sMethod.Asimilarmethodis Simpsonsmethod ,in whichweuseparabolasinsteadofstraightlinesforapproximation.For simplicity,letusnowassumethatthepoints xisplittheinterval[ a,b ] into n equalparts,thatis, xiŠ xi Š 1=( b Š a ) /n .Inordertosimplify thenotation,letusset yi= f ( xi). Foranyinteger i [1 ,n Š 1],considerthepoints( xi Š 1,yi Š 1),( xi,yi), and( xi +1,yi +1).Thereisexactlyoneparabola pioftheform y = Ax2+ Bx + C thatcontainsthesethreepoints.Itcanthenbeproved

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647.METHODSOFINTEGRATION thattheareaunderthatparabola„moreprecisely,theareaofthe domain Piborderedbythehorizontalaxis,theverticallines x = xi Š 1and x = xi +1and pi„isequalto (7.19) b Š a 3 n ( yi Š 1+4 yi+ yi +1) Ifwesummed(7.19)overallpossiblevaluesof i ,wewouldnotgeta goodestimate,sincemostpointsofthedomainunderthecurvewould bepartof two ofthe Pi.Forinstance,apointwithahorizontalcoordinatebetween xiand xi +1ispartofboth Piand Pi +1.Therefore, wesumthelastdisplayedequationoverall even valuesof i ,and,accordingly,westipulatethat n bean even number.Thisleadstothe followingestimate. Theorem 7.1(SimpsonsMethod) Let n bean even positiveinteger.Then b af ( x ) dx b Š a 3 n ( y0+4 y1+2 y2+4 y3+ +2 yn Š 2+4 yn Š 1+ yn) Example 7.30 UseSimpsonsmethodwith n =4 toapproximate 1 0sin( x2) dx Figure7.8. Simpsonsmethod.

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47.APPROXIMATEINTEGRATION65 Solution: Wehave y0=0, y1=sin(1 / 16), y2=sin(1 / 4), y3= sin(9 / 16),and y4=sin1.SoSimpsonsmethodyields 1 0sin x2 dx 1 12 (4sin(1 / 16)+2sin(1 / 4)+4sin(9 / 16)+sin1) 0 31 Notethatthisresultcon“rmsourintuitioninthatif x [0 1],then sin( x2) sin x ,andso 1 0sin x2 dx 1 0sin x =1 Š cos1 0 4597 .47.3.BoundsontheErrorTerm.Theerrorterm E ofanapproximation isthedierencebetweenthenumberobtainedbytheapproximation andtheactualvalueofthequantitythatwasapproximated.(Inthis section,thatactualvalueisthevalueof b af ( x ) dx .)Itgoeswithout sayingthatthesmallertheabsolutevalueoftheerrorterm,thebetter theapproximationis. The“eldof numericalanalysis studiestheerrortermsofapproximationmethods.Thetechniquesofthat“eldyieldvariousbounds onerrorterms.Wecollectedsomeoftheseboundsinthefollowing theorem. Theorem 7.2 Let f beatwice-dierentiablefunctionon [ a,b ] such that | f( x ) | M if x [ a,b ] .Thenthefollowingholdfortheapproximationmethodsusedtocompute b af ( x ) (a) If ETistheerrortermofthetrapezoidmethod,then | ET| M ( b Š a )3 12 n2. (b) If EMistheerrortermofthemidpointmethod,then | EM| M ( b Š a )3 24 n2. (c) If,forall x [ a,b ] ,thenumber f(4)( x ) isde“nedandisat mostaslargeastheconstant K ,and ESistheerrortermof Simpsonsmethod,then | ES| K ( b Š a )5 180 n4. Comparingtheformulasofparts(a)and(b)oftheprevioustheorem,wecanconcludethatthe worst-case scenarioofthemidpoint methodisbetterthantheworst-casescenarioofthetrapezoidmethod.

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667.METHODSOFINTEGRATION This,ofcourse,doesnotmeanthatthemidpointmethodis always betterthanthetrapezoidmethod. Example 7.31 FindanupperboundfortheapproximationobtainedinExample7.29. Solution: Weapplypart(a)ofTheorem7.2.Wehave f ( x )= ex2, so f( x )= ex2(4 x2+2).Thisisanincreasingfunctionon[0 1],soits maximumistakenat x =1,showingthat | f( x ) | 6 e .AsinExample 7.29,wechose n =4,sotheprevioustheoremyields | ET| M ( b Š a )3 12 n2= 6 e 12 16 0 085 47.4.Exercises.(1)Use n =4andthemidpointmethodto“ndtheapproximate valueof 1 0ex2dx (2)Use n =4andtheleft-endpointmethodto“ndtheapproximatevalueof 1 0ex3dx (3)Use n =4andtheright-endpointmethodto“ndtheapproximatevalueof 1 0ex4dx (4)Use n =6andthemidpointmethodto“ndtheapproximate valueof 1 0 1+ xdx (5)Use n =4andthetrapezoidmethodto“ndtheapproximate valueof 1 0eŠ x2dx (6)Use n =4andthetrapezoidmethodto“ndtheapproximate valueof 1 0sin( ex) dx (7)Use n =4andthetrapezoidmethodto“ndtheapproximate valueof 2 1 sin x xdx (8)Use n =4andthetrapezoidmethodto“ndtheapproximate valueof 1 0 1 1+ x5dx (9)Use n =4andSimpsonsmethodto“ndtheapproximate valueof 1 0sin x3dx (10)Use n =4andSimpsonsmethodto“ndtheapproximate valueof 2 1 ln x 1+ x2dx (11)Use n =4andSimpsonsmethodto“ndtheapproximate valueof 1 0esin xdx (12)Use n =4andSimpsonsmethodto“ndtheapproximate valueof 1 0eexdx .

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48.IMPROPERINTEGRALS67 (13)Useyourfavoriteapproximationmethodwith n =4tocompute 1 0sin xdx .Thencomputethesameintegralby“nding theantiderivativeofsin x andcomparethetworesultsyou obtain. (14)Useyourfavoriteapproximationmethodwith n =4tocompute 1 0e xdx .Thencomputethesameintegralby“ndingthe antiderivativeof e xandcomparethetworesultsyouobtain. (15)Useyourfavoriteapproximationmethodwith n =4tocompute 1 0ln( x2+9) dx .Thencomputethesameintegralby “ndingtheantiderivativeofln( x2+9)andcomparethetwo resultsyouobtain. (16)Whatvalueof n shouldbeusedineachofexercises1,5,and 9togetanerrortermthatislessthan10Š 6? (17)Letusassumethatthepoints x1,x2,...,xn Š 1usedforapproximateintegrationsplittheinterval[ a,b ]into n equalsegments. Provethattheresultobtainedbythetrapezoidmethodwillbe theaverageoftheleft-endpointmethodandtheright-endpoint method. (18)Howlargeistheerrortermoftheapproximationinexercise1 intheworstcase? (19)Howlargeistheerrortermoftheapproximationinexercise7 intheworstcase? (20)Findanupperboundfortheerrortermoftheapproximation showninthesolutionofExample7.30. 48.ImproperIntegrals Inourstudiesofintegration,wehavenotdealtwithde“niteintegralsoverin“niteintervals,nordidweintegratefunctionsoveran intervalifthefunctionwasnotde“nedineverypointofthatinterval. Inthissection,wewillconsiderde“niteintegralsofthesekinds,which arecalled improperintegrals .48.1.InniteIntervals.For“niteintervals,wehaveidenti“ed b af ( x ) dx withtheareaofthedomainlimitedbythegraphof f ,theverticallines x = a and x = b ,andthehorizontalaxis.Welearnedthat,bythefundamentaltheoremofcalculus,theequality b af ( x ) dx = F ( b ) Š F ( a ) holds,where F isanantiderivativeof f Nowletusconsidertheintegralof f overthe in“nite interval[ a, ). Recallingthat b af ( x ) dx isequaltoacertainarea,weintuitivelywant af ( x ) dx toequaltheareaofthedomainborderedbytheline x = a ,

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687.METHODSOFINTEGRATION Figure7.9. Areaunderthecurve y = f ( x )from x = a to x = b thehorizontalaxis,andthegraphof f .Notethatthisareamaybea “nitenumber,evenifitisnotsqueezedbetweentwoverticallines.One exampleofthisiswhen f ( x )=0if x>N forsomerealnumber N Timehascometoformallyde“ne af ( x ) dx Definition 7.1 Let f beafunction.Iftheintegral b af ( x ) dx existsforall b>a and limb b af ( x ) dx = L existsasa(“nite)real number,thenwesaythattheintegral af ( x ) dx is convergent ,and wewrite af ( x ) dx = L If limb b af ( x ) dx doesnotexistorisin“nite,thenwesaythat af ( x ) dx is divergent Notethatif F isanantiderivativeof f ,then limb b af ( x ) dx =limb ( F ( b ) Š F ( a )) =(limb F ( b )) Š F ( a ) Therefore,theintegral af ( x ) dx isconvergentifandonlyif limb F ( b )existsandis“nite. Example 7.32 Let f ( x )= xŠ 2.Compute 1f ( x ) dx .

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48.IMPROPERINTEGRALS69 Figure7.10. Areaunderthecurve y = f ( x )from1to Solution: Wehave 1xŠ 2dx =limb b 1xŠ 2dx =limb Š xŠ 1 b 1=limb Š 1 b +1 =1 Inparticular, 1f ( x ) dx isconvergent. Encouragedbythesimplesolutionofthelastexample,wearegoing tocomputethemoregeneralintegral 1xrfor any realnumber r Example 7.33 Let f ( x )= xr.Compute 1f ( x ) dx Solution: Letus“rstassumethat r = Š 1.Thenwehave 1xrdx =limb b 1xrdx =limb 1 r +1 xr +1 b 1. If r> Š 1,then r +1 > 0andlimx xr +1= ,sothelimitinthe lastdisplayedrowisin“nite,andhence 1xrdx isdivergent. If r< Š 1,then r +1 < 0andlimx xr +1=0,sothelimitinthe lastdisplayedrowisequalto1 r +1,andhence 1xrdx isconvergent.

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707.METHODSOFINTEGRATION If r = Š 1,thenweneedtocompute 1xrdx dierently,since,in thatcase, xadx =xa +1 a +1.Instead,wehave 1xŠ 1dx =limb b 1xŠ 1dx =limb ln x b 1=limb ln b = Therefore, 1xŠ 1dx isdivergent. Notethattheresultsofthepreviousexampleprovethefollowing importanttheorem. Theorem 7.3 Let r bearealnumber. (i) If r Š 1 ,then 1xrdx isdivergent. (ii) If r< Š 1 ,then 1xrdx isconvergent. Thefollowingde“nitionisnotverysurprising.Itisthecounterpart ofDe“nition7.1. Definition 7.2 Let f beafunctionandlet b bearealnumber suchthat,forallrealnumbers a
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48.IMPROPERINTEGRALS71 Figure7.11. m Šf ( x ) dx isblue,while mf ( x ) dx isorange. Figure7.12. 0 ŠeŠ xdx Solution: Weset m =0andapplyDe“nition7.3.Wegetthat ŠeŠ xdx isconvergentifbothof 0 ŠeŠ xdx and 0eŠ xdx areconvergent.However, 0 ŠeŠ xdx =lima ŠŠ eŠ x 0 a=1+ isdivergentandthereforesois ŠeŠ x. Figure7.12showsthedomainwhoseareaisequalto 0 ŠeŠ xdx Thereadercouldaskhowweknewthatweneededtoselect0,and notsomeotherrealnumber,fortheroleof m ,thatis,tosplitthe realnumberlineintotwoparts.Theansweristhatwedidnot,and otherchoicesof m wouldhavegiventhesameresultsincetheintegrand

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727.METHODSOFINTEGRATION convergestoin“nityas x goestonegativein“nity.Wechose m =0 becauseitwasconvenienttodoso. Notethatallimproperintegralsdiscussedinthissectionarecalled Type1improperintegrals .48.2.VerticalAsymptotes.Sometimeswemaywanttocomputethe integralofafunction f ona“niteinterval[ a,b ]sothatinsomepoint c [ a,b ],thefunction f hasaverticalasymptote.Anexampleisthe function f ( x )=1 / ( x2Š 4)ontheinterval[1 3].Inthiscase,weuse thetechniqueoflimitstoformallyde“ne b af ( x ) dx ,aswedidinthe previoussection. Definition 7.4 Let f beafunctionthatiscontinuouson [ a,b ] exceptforonepoint c [ a,b ] .Thenweset (7.20) c af ( x ) dx =limt cŠt af ( x ) dx and (7.21) b cf ( x ) dx =limt c+b tf ( x ) dx. Furthermore,ifbothofthetwolimitsdisplayedaboveexistandare “nite,weset b af ( x ) dx = c af ( x ) dx + b cf ( x ) dx. Notethatiftheonlypoint c inwhich f isnotcontinuousisone oftheendpointsof[ a,b ],thenweonlyhavetocomputeoneof(7.20) and(7.21),sincetheotherintegralistakenoveratrivialintervaland ishencezero. Example 7.35 Compute 1 0xŠ 1 / 2dx Solution: Astheonlypointin[0 1]inwhich f ( x )= xŠ 1 / 2isnot continuousis0,weuseformula(7.21)with c =0and b =1.

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48.IMPROPERINTEGRALS73 Figure7.13. 1 0xŠ 1 / 2dx Weget 1 0xŠ 1 / 2dx =limt 0+1 txŠ 1 / 2dx =limt 0+2 x1 / 2 1 t=2 Š limt 0+t1 / 2=2 Š 0 =2 Sotheintegral 1 0xŠ 1 / 2dx isconvergent. Example 7.36 Compute 4 Š 1xŠ 2dx Solution: WeapplyDe“nition7.4sincetheinterval[ Š 1 4]hasone point, c =0,wheretheintegrandisnotcontinuous.Therefore, 4 Š 1xŠ 2dx = 0 Š 1xŠ 2dx + 4 0xŠ 2dx =limt 0Št Š 1xŠ 2dx +limt 0+4 txŠ 2dx =limt 0ŠŠ xŠ 1 t Š 1+limt 0+Š xŠ 1 4 t= + = .

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747.METHODSOFINTEGRATION Figure7.14. 4 Š 1xŠ 2dx Sotheintegralinquestionisdivergent. Figure7.14showsthedomainwhoseareaisequalto 4 0xŠ 2dx and thecorrectwayofbreakingthatintervaluptotwoparts. Notethatwewouldhavereachedthe wrong conclusionifwehad disregardedthefactthat xŠ 2isnotcontinuousat x =0andtriedto applythefundamentaltheoremofcalculus.Indeed,inthatcase,we wouldhaveobtainedthe wrong result: Š xŠ 1 4 Š 1=Š 1 4Š 1= Š5 4.This resultisincorrect,andtheincorrectstepwastoapplythefundamental theoremofcalculusforafunctionthatisnotcontinuousintheentire intervalofintegration. Theintegralsthatwehavediscussedinthissectionarecalled Type 2improperintegrals .48.3.FurtherRemarks. 48.3.1.ImproperIntegralsofMixedType.Therearesomeintegralsthat areimproperfortworeasons.Theyaretakenoveranin“niteinterval, andthatintervalcontainsapointinwhichthefunctionisnotcontinuous.Inthatcase,wesplituptheintervalofintegrationsothatnow wehavetwointegrals,oneofwhichisofType1andtheotherofwhich isType2. Example 7.37 Compute 0 1 ( x Š 2)2dx Solution: Webreakuptheinterval[0 )totheunionofthetwo intervals[0 2]and[2 ),getting 01 ( x Š 2)2dx = 2 01 ( x Š 2)2dx + 21 ( x Š 2)2dx.

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48.IMPROPERINTEGRALS75 Figure7.15. 0 1 ( x Š 2)2dx Figure7.16. 3 1 x2ln xdx The“rsttermontheright-handsideisanimproperintegralofType 2,andthesecondtermontheright-handsideisanimproperintegral ofType1.Wecancomputebothbythemethodspresentedearlierin thissection. 48.3.2.ComparisonTest.Comparisontestsforimproperintegralswork verysimilarlytothoseforproperintegrals. Theorem 7.4 Letusassumethat,forall x a ,thechainof inequalities 0 f ( x ) g ( x ) holds. (i) If af ( x ) dx isdivergent,thensois ag ( x ) dx (ii) If ag ( x ) dx isconvergent,thensois af ( x ) dx Example 7.38 Showthat 3 1 x2ln xdx isconvergent.

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767.METHODSOFINTEGRATION Solution: If x 3,thenln x> 1,so x2ln x>x2,andtherefore theintegrandislessthan1 x2.SeeFigure7.16foranillustration.On theotherhand,weknowthat 31 /x2dx isconvergent,soourclaim followsfromthecomparisontest. 48.4.Exercises.(1)Is 1sin xdx convergentordivergent? (2)Is 0 5xŠ 1 5dx convergent? (3)Is ŠxŠ 2dx convergent? (4)Is 0xeŠ xdx convergent? (5)Is ŠxeŠ x2dx convergent? (6)Is 0 1 ex+ xdx convergent? (7)Is 2 1 x ln xdx convergent? (8)Is 4 0 1 x2Š 8 x +7dx convergent? (9)Compute 0eŠ axdx ,where a isa“xednonnegativerealnumber. (10)Compute 3 0ln xdx (11)Compute 3 0(ln x )2dx (12)Compute 3 0 ln x xdx (13)Compute 1 Š 3 1 x +3dx (14)Compute 0 1 x2+1dx (15)Compute 0 ex e2 x+9dx (16)Compute 1 0 1 1 Š x2dx (17)Istherearealnumber a suchthat 0xadx isconvergent? (18)Provethatif f isan even function,and 0f ( x ) dx isconvergent,then Šf ( x ) dx isconvergentand Šf ( x ) dx = 2 0f ( x ) dx (19)Stateandsolvetheanalogousversionofthepreviousexercise forthecaseof odd functions f (20)Showanexampleofafunction f forwhich Šf ( x ) dx =limRR Š Rf ( x ) dx.

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CHAPTER8 SequencesandSeries 49.InniteSequences Asequencecanbethoughtofasanorderedlistofnumbers a1,a2, ...,an,an +1,... .Thesubscript n indicatesthepositionofanumber anin thesequence;forexample, a1isthe“rstelement, anisthe n thelement, andsoon. Definition 8.1(Sequence) A sequence isafunction f de“nedon thesetofallpositiveintegers;thatis,itisarulethatassignsanumber toeachpositiveinteger.If f ( n )= anfor n =1 2 ,... ,itiscustomary todenotetherangeof f bythesymbol { an} or { an} 1. Soasequencecanbede“nedbyspecifyingtherule an= f ( n )to calculatethe n thtermfromaninteger n .Forexample, an= n n +1 n n +1 1= 1 2 2 3 3 4 ,... an= ( Š 1)n n ( Š 1)n n 1= Š 1 1 2 Š 1 3 1 4 ,... an= qn{ qn} 0= { 1 ,q,q2,q3,... } Sequencescanalsobede“nedrecursively,thatis,byarelationthatallowsusto“nd anif am, m
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788.SEQUENCESANDSERIES Figure8.1. Setofpointsonthegraph y = x/ ( x + 1)correspondingtointegervalues x = n .Forlarge x x/ ( x +1)approaches1frombelow,andhence n/ ( n +1)= 1 / (1+1 /n ) 1as n .Thedierence1 Š n/ ( n +1)= 1 / ( n +1)canbemadesmallerthanany(small)number > 0for all n>N andsomeinteger N decreaseswithincreasing n andhencecanbemadesmallerthanany preassignedpositivenumber forall n>N ,where N dependson Forexample,put =10Š 2.Thenthecondition1 Š an< impliesthat 1 / ( n +1) < or1 / Š 1 99.If =10Š 4,then1 Š an< 10Š 4forall n>N =9999.In otherwords,nomatterhowsmall is,thereisonlya “nite numberof elementsofthesequencethatlieoutsidetheinterval(1 Š 1+ ).In thiscase,thesequenceissaidtoconvergetothelimitvalue1. Definition 8.2(LimitofaSequence) Asequence { an} hasthe limit a if,forevery > 0 ,thereisacorrespondinginteger N suchthat | anŠ a | < forall n>N .Inthiscase,thesequenceissaidtobe convergent ,andonewrites limn an= a or an a as n Ifasequencehasnolimit,itiscalled divergent Onecansaythat asequence { an} convergestoanumber a ifand onlyifeveryopenintervalcontaining a hasallbut“nitelymanyofthe elementsof { an} Theorem 8.1(UniquenessoftheLimit) Thelimitofaconvergent sequenceisunique: limn an= a andlimn an= a= a = a.

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49.INFINITESEQUENCES79 Figure8.2. De“nitionofthelimitofasequence.The dotsindicatenumericalvalues an(verticalaxis).The integer n increasesfromlefttoright(horizontalaxis). Theconvergenceof antoanumber a meansthat,for anysmall > 0,thereisaninteger N suchthat all the numbers an, n>N ,lieintheinterval( a Š ,a + ). Itisclearthat N dependson .Generally,asmaller requiresalarger N Proof. Fix > 0.Then,bythede“nitionofthelimit,thereare numbers N and Nsuchthat | anŠ a | if n>N and | anŠ a| if n>N.Hence,bothinequalitiesholdfor n> max( N,N)andforall such n : 0 | a Š a| = | a Š an+ anŠ a|| anŠ a | + | anŠ a| < 2 Thisinequalityshowsthatthe nonnegative number | a Š a| issmaller than any preassignedpositivenumber,whichmeansthat | a Š a| =0 or a = a. Sinceasequenceisafunctionde“nedonallpositiveintegers,there isagreatdealofsimilaritybetweentheasymptoticbehaviorofafunction f ( x )as x andasequence an= f ( n ). Theorem 8.2(LimitsofSequencesandFunctions) Let f bea functionon (0 ) .Supposethat limx f ( x )= a .If an= f ( n ) where n isaninteger,then limn an= a Thevalidityofthetheoremfollowsimmediatelyfromthede“nition ofthelimitlimx f ( x )= a (i.e.,given > 0,thereisacorresponding number M suchthat | f ( x ) Š a | < forall x>M )bynotingthatthe rangeof f ( x )containsthesequence an= f ( n ).

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808.SEQUENCESANDSERIES Example 8.1 Findthelimitofthesequence an=ln n/n ifitexists orshowthatthesequenceisdivergent. Solution: Considerthefunction f ( x )=ln x/x sothat an= f ( n )for allpositiveintegers.Hence, limn ln n n =limx ln x x =limx 1 /x 1 =0 wheretheindeterminateform arisingfromln x/x as x has beenresolvedbymeansoflHospitalsrule.NotethatlHospitalsrule appliesnottosequencesbuttofunctionsofarealvariable. Followingtheanalogybetweenthelimitsofsequencesandfunctions,onecanselectaparticularclassofdivergentsequences. Definition 8.3(In“niteLimits) Thelimit limn an= means that,foreverypositivenumber M ,thereisacorrespondinginteger N suchthat an>M forall n>N .Similarly,thelimit limn an= Š meansthat,foreverynegativenumber M ,thereisacorresponding integer N suchthat anN Example 8.2 Analyzetheconvergenceofthepowersequence an= 1 /np,where p isreal. Solution: Put f ( x )=1 /xpfor x> 0.Then an= f ( n )andtherefore limn 1 np=limx 1 xp= 0if p> 0 1if p =0 if p< 0 Example 8.3 Analyzetheconvergenceofthesequence an= qn, n =0 1 ,... ,where q isreal. Solution: Suppose q> 0.Put f ( x )= qx= ex ln q.Fromtheproperties oftheexponentialfunction,itfollowsthat eax if a =ln q> 0, eax=1if a =ln q =0,and eax 0if a =ln q< 0.Therefore, an if q> 1, an=1 1if q =1,and an 0if0
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49.INFINITESEQUENCES81 Š asitoscillatestakingever-increasingpositiveandnegativevalues. Thus, limn qn= 0if q ( Š 1 1) 1if q =1 if q> 1 andthesequencedoesnotconvergeif q Š 1. Example 8.4 Findthelimitofthesequence an=(1+ q/n )1 /n. Solution: Let f ( x )=(1+ q/x )xfor x> | q | tomake1+ q/x positive forany q .Then limn an=limx 1+ q x x=limu 0+(1+ qu )1 /uwherethesubstitution x =1 /u hasbeenmade.To“ndthelatterlimit, considerln f ( x )=ln(1+ qu )1 /u=ln(1+ qu ) /u .UsinglHospilalsrule, limu 0+ln(1+ qu ) u =limu 0+q/ (1+ qu ) 1 =limu 0+q 1+ qu = q. Itfollowsfromln f ( x ) q that f ( x ) eqas x and,hence, limn 1+ q n n= eqforallreal q 49.2.Subsequences.Givenasequence { an} ,considerasequence { nk} ofpositiveintegerssuchthat n1
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828.SEQUENCESANDSERIES tonumbers a and b ,respectively,and c isaconstant,then limn ( an+ bn)=limn an+limn bn= a + b, limn ( can)= c limn an= ca, limn ( anbn)=limn anlimn bn= ab, limn an bn= limn an limn bn= a b if b =0 limn ( an)p=(limn an)p= apif p> 0and an> 0 Itshouldbeemphasizedthatthe convergence ofthesequences anand bniscrucialfortheserelationstohold.Forexample,let bn= n and an= q Š n where q isanumber.Thenboththesequencesdiverge.In particular, bn and anŠ as n .However, an+ bn= q and,hence, an+ bn q as n .Evidently,therelation q = Š Žmakesnosense.Furthermore,take c =0sothatthesequence can=0 n =0isaconstantsequenceand,hence,convergesto0.The relation0=0 Žwouldnotmakeanysenseeither.Similarly,put an= n and bn= q/n .Then andiverges, an ,whilethesequence bnconverges bn 0as n 0.Howeverthesequenceoftheproducts isaconstantsequence anbn= q and,hence,converges.Therelation  q =0 Žmakesnosense.If an= pn q> 0,and bn= n ,then an and bn as n ,wherethesequenceoftheratios converges: an/bn= q .Therelation q = Žismeaningless. Thesqueezetheoremalsoappliestosequences(seeFig.8.3). Theorem 8.4(SqueezeTheorem) If cn an bnfor n>N and limn bn=limn cn= a ,then limn an= a ,where a canalsobe + or Š Example 8.5 Findthelimitof an=sin( / n ) Solution: Since Š x sin x x if x 0,onehas cn= Š / n an / n = bn,where cn 0and bn 0as n (seeExample 8.2).Bythesqueezetheorem,sin( / n ) 0as n Theorem 8.5 If limn | an| =0 ,then limn an=0 Thistheoremfollowsdirectlyfromthede“nitionofthelimitofa sequencewhere a =0.Thenextresultprovidesaconvenienttoolto calculatelimitsofsequencesusingcontinuousfunctions. Theorem 8.6 If an a as n andthefunction f iscontinuousat a ,then limn f ( an)= f ( a ) .

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49.INFINITESEQUENCES83 Figure8.3. Thesqueezetheorem.Thedotsindicatenumericalvalues(verticalaxis)ofthesequences bn(blue), cn(black),and an(red).Theinteger n increases fromlefttoright(horizontalaxis).Thesequences bnand cnconvergetoanumber a .Thismeansthatthedierences | bnŠ a | and | cnŠ a | canbemadearbitrarilysmallfor all n N andsomeinteger N .Since cn an bn,the dierence | anŠ a | isalsoarbitrarilysmallforall n N Bythede“nitionofthelimit,thesequence anmustconvergeto a ,too. Thistheoremassertsthatifacontinuousfunctionisappliedtothe termsofaconvergentsequence,theresultisalsoconvergent. Proof. Thecontinuityof f at a meansthatlimx af ( x )= f ( a ) or,bythede“nitionofthislimit,forany > 0,thereisacorresponding > 0suchthat | f ( x ) Š f ( a ) | < whenever | x Š a | < .Havingfound such ,put = and,bythede“nitionofthelimitlimn an= a foranysuch > 0,thereisacorrespondinginteger N suchthat | anŠ a | <= if n>N .Therefore,forany > 0,onecan“nda correspondinginteger N suchthat | f ( an) Š f ( a ) | < forall n>N whichmeansthatlimn f ( an)= f ( a ). Example 8.6 Findthelimitofthesequence an=exp(1 /n2) Solution: Considerthesequence bn=1 /n2.Then limn bn=limx 1 x2=0 Put f ( x )= ex.Then an= f ( bn).Bycontinuityoftheexponential function, limn an=exp(limn bn)= e0=1

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848.SEQUENCESANDSERIES 49.4.Exercises.(1)Findaformulaforthegeneralterm anofthesequences: { an} = 1 Š 1 3 1 5 Š 1 7 1 9 Š 1 11 ,... { an} = 1 1 2 1 4 1 8 1 16 ,... { an} = Š 1 4 2 7 Š 3 10 4 13 Š 5 16 ,... (2)Let { an} convergetoanumber a .Provethatthesequences bn= an + k,where k isa“xedpositiveinteger, cn= an2,and dn= a2nconvergeand“ndtheirlimits. (3)Showthatthesequence an=(2 n Š 1) / ( n +1)convergesto 2.Determinehowmanytermsof anlieoutsidetheinterval (2 Š 2+ )if =0 1and =0 01. (4)Showthatthesequence an=( n2Š 2( Š 1)nn +1) / ( n2+1) convergesto1.Given2 >> 0,“ndthenumberofterms ofthesequencethatlieoutsidetheinterval(1 Š 1+ )asa functionof Inthefollowingexercises,determinewhetherthesequence anconvergesordivergesforthespeci“edrangesofparameters (iftherangeofaparameterisnotgiven,assumethatitcan beanyrealnumber).Ifthesequenceconverges,“nditslimit. (5) an=2n. (6) an=2nŠ ( Š 1)n2n. (7) an=(3 Š 5 n2) / (1+ n2). (8) an= Pm( n ) /Qk( n ),where Pmand Qkarepolynomialsofdegree m and k ,respectively. (9) an=1 / n2+ n +1. (10) an=( n2+1)1 / 3/ ( n3+1)1 / 2. (11) an=( n +2)3/ ( n6+ n2+1)1 / 2. (12) an=3 ( n3+1) / (8 n3+4 n2+2 n +1). (13) an=[ Pm( n )]q/ [ Qk( n )]p,where q and p arepositivenumbers and Pmand Qkarepolynomialsofdegrees m and k ,respectively. (14) an=(2n+3n) eŠ n. (15) an=(2n+3n) eŠ 2 n. (16) an=(2n+3n+5n) / (2np +3n+5np ),where q and p arepositive numbers. (17) an=(2Š n+3Š n+5Š n) / (2Š np +3Š n+5Š nq ),where q and p arepositivenumbers.

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50.SPECIALSEQUENCES85 (18) an= Pm( pn) /Qk( qn),where Pmand Qkarepolynomialsof degree m and k ,respectively,and0 0. (33) an= n2n e Š n2Š n (34) an= bnsin( p/bn),where bn as n (35) an=( bnŠ tan bn) / ( bn)p,where p> 0and bn 0as n (36) an= f ( bn) /g ( bn),wherethefunctions f and g aretwicedifferentiableandhavearoot b suchthat f( b )= g( b )=0, g( b ) =0,and bn b as n (37)Provethatasequencethatisthesumofconvergentanddivergentsequencesdiverge. 50.SpecialSequences Theorem 8.7(SpecialSequences) Let p and q berealnumbers. limn n p =1if p> 0 (8.1) limn n n =1 (8.2) limn nq pn=0if p> 1 (8.3) limn n nn=0 (8.4) limn qn n =0 (8.5)

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868.SEQUENCESANDSERIES Proof. (8.1) .If p> 1,put an=n p Š 1.Then an> 0and,bythebinomial theorem, p =(1+ an)n=1+ nan+ n ( n Š 1) an/ 2+ + nan Š 1 n+ an n. Alltermsintherightsidearepositive.Thereforebyretainingonlythe “rsttwoterms,asmallernumberisobtained: 1+ nan (1+ an)n= p. Itfollowsfromthisinequalitythat 0 1. (8.2) .Put an=n n Š 1.Then an 0and,bythebinomialtheorem, n =(1+ an)n n ( n Š 1) 2 a2 n. Hence,for n 2, 0 an 2 n Š 1 Bythesqueezetheorem, an 0orn n = an+1 1as n 0. (8.3) .Considerthefunction f ( x )= xqeŠ cx,where c> 0.Bythe asymptoticpropertyoftheexponentialfunction, f ( x ) 0as x forany q ;theexponentialgrowsfasterthananypowerfunction(which hasbeenprovedinCalculusI).Since an= f ( n )for c =ln p> 0if p> 1,oneconcludesthat limn nq pn=limn nqeŠ n ln p=limx xqeŠ x ln p=0 (8.4) .Thefollowinginequalityholds: an= n nn= 1 2 3 n n n n n = 1 n 2 n 3 n n n 1 n 0 0,thenthereisapositiveinteger k suchthat k Š 1 q
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50.SPECIALSEQUENCES87 Figure8.4. Thesequenceintheleftpanelismonotonicallyincreasing,andthesequenceintherightpanelis monotonicallydecreasing. thatis, k isthesmallestpositiveintegersuchthat q/k< 1.The followinginequalityholds: an= qn n = q 1 q 2 q k Š 1 q k q n qk Š 1q k q n Š 1 q n qk Š 1q n = qk n Bythesqueezetheorem,0 1 ,ifitexists orshowthatthesequencediverges. Solution: limn n nq=limn (n n )q=(limn n n )q=1q=1 by(8.2)andthebasiclimitlaws. 50.1.MonotonicSequences.Definition 8.4(MonotonicSequences) Asequence anissaidto be monotonicallyincreasingif an an +1, monotonicallydecreasingif an an +1forall n =1 2 ,... Theclassof monotonic sequencesconsistsoftheincreasingandthe decreasingsequences. Example 8.8 Showthatthesequence an= n/ ( n2+1) ismonotonicallydecreasing.

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888.SEQUENCESANDSERIES Figure8.5. Aboundedsequence.Thedotsindicate numericalvaluesof an(verticalaxis).Theinteger n increasesfromlefttoright(horizontalaxis).Allthenumbers anlieintheinterval: m an M Solution: Theinequality an an +1mustbeestablished.Itisequivalenttothefollowinginequalitiesobtainedbycross-multiplication: n +1 ( n +1)2+1 n n2+1 ( n +1)( n2+1) n [( n +1)2+1] n3+ n2+ n +1 n3+2 n2+2 n 1 n2+ n. Thelatterinequalityistruefor n 1.Therefore, an +1 an(infact, thestrictinequality an +1
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50.SPECIALSEQUENCES89 Completenessaxiomforthesetofrealnumbers.The completenessaxiom forthesetofrealnumberssaysthatif S isanonemptysetofreal numbersthathasanupperbound M ( x M forall x S ),then S hasa leastupperbound .Byde“nition,thenumber a isaleastupper upperboundof S if,forany > 0, a Š isnotanupperboundof S .Theleastupperboundiscalledthe supremum of S anddenoted sup S .Naturally,sup S M foranyupperbound M of S .If S has alowerbound m ,thenitalsohasthe greatestlowerbound ,denoted inf S (the in“mum of S ).Thenumberinf S isalowerboundof S such thatinf S + isnotalowerboundof S foranypositive > 0;thatis, m inf S foranylowerbound m of S .Thecompletenessaxiomisan expressionofthefactthatthereisnogaporholeintherealnumber line. Theorem 8.8(MonotonicSequenceTheorem) Suppose { an} is monotonic.Then { an} convergesifandonlyifitisbounded. Proof. Suppose an an +1(theproofisanalogousintheother case).Let S betherangeof { an} .If { an} isbounded,let a =sup S betheleastupperboundof S (itexistsbythecompletenessaxiom). Then an a forall n 1.Bythede“nitionofsup S ,forevery > 0, theinterval( a Š ,a ]shouldcontainanelementof S (otherwise, a Š wouldbeanupperboundof S ).Thereforethereshouldexistaninteger N suchthat a Š
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908.SEQUENCESANDSERIES Figure8.6. Monotonicsequencetheorem.Abounded monotonicsequencewithnumericalvaluesindicatedby dots(verticalaxis).Theinteger n increasesfromleftto right.If a =sup { an} istheleastupperboundofall an, then,for any number > 0, a Š isnotanupperbound ofthesequence.Since anincreasesmonotonically,there isaninteger N suchthat all thenumbers an, n N ,are greaterthan a Š andhencelieintheinterval( a Š ,a ]. Thismeansthat anconvergesto a with n =1,onecanestablishthestatementfor n =2, n =3,ansoon. Thisisthebasicideaofmathematicalinduction.Usingtherecurrence relation, ak +1>ak= 1 2 ( ak +1+3) > 1 2 ( ak+3)= ak +2>ak +1. Thus,thesequenceisindeedmonotonicallyincreasing.Ifithappensto bebounded,thenitconverges.Again,mathematicalinductionturns outtobehelpful.The“rsttermssuggestthat an< 3.Thisistruefor n =1.Supposetheinequalityistruefor n = k .Letustrytoprove thatthishypothesisimpliesthattheinequalityholdsfor n = k +1. Usingtherecurrencerelation, ak< 3= 1 2 ( ak+3) < 1 2 (3+3)= ak +1< 3 Thus,thesequenceismonotonicandboundedandhenceconverges.If thesequence anconvergesto a ,thensodoesthesequence an + kforany integer k (thesequence bn= an + kisasubsequenceoftheconvergent sequence anandhence bnconvergesto a byTheorem8.3).Sincethe existenceofthelimithasbeenestablished,thelimitsofbothsidesof therecurrencerelationmustcoincidebythebasiclimitlaws: limn an +1= 1 2 (limn an+3)= a = 1 2 ( a +3)= a =3 .

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50.SPECIALSEQUENCES91 Thus, an 3as n Example 8.10 Investigatetheconvergenceofthesequencede“ned bytherecurrencerelation a1= 2 and an +1= 2+ an. Solution: The“rstfewtermsofthesequencesuggestthatthesequenceisincreasing: a1= 2 0.

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928.SEQUENCESANDSERIES (15) an=n Pm( n ),where Pmispositivepolynomialofdegree m> 1. (16)Provethateveryconvergentsequenceisbounded. In(17)…(22),determinewhetherthesequenceismonotonicornot monotonic.Isthesequencebounded? (17) an=( Š 2)n. (18) an=( Š 1)nn (19) an= neŠ n. (20) an= n +1 n. (21) an=sin( qn ) /n ,where q isreal. (22) an=(1+ q/n )n,where q isreal. In(23)…(27),“ndthelimitofthesequenceorshowthatitdoesnot exist. (23) a1=1and an +1=4 Š an. (24) a1=1and an +1=1 / (1+ an). (25) a1=1and an +1=3 Š 1 /an. (26) a1=2and an +1=1 / (3 Š an). (27) a1=1and an +1=1+1 / (1+ an). (28)Thesizeofanundisturbed“shpopulationhasbeenmodeled bytheformula pn +1= bpn/ ( a + pn),where pnisthe“shpopulationafter n yearsand a and b arepositiveconstantsthat dependonthespeciesandtheenvironment.Supposethat p0> 0.Showthat pn +1< ( b/a ) pn.Thenprovethat pn 0 if a>b ;thatis,thepopulationdiesout.Finally,showthat pn b Š a if b>a Hint: Showthat pnisincreasingand bounded,0 b Š a ,then pnisdecreasingandbounded, pn>b Š a 51.Series51.1.BasicDenitionsandNotation.Withasequence { an} ,onecan associateasequence { sn} ,where sn=nk =1ak= a1+ a2+ + an. Thesymbol (8.6)n =1an= a1+ a2+ a3+

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51.SERIES93 iscalledan in“niteseries ,orjusta series .Thenumbers snarecalled the partialsums oftheseries(8.6).Thelimitsofsummationareoften omittedtodenoteaseries.Thesymbol analsostandsforanin“nite series n =1an.If { sn} convergesto s ,thentheseriesissaidto converge andonewritesn =1an= s orlimn nk =1ak= s. Thenumber s iscalledthe sumoftheseries .Ifthesequenceofpartial sums { sn} diverges,theseriesissaidtodiverge.Itshouldbeunderstoodthat s is thelimitofasequenceofsums ,anditisnotobtained merelybyaddition. Forexample,thesequenceofpartialsumsfortheseries ( Š 1)nis s1= Š 1, s2= Š 1+1=0, s3= s2Š 1= Š 1,or,generally, sn=(( Š 1)nŠ 1) / 2.Thissequencedivergesasithastwosubsequences s2 n=0and s2 n Š 1= Š 1,whichconvergetodierentnumbers,0 = Š 1. Ifonesimplyusesaddition,dierentvaluesforthesumoftheseries maybeobtained:n =1an=( a1+ a2)+( a3+ a4)+( a5+ a6)+ =0+0+ =0 ,n =1an=( a1+ a2+ a3)+( a4+ a5+ a6)+ = Š 1 Š 1 Š = Š ,n =1an= a1+( a2+ a3)+( a4+ a5)+ = Š 1+0+0+ = Š 1 Generally,bygroupingtermsinthesumindierentways(according totheassociativityofaddition),thesumisfoundtobe any integer! Thereaderisadvisedtoverifythis.Thus,theadditionrulescannot generallybeappliedtoevaluatethesumofaseries.51.2.GeometricSeries.Takeapieceofropeoflength1m.Cutitin half.Keeponehalfandcuttheotherhalfintwopiecesofequallength. Keepdoingthis,thatis,keepingonehalfandcuttingtheotherhalfin twoequal-lengthpieces.Thetotallengthoftheretainedpiecesis 1 2 + 1 4 + 1 8 + = 1 2 1+ 1 2 + 1 4 + = 1 2n =01 2n. Thisseriesmustconverge.Thepartialsum snhereisthetotallengthof retainedpieces.Thesequence { sn} ismonotonicallyincreasing(after eachcutpieceofropeisadded)andboundedbythetotallength1.So

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948.SEQUENCESANDSERIES itconverges.Fromthegeometry,itisalsoclearthat1 Š sn=1 / 2n, where n isthenumberofcuts,andhence sn 1asonewouldexpect (thetotallengthoftherope).Soitisconcludedn =01 2n=2 Thisseriesisanexampleofthe geometricseries : 1+ q + q2+ q3+ =n =0qn, where q isanumber.Thegeometricseriesdoesnotconvergeforany valueof q Theorem 8.9(ConvergenceofaGeometricSeries) Ageometric series n =1qnconvergesif | q | < 1 ,and,inthiscase,n =0qn= 1 1 Š q | q | < 1 andtheseriesdivergesotherwise. Proof. If q =1,thesequenceofpartialsumsobviouslydiverges. If q =1,onehas sn=1+ q + q2+ + qn Š 1= qsn= q + q2+ q3+ + qn. Subtractingtheseequations,oneinfers snŠ qsn=1 Š qn= sn= 1 Š qn 1 Š q Therefore, limn sn=limn 1 Š qn 1 Š q = 1 1 Š q Š 1 1 Š q limn qn. Ithasbeenfound(Example8.3)thatthesequence an= qnconverges onlyif | q | < 1,and,inthiscase, qn 0as n .If | q | 1,the geometricseriesdiverges. Example 8.11 Analyzetheconvergenceoftheseries 4 Š8 3+16 9Š32 27+ Solution: Theseriescanbewrittenintheform4 q0+4 q1+4 q2+4 q3+ ,where q = Š 2 / 3.Soitspartialsumsarefourtimesthepartial sumofthegeometricserieswith q = Š 2 / 3.Therefore,n =14 Š 2 3 n=4n =1 Š 2 3 n= 4 1 Š ( Š2 3) = 12 5

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51.SERIES95 Whenrealnumbersarepresentedindecimalfrom,oneoftenencountersasituationwhenanumberhasarepeatedpatternofdecimal places.Take,forexample,thenumber1 2131313 ... ;thatis,thecombination13repeatsitselfinalldecimalplacesstartinginthesecond decimalplace. Example 8.12 Isthenumber 1 2131313 ... rationalorirrational? Ifitisrational,writeitasaratioofintegers. Solution: Byde“nitionofthedecimalrepresentation, 1 2131313 ... =1 2+ 13 103+ 13 105+ 13 107+ =1 2+ 13 103 n =0 1 100 n=1 2+ 13 103100 99 = 12 10 + 13 990 = 1201 990 51.3.NecessaryConditionforaSeriestoConverge.Thefollowingtheoremfollowsfromthelimitlawsappliedtothesequencesofpartial sums. Theorem 8.10(PropertiesofSeries) Supposethattheseries anand bnareconvergentandtheirsumsare s and t ,respectively.Let c beanumber.Thentheseries ( an+ bn) and canconvergeand ( an+ bn)= an+ bn= s + t ( can)= c an= cs. Indeed,if { sn} and { tn} arethesequencesofpartialsumsofthe series anand bn,respectively,thenthepartialsumsoftheseries ( an+ bn)and ( can)are sn+ tnand csn,respectively.Bythelimit laws, sn+ tn s + t and csn cs Notethattheconvergenceoftheseries ( an+ bn)does not imply theconvergenceof anand bn.Forexample,put an=1and bn= Š 1.Then sn= n and tn= Š n .Thesequence sn+ tn=0 convergesto0.Therefore ( an+ bn)=0.However,thesequences sn= n and tn= Š n divergeandsodotheseries an= 1and bn= ( Š 1).Theequality ( an+ bn)= an+ bnbecomes meaningless(0= Š Ž).Recallthatthebasiclimitslawshold onlyfor convergent sequences(Section49.3).Thisshowsthattherules ofalgebrafor“nitesumsare not generallyapplicabletoseries.Only seriesfromaspecialclassof absolutelyconvergent series,discussedlater, behaveprettymuchas“nitesums.

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968.SEQUENCESANDSERIES Everytheoremaboutsequencescanbestatedintermsofseriesand viceversa.Thesequence { an} canbeexpressedviathepartialsums oftheseries anbyputting a1= s1and an= snŠ sn Š 1,n> 1 Inparticular,iftheseriesconverges,meaningthat sn s as n thenalso sn Š 1 s .Bythebasiclimitlaws,thelimitonbothsidesof theaboverecurrencerelationmaybetaken: limn an=limn ( snŠ sn Š 1)=limn snŠ limn sn Š 1= s Š s =0 Thus,foraconvergentseries an,thesequence { an} necessarilyconvergesto0. Theorem 8.11(NecessaryConditionforaSeriestoConverge) If theseries anconverges,then limn an=0 Theconverseisnotgenerallytrue.Theconditionlimn an=0is notsucientforaseriestoconverge.However,itcanstillbeusedas atestfordivergenceofaseries. Corollary 8.1(TestforDivergenceofaSeries) Ifthelimit limn andoesnotexistorif limn an =0 ,thentheseries andiverges. Example 8.13 Showthattheseries n3/ (3 n3+1) diverges. Solution: limn an=limn n3 3 n3+1 =limn 1 3+1 n3= 1 3 =0 sotheseriesdiverges. Ifthenecessaryconditionissatis“ed,theseriesmayconvergeor diverge.Thesequenceofpartialsumshastobeanalyzed. Example 8.14 Findthesumoftheseries n =1 1 n ( n +1)ifitexists orshowthatitdoesnotexist.

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51.SERIES97 Solution: Thenecessaryconditionforconvergenceisevidentlysatis“ed.Sothesequenceofpartialsumshastobeanalyzedforconvergence: sn=nk =11 k ( k +1) =nk =1 1 k Š 1 k +1 = 1 Š 1 2 + 1 2 Š 1 3 + 1 3 Š 1 4 + + 1 n Š 1 n +1 =1 Š 1 n +1 1as n Sothesequence { sn} convergesto1andhencen =11 n ( n +1) =1 Thisexampleisaparticularcaseofa telescopicseries Theorem 8.12(ConvergenceofaTelescopicSeries) Atelescopic series n =1( anŠ an +1) convergesif limn an= a ,and,inthiscase,n =1( anŠ an +1)= a1Š a, Theproofisanalogoustotheaboveexampleandbasedonthefact thatthesequenceofpartialsumsofatelescopicseries sn= a1Š an +1convergesto a1Š a .Thedetailsarelefttothereaderasanexercise.51.4.Exercises.In(1)…(4),“ndthesequenceofpartialsumsoftheseriesandinvestigate itsconvergence. (1)1 Š 1 2 + 1 4 Š 1 8 + + ( Š 1)n Š 1 2n Š 1+ (2) 1 2 + 1 3 + 1 4 + 1 9 + + 1 2n+ 1 3n + (3) 1 2 + 2 4 + 3 8 + + n 2n+ (4) 1 2 + 3 22+ 5 23+ + 2 n Š 1 2n+ Hint: Put n =1+1+ +1(thesumof n units)in(3)and(4),when calculatingthepartialsum,thenusepartialsumsofageometricseries.

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988.SEQUENCESANDSERIES In(5)…(9),determinewhetherthe(geometric)seriesconvergesordiverges( p> 0and q> 0).Findthesumoftheseriesifitexists. (5)n =0n 3n +1(6)n =1en 3n Š 1(7)n =0( Š 5)n 32 n(8)n =1qn+ pn ( p + q )n(9)n =01+2n+3n+4nŠ 5n 5nIn(10)…(14),determinewhethertheseriesconvergesordivergesby expressingitasatelescopicseries.Findthesumoftheseriesifit exists. (10)n =21 n2Š 1 (11)n =13 n2+3 n +2 (12)n =1ln n n +1 (13)n =11 (3 n Š 2)(3 n +1) (14)n =1 n +2 Š 2 n +1+ n In(15)…(20),determinewhethertheseriesconvergesordiverges.Ifit converges,“nditssum.Here p isapositivenumber, p> 0. (15)n =1k2 k2+ k +1 (16)n =12 Š 3n 5n(17)n =1n p (18)n =1(sin p )n(19)n =1en np(20)n =1 eŠ 2 n+ 4 n ( n +1) In(21)and(22),expressthenumberasaratioofintegers (21)1 23232323 .... (22)1 53525252 .... In(23)…(25),“ndthevaluesof x forwhichtheseriesconverges.Find thesumoftheseriesforthosevaluesof x (23)n =1xn 2n(24)n =1sinnx 3n(25)n =1( x Š 5)nIn(26)and(27),solvetheequation. (26)n =2(1+ x )Š n=3(27)n =0enx=9

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52.SERIESOFNONNEGATIVETERMS99 (28)Supposethat n =1= a .Let1
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1008.SEQUENCESANDSERIES obtained: sn f (1)+ 2 1f ( x ) dx + + n n Š 1f ( x ) dx = f (1)+ n 1f ( x ) dx. Similarly,takingthesumover k =1 2 ,...,n oftherightinequalityin (8.7),thelowerboundisderived: sn 2 1f ( x ) dx + 3 2f ( x ) dx + + n +1 nf ( x ) dx = n +1 1f ( x ) dx, Therefore (8.8) n +1 1f ( x ) dx sn f (1)+ n 1f ( x )forall n 1 Thisinequalityshowsthatthefollowingtheoremholds. Theorem 8.13(IntegralTest) Suppose f isacontinuous,positive,decreasingfunctionon [1 ) andlet an= f ( n ) .Thentheseries n =1anconvergesifandonlyiftheimproperintegral 1f ( x ) dx converges.Inotherwords, 1f ( x ) dx converges= n =1f ( n )converges 1f ( x ) dx diverges= n =1f ( n )diverges Proof. Iftheimproperintegralconvergestoanumber If,then by(8.8)thesequenceofpartialsumsisbounded, sn f (1)+ If,and monotonicallyincreases, sn sn+ f ( n +1)= sn +1.Therefore,itis convergentbyTheorem8.8.Iftheimproperintegraldiverges,then,for anynumber M> 0,thereisaninteger N suchthat n +1 1f ( x ) dx M forall n>N .Bytheleftinequalityof(8.8), M snforall n>N ; thatis, { sn} isamonotonicallyincreasing,unboundedsequenceand henceitdiverges. Figure8.7. Integraltest.Anillustrationofinequality(8.7).

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52.SERIESOFNONNEGATIVETERMS101 Remark. Supposethat an= f ( n ),where f ( x )isafunctionon [1 ),suchthatitiscontinuous,positive,anddecreasingon[ N, ), where N 1isaninteger.Then (8.9)n =1anconverges Nf ( x ) dx converges; thatis,theintegraltestappliesevenifthesequence anbecomesmonotonicallydecreasingonlyfor n N 1.Thisiseasytounderstand byisolatingthe“rst N Š 1termsintheseriesn =1an= a1+ a2+ + aN Š 1+n = Nan= a1+ a2+ + aN Š 1+n =1bn, where bn= aN + n Š 1.Convergenceof bnimpliesconvergenceof anandviceversaastheydierbya number .Put bn= g ( n ),where g ( x )= f ( x + N Š 1),whichisacontinuous,positive,decreasingfunction on[1 ),and 1g ( x ) dx = 1f ( x + N Š 1) dx = Nf ( u ) du bychangingtheintegrationvariable u = x + N Š 1.52.2.SpecialSeriesofNonnegativeTerms.Theorem 8.14(Convergenceofp-series) The p-seriesn =11 npconvergesif p> 1 anddivergesif p 1 Proof. If p 0,theseriesdivergesbecausethenecessaryconditionforconvergenceisnotful“lled, an if p< 0and an=1 =0if p =0.For p> 0,considerthefunction f ( x )= xŠ p,whichispositive, continuous,anddecreasingon[1 ),and a 1dx xp= 1 p Š 1 1 Š1 ap Š 1 if p =1 ln a if p =1 So,bytheintegraltest,theseriesconvergesif p> 1becausethe improperintegraldivergesif0

1(the limit a existsonlyif p> 1). Harmonicseries.Theseries nŠ pdivergesforall0


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1028.SEQUENCESANDSERIES Riemann’szetafunction.Thesumofa p -series ( p )= nŠ pdepends onthevalueof p> 1;thatis,thisseries de“nes afunctionon(1 ). Thisfunctioniscalled Riemannszetafunction Example 8.15 Investigatetheconvergenceoftheseries n =1( n +2)Š 3 / 2. Solution: Theseriescanbewrittenasn =11 ( n +2)3 / 2=n =31 n3 / 2= Š 1 Š 1 23 / 2+n =11 n3 / 2. Thelatterseriesisa p -seriesthatconvergesfor p =3 / 2 > 1. Theorem 8.15 Theseriesn =21 n (ln n )pconvergesif p> 1 ,anditdivergesif p 1 Proof. Considerthefunction g ( x )= x (ln x )pfor x> 1.Itsderivativereads g( x )=(ln x )p Š 1( p +ln x ).If p 0,then g( x ) > 0for all x> 1and g ( x )increases,whileitsreciprocal f ( x )=1 /g ( x )should decrease.If p< 0,then g( x ) > 0forall x>eŠ pandhence g ( x )increases,while f ( x )=1 /g ( x )decreasesif x>eŠ p> 1.Thus,forany p thereisaninteger N suchthatthefunction f ( x )=1 / [ x (ln x )p]iscontinuous,positive,anddecreaseson[ N, ).Bytheintegraltest(8.9), theseriesinquestionconvergesifandonlyiftheimproperintegral Ndx x (ln x )p= ln Ndu upconverges,wheretheintegrationvariablehasbeenchanged, u =ln x du = dx/x .Thisintegraldivergesif p 1andconvergesif p> 1,and theconclusionofthetheoremfollows. Ontheuseofthecomparisontestforimproperintegrals.Theimproper integralofthefunction f ( x )intheintegraltestoftencannotbeexplicitlyevaluated.Thecomparisontest(Theorem7.4)maybeusedto assesstheconvergenceoftheimproperintegral. Example 8.16 Investigatetheconvergenceoftheseries anwhere an=( n +1 Š n Š 1) /np, p> 0 .

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52.SERIESOFNONNEGATIVETERMS103 Solution: Let f ( x )= x +1 Š x Š 1 xp= x +1 Š x Š 1 xp x +1+ x Š 1 x +1+ x Š 1 = 2 xp( x +1+ x Š 1) > 0 for x 1.Theideaisto“ndlowerandupperboundsfor f ( x )asfunctionssuchthattheconvergenceoftheirimproperintegralsiseasyto analyze.Itfollowsfromfromtheobviousinequality x +1 > x Š 1 that 2 2 xp x +1 x/ 2forall x> 2. Thus, 1 xp +1 / 2 2 2.ByTheorem7.4thisinequalityimpliesthattheimproper integralof f ( x )converges ifandonlyif theimproperintegralofthe powerfunction xŠ p Š 1 / 2converges.Thelatteristhecaseif p +1 / 2 > 1 or p> 1 / 2.Thus,theseriesinquestionconvergesif p> 1 / 2and divergesotherwise. Example 8.17 Investigatetheconvergenceoftheseries n2Š npwhere p 1 Solution: Put f ( x )=2Š xp> 0if x 1.Bymonotonicityofthe exponentialfunctionandbytheobviousinequality xp x forall x 1 and p 1,thefunction f isboundedfromaboveby g ( x )=2Š x= eŠ x ln2.Itfollowsfromtheinequality0
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1048.SEQUENCESANDSERIES anconvergestoanumber s ,then n +1f ( x ) dx s Š sn nf ( x ) dx, where { sn} isthesequenceofpartialsums. Proof. The“rstinequalityisobtainedbytakingthelimit n in(8.8)withtheresult (8.10) 1f ( x ) dx n =1an f (1)+ 1f ( x ) dx, whichisalegitimateoperationbecause(8.8)holdsforall n andthe seriesconverges(andsodoestheimproperintegralbytheintegral test).Theremainderestimateisobtainedbysubtracting(8.8)from (8.10).Notethevalueoftheimproperintegraldoes not coincidewith thesum;itonlydeterminesaninterval(8.10)inwhichthesumofa serieslies. Example 8.18 Testtheseries n =1( n2+1)Š 1forconvergenceor divergence.Ifitconverges,estimateitssum. Solution: Put f ( x )=( x2+1)Š 1,whichisacontinuous,positive,decreasingfunctionon[1 ),suchthattheseriesinquestionis f ( n ). Therefore,theintegraltestapplies,andtheseriesconvergesbecause 1dx x2+1 =lima tanŠ 1x a 1=lima tanŠ 1a Š 4 = 2 Š 4 = 4 By(8.10),itssumliesintheinterval 4 s f (1)+ 4=1 2+ 4. Example 8.19 Testtheseries n =1neŠ nforconvergenceordivergence.Ifitconverges,estimateitssum. Solution: Considerthefunction f ( x )= xeŠ x.Since f( x )= eŠ xŠ xeŠ x=(1 Š x ) eŠ x 0if x 1,thefunctiondecreaseson[1 ),and theintegraltestappliestoassessconvergenceoftheseries f ( n ): 1xeŠ xdx = Š 1xdeŠ x= Š lima xeŠ x a 1+ 1eŠ xdx = 1 e + 1 e = 2 e wheretheintegrationbypartshasbeenusedtoevaluatetheintegral. Theseriesconvergestoanumber s thatliesintheinterval2 eŠ 1 s f (1)+2 eŠ 1=3 eŠ 1. Example 8.20 EstimatevaluesofRiemannszetafunction ( p ) Howmanytermsdoesoneneedtoretaininthepartialsum sntoapproximate ( p ) sothattheerrorislessthan 10Š N?

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52.SERIESOFNONNEGATIVETERMS105 Solution: Riemannszetafunctionisde“nedbythesumoftheseries ( p )= n =1nŠ p.For p> 1, 1dx xp=lima x1 Š p 1 Š p a 1=lima a1 Š p 1 Š p Š 1 1 Š p = 1 p Š 1 Since f (1)=1,by(8.10), 1 p Š 1 ( p ) p p Š 1 ByCorollary8.2, 0 ( p +1) Š sn ndx xp +1= 1 pnp. If ( p +1), p> 0,istobeapproximatedwithanerrornotexceeding 10Š N,thentheremaindershouldbelessthan10Š N,whichyieldsthe conditiononthenumberofterms:1 pnp< 10Š Nor np> 10N/p or n> 10N/p/p1 /p.Thesmallest n satisfyingthisinequalityis1plusthe integerpartof10N/p/p1 /p.Forexample,take p =1and N =2.Then (2) Š sn< 0 01forall n 101. 52.4.Exercises.In(1)…(15),determinewhethertheseriesconvergesordiverges. (1)n =1n2eŠ n(2)n =11 n ( n +4) (3)n =1eŠ n(4)n =1n2eŠ n(5)n =12Š3 n(6)n =11 n ( n +1)( n +2) (7)n =11 n9 / 8(8)n =2(ln n )4 n (9)n =21 Š n ln n n2(10)n =1n2Š 2 n Š 5 3 n7 / 3(11)n =11 n2Š 4 n +5 (12)n =1n n4+1 (13)n =1e1 /n n2(14)n =12 n +1 n ( n +1) (15)n =1tanŠ 1n n2+1

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1068.SEQUENCESANDSERIES In(16)…(26),determinethevaluesof p forwhichtheseriesisconvergent. (16)n =1p pn +100 ,p> 0(17)n =1 n + | p |Š n Š| p | np(18)n =1exp( Š np)(19)n =1aŠ np,a> 1(20)n =1| p |ln n(21)n =31 n ln n (ln(ln n ))p(22)n =1n (1+ n2)p(23)n =1npeŠ n(24)n =1 p n Š 1 n +1 (25)n =1npnŠ ln n(26)n =1nŠ (ln n )p(27)HowmanytermsoftheseriesinTheorem8.15doesonehave toaddtoapproximateitssumwithanerrorlessthan10Š N? In(28)…(30),estimatethesumoftheseriesand,in(28)and(29),also determinehowmanytermshastobeaddedtoapproximateitssum withanerrorlessthan0 01. (28)n =12Š n(29)n =12 n2+5 n4+5 n2+4 (30)n =21 n4Š 1 (31)Showthatthesequence an=1+ 1 2 + 1 3 + + 1 n Š ln n converges.Thelimitlimn an= iscalledthe Eulernumber Hints :(i)Use(8.8)toshowthatif snisthepartialsum oftheharmonicseries,then sn 1+ln n andhence an 1 (i.e.,thesequence { an} isbounded).(ii)Interpret anŠ an +1asadierenceofareastoshowthat { an} ismonotonic. 53.ComparisonTests Considertheseriesn =1an=n =11 n4+1 Ithastermssmallerthanthecorrespondingtermsoftheconvergent p -series:n =1bn=n =11 n4

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53.COMPARISONTESTS107 because an 0isarationalfunctionof n ,aconvenientchoice ofaseriesinthecomparisontestisa p -series: 0 < 2 n +1 3 n3+ n2+1 < 2 n +1 3 n3= 2 3 1 n2+ 1 3 1 n3. Theseriesn =12 n +1 3 n3= 2 3n =11 n2+ 1 3n =11 n3convergesasthesumoftwoconvergent p -series. Example 8.22 Testtheseries n =1( n +1 Š n ) forconvergence.

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1088.SEQUENCESANDSERIES Solution: Onehas an= n +1 Š n = ( n +1 Š n )( n +1+ n ) n +1+ n = 1 n +1+ n 1 2 n + n = 1 1+ 2 1 n = bn. The p -series 1 / n divergesandsodoestheseriesinquestionbythe comparisontest. Theorem 8.17(LimitComparisonTest) Supposethat anand bnareserieswithpositiveterms.Let c =limn ( an/bn) € If c =0 and bnconverges,then anconverges. € If 0 N bythede“nitionofthelimit.Hence, anN .If bnconverges,then anconvergesbythecomparison test.If c (0 ),then,bythede“nitionofthelimit,foranynumber c>> 0,thereisaninteger N suchthat c Š an bn < m = c Š < an bnN. Therefore, mbnN. Bythecomparisontest,convergenceof bnimpliesconvergenceof anduetotheinequality an 0,thereisaninteger N suchthat an/bn>M when n>N .Theinequality an>Mbnshowsthatdivergenceof bnimpliesdivergenceof anbythecomparisontest. Itisoftenhelpfultoinvestigatetheasymptoticbehaviorof anas n toidentifyasuitable bninthelimitcomparisontest(Recall thenotionofaslantasymptoteofafunction f ( x )inCalculusI). Example 8.23 Testtheseries n =1(2 n3 / 2+ n ) / n6+ n4+1 for convergence. Solution: Letus“ndtheasymptoticbehaviorof anas n .For large n ,thetopoftheratiois n3 / 2(2+ nŠ 1 / 2) 2 n3 / 2becausetheterm nŠ 1 / 2maybeneglectedascomparedto2.Theasymptoticbehavior

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53.COMPARISONTESTS109 willbedenotedby ,inparticular,2 n3 / 2+ n 2 n3 / 2.Thebottomof theratiobehavesas ( n6)1 / 2= n3.Therefore, an= 2 n3 / 2+ n n6+ n4+1 = 2 n3 / 2(1+1 2 n) n3 1+1 n2+1 n6 2 n3 / 2 n3=2 nŠ 3 / 2= bnintheasymptoticregion n .Thisshowsthat limn an bn=limn 1+1 2 n 1+1 n2+1 n6=1= c. Bythelimitcomparisontest,theseries anconvergesbecausethe p -series bn=2 nŠ 3 / 2converges. Example 8.24 Testtheseries n =1( n5+3n) / n3+5nforconvergence. Solution: Recallthatthepowerfunctionincreasesmoreslowlythan theexponentialfunction,thatis, npqŠ n 0as n forany q> 1 andany p .Hence,theasymptoticbehaviorof anis an= 3n(1+ n53Š n) 5n/ 2(1+ n35Š n)1 / 2 3n 5n/ 2= 3 5 n= bn. Thisshowsthattheratio an/bnconvergesto c =1as n .Bythe limitcomparisontest, andivergesbecausethegeometricseries bndiverges( q =3 / 5 > 1). 53.1.EstimatingSums.Let n =1bnbeaconvergentseriesofnonnegativeterms.Supposethat0 an bnforall n .Thenbythe comparisontest,theseries n =1anconverges.Thesumof ancan beestimatedbycomparingremaindersfortheseries bn.Indeed, put an= s and bn= t .Let { tn} and { sn} bethesequencesof partialsumsfor bnand an,respectively.Theremainderssatisfy theinequality: s Š sn= an +1+ an +2+ bn +1+ bn +2+ = t Š tn. Sotheaccuracyoftheapproximation s snisthesameorhigherthan thatoftheapproximation t tn.If,forexample,one“ndsthat n = N issucientfortheequality t = tNtobecorrectwithinaspeci“ederror, then s = sNisalsocorrectwithinthesameorevensmallererror.The remainderiseasytoestimateif bn= f ( n )andthefunction f issimple enoughtointegrate, t Š tn nf ( x ) dx (Corollary8.2).

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1108.SEQUENCESANDSERIES Example 8.25 Determinehowmanytermsareneededtoestimate thesumoftheseries n =1tanŠ 1( n2) / ( n3+1) correctwithinanerror notexceeding 10Š 4. Solution: ThefunctiontanŠ 1x ismonotonicallyincreasingfor x> 0 approachingasymptoticallythevalue / 2.Therefore, an= tanŠ 1( n2) n3+1 2 1 n3+1 2 1 n3= bn,n 1 Hence, s Š sn t Š tn 2 ndx x3= 4 n2 10Š 4 n 2 102 157 1 Sotheneededaccuracyisguaranteedif n =158orlarger. 53.2.Exercises.In(1)…(15),determinewhethertheseriesconvergesordiverges. (1)n =11 n2+3 n +2 (2)n =1 n +1 n2+3 n +2 (3)n =12n n100+2n/ 2(4)n =1n n5 / 3+ n1 / 3+1 (5)n =2n (ln n )4 n2+1 (6)n =2cos2( n ) n2(7)n =11+( Š 1)n n3 / 2+1 (8)n =11 3 n3+ n +1 (9)n =11+2n n2+2n(10)n =1e1 / n n +1 (11)n =1sin2 1 n (12)n =1n nn(13)n =11 n1+1 /n(14)n =1(n n Š 1)n(15)n =1n2 n In(16)…(20),determinethevaluesoftheparametersforwhichthe seriesconverges. (16)n =1sin2( xn ) n2(17)n =1( np+ n +1)q(18)n =1np 1+ q n n(19)n =1( n +1 Š n )p(20)n =1( n +1 Š n )pln n Š 1 n +1 (21)Assume an= PN( n ) /QM( n ),where PNand QMarepolynomialsofdegree N and M ,respectively,and QM( n ) > 0forall n .Investigatetheconvergenceoftheseries an.

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54.ALTERNATINGSERIES111 In(22)and(23),determinehowmanytermsoneneedsinthepartial sumtoestimatethesumoftheseriescorrectwithinanerrorlessthan 0 001. (22)n =1sin3n n3+ n (23)n =12n n +3n(24)Considerasequence { an} ,where ancantakeanyvaluefrom theset { 0 1 2 ,...,p Š 1 } and p> 1isaninteger.Themeaning oftherepresentation0 .a1a3a3... ofanumberwithbase p is that (8.11)0 .a1a3a3... = a1 p + a2 p2+ a3 p3+ When p =10,thedecimalsystemisobtained.Thebinary representationcorrespondsto p =2.TheMayaused p =20 (thenumberof“ngersandtoes).TheBabyloniansused p = 60.Showthattheseries(8.11)alwaysconverges. (25)Showthatif an> 0and anconverges,then ln(1+ an) converges,too. (26)Provethatiftheseries a2 nand b2 nconverge,thenthe series | anbn| and ( an+ bn)2converge. (27)Provethattheconvergenceof an,where an> 0,impliesthe convergenceof an/n (28)If anconvergesandifthesequence { bn} ismonotonicand bounded,provethat anbnconverges. (29)Provethatiflimn nan= c =0,thentheseries andiverges. (30)Lettheseriesofnonnegativeterms, anand bn,converge.Whatcanbesaidabouttheconvergenceoftheseries max( an,bn)and min( an,bn),wheremin( p,q )and max( p,q )arethesmallestandlargestnumbersinthepair ( p,q ),respectively,andmin( p,p )=max( p,p )= p ? 54.AlternatingSeries Definition 8.6(AlternatingSeries) Let { bn} beasequenceof nonnegativeterms.Theseriesn =1( Š 1)n Š 1bn= b1Š b2+ b3Š b4+ b5Š iscalledanalternatingseries.

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1128.SEQUENCESANDSERIES Forexample,theseries (8.12)1 Š 1 2 + 1 3 Š 1 4 + 1 5 + =n =1( Š 1)n Š 1 n isanalternatingseries.Itiscalledthe alternatingharmonicseries Theorem 8.18(AlternatingSeriesTest) Ifasequenceofpositive terms { bn} ismonotonicallydecreasingand limn bn=0 ,thenthe alternatingseries n =1( Š 1)n Š 1bnconverges: ( i ) bn +1 bnforall n ( ii )limn bn=0 = n =1( Š 1)n Š 1bnconverges Proof. Theconvergenceofthesequenceofpartialsums { sn} is tobeestablished.Considerasubsequenceofevenpartialsums { s2 k} Onehas s2= b1Š b2 0, s4= s2+( b3Š b4) s2,and,ingeneral, s2 k= s2( k Š 1)+( b2 k Š 1Š b2 k) s2( k Š 1) s2( k Š 2) s2 0 bythemonotonicityofthesequence { bn} .Thus,thesubsequence { s2 k} ismonotonicallyincreasing.Byregroupingthetermsinadierentway, onecanseethat s2 k= b1Š ( b2Š b3) Š ( b4Š b5) ŠŠ ( b2 k Š 2Š b2 k Š 1) Š b2 k b1becauseallnumbersinparenthesesarenonnegativebyhypothesis(i), whichshowsthat { s2 k} isalsobounded.Therefore,itconvergesbythe monotonicsequencetheorem: limk s2 k= s. Forthesubsequenceofoddpartialsums s2 k +1= s2 k+ b2 k +1,oneinfers bythelimitlawsandhypothesis(ii)that limk s2 k +1=limk s2 k+limk b2 k +1= s +0= s. Theconvergenceoftwoparticularsubsequencesofasequencetothe samenumber s doesnotgenerallyguaranteethatthesequenceconvergesto s (allitssubsequencesshouldconvergeto s ).Byde“nition, thelimitsof { s2 k} and { s2 k +1} meanthat,givenanynumber > 0, therearepositiveintegers N1and N2suchthat | s2 kŠ s | < if k>N1and | s2 k +1Š s | < if k>N2.Put N =max(2 N1, 2 N2+1).Then | snŠ s | < forall n>N ,whichmeansthat sn s as n Bythistest, thealternatingharmonicseries(8.12)converges becausethesequence bn=1 /n ismonotonicallydecreasingandconverges to0.

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54.ALTERNATINGSERIES113 Figure8.8. Alternatingseriestest.Anillustrationof itsproofwheretwosubsequences, s2 kand s2 k Š 1,ofthe sequence snofpartialsumsareanalyzedforconvergence. Example 8.26 Testtheseries n =1sin( n/ 2) /n forconvergence. Solution: Onehassin( n/ 2)=1 0 Š 1 0 1 ,... for n =1 2 3 4 5 ,... respectively,or,ingeneral,forodd n =2 k Š 1,sin( n/ 2)=( Š 1)k Š 1, whileforeven n =2 k ,sin( n/ 2)=sin( k )=0.Thus,theseriesin questionisanalternatingseries:n =1sin( n/ 2) n =n =1( Š 1)n Š 1 2 n Š 1 =n =1( Š 1)n Š 1bn,bn= 1 2 n Š 1 Thesequence { bn} ismonotonicallydecreasingand bn 0as n Sotheseriesconvergesbythealternatingseriestest. Scopeofthealternatingseriestest.Theorem8.18providesonlya sucient conditionforanalternatingseriestoconverge.Thereare convergentanddivergent alternatingseriesthatdo not satisfythehypothesis (i)ofTheorem8.18.Asanexampleofaconvergentseries,considerthe alternatingseriesn =1( Š 1)n Š 1bn=n =1( Š 1)n Š 11 Š ( Š 1)n n2=n =1 ( Š 1)n Š 1 n2+ 1 n2

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1148.SEQUENCESANDSERIES Thesequence { bn} isnotmonotonicbecause b2 k=0while b2 k +1> 0. Nevertheless,theseriesconvergesbyTheorem8.10becausetheseries ( Š 1)n Š 1/n2and 1 /n2converge.Theformerseriesconvergesby thealternatingseriestest,whilethelatteristhe p Š serieswith p =2. Ontheotherhand,considerthealternatingseriesn =1( Š 1)n Š 1bn=n =1( Š 1)n Š 11 Š ( Š 1)n n =n =1 ( Š 1)n Š 1 n + 1 n Hereagainthesequence { bn} ofnonnegativetermsisnotmonotonic. If snisapartialsumofthisseries,then sn= tn+ hnwhere tnisthe partialsumofthealternatingharmonicseries,whichhasbeenshown toconverge,and hnisthepartialsumoftheharmonicseries,whichis knowntodiverge.Thus,thesequence { sn} isthesumofconvergent anddivergentsequencesand,hence,itdiverges. Remark. Hypothesis(i)ofTheorem8.18maybeweakened (i) bn +1 bnforall n N forsomeinteger N 1.Indeed,n =1( Š 1)n Š 1bn= b1Š b2+ b3ŠŠ bN Š 1+n = N( Š 1)n Š 1bn= b1Š b2+ b3ŠŠ bN Š 1+( Š 1)N Š 1 n =1( Š 1)n Š 1cn, where cn= bn + N Š 1.Theseries ( Š 1)n Š 1bnand ( Š 1)n Š 1cndier byanumber,andthereforetheconvergenceofoneofthemimplies theconvergenceoftheother.Theseries ( Š 1)n Š 1cnconvergesby Theorem8.18as cn +1 cnforall n andlimn cn=limn bn + N Š 1= 0. Example 8.27 Testtheseries n =1( Š 1)n Š 1np/ ( n +1) forconvergenceif p< 1 Solution: Here bn= np/ ( n +1)and,for p< 1, limn bn=limn np n +1 =limn np Š 1 1+1 n=limn np Š 1=0 Sohypothesis(ii)ofTheorem8.18isful“lled.However,themonotonicityof { bn} isnotobvious.Toinvestigateit,considerthefunction f ( x )= xp/ ( x +1),where x 1.If f ( x )monotonicallydecreases,then sodoesthesequence bn= f ( n ).Thecondition f( x ) 0hastobe

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54.ALTERNATINGSERIES115 veri“ed: f( x )= pxp Š 1( x +1) Š xp ( x +1)2 0 ( p Š 1) xp+ pxp Š 1 0 p x (1 Š p ) If p 0,thisistrueas x 1.If0 N .Sotheseriesconverges forall p< 1. 54.1.EstimatingSumsofAlternatingSeries.Apartialsum snofany convergent alternatingseriescanbeusedasanapproximationofthe totalsum s ,butthisisnotofmuchuseunlesstheaccuracyofthe approximationisassessed.Thefollowingtheoremassertsthattheabsoluteerroroftheapproximation s sndoesnotexceedthevalueof bn +1. Theorem 8.19(AlternatingSeriesSumEstimation) If s = ( Š 1)n Š 1bnisthesumofanalternatingseriesthatsatis“es (i)0 bn +1 bnforall n and(ii)limn bn=0 then | s Š sn| bn +1. Proof. Intheproofofthealternatingseriestest,itwasfoundthat thesubsequence { s2 k} approachesthelimitvalue s frombelow, s2 k s Ontheotherhand,thesubsequence { s2 k Š 1} approachesthelimitvalue s fromabove(seeFig.8.8).Indeed, s1= b1, s3= s1Š b2+ b3 s1because b3 b2,and,ingeneral, s2 k +1= s2 k Š 1Š b2 k+ b2 k +1 s2 k Š 1;that is, { s2 k +1} ismonotonicallydecreasing.Thisshowsthatthesequence ofpartialsums snoscillatesaround s sothatthesum s alwayslies betweenanytwoconsecutivepartialsums snand sn +1asdepictedin Figure8.8.Hence, | s Š sn|| sn +1Š sn| = bn +1. Example 8.28 Estimatethenumberoftermsinapartialsum snneededtoapproximatethesumofthealternatingharmonicseries correctwithintheabsoluteerrornotexceeding 10Š N. Solution: Here, bn=1 /n .Hence,theapproximation s snhasthe neededaccuracyif | s Š sn| bn +1 10Š Nor1 / ( n +1) 10Š Nor n 10NŠ 1.

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1168.SEQUENCESANDSERIES Ifthemonotonicitycondition bn +1 bnholdsonlyif n N ,the conclusionofTheorem8.19alsoholdsonlyif n N .Indeed,inthenotationfromRemarkaboveExample8.27,put t = ( Š 1)n Š 1cn,where cn= bn + N Š 1.Let tnbeapartialsumfortheseries ( Š 1)n Š 1cn.Then s = sN Š 1+( Š 1)N Š 1t and sn= sN Š 1+( Š 1)N Š 1tn Š N +1for n N Therefore, | s Š sn| = | t Š tn Š N +1| cn Š N +2= bn +1forall n N.54.2.Exercises.In(1)and(2),provetheconvergenceoftheseriesand“nditssum. (1)1+ 1 2 Š 1 4 + 1 8 + 1 16 Š 1 32 + (2)1 Š 3 2 + 5 4 Š 7 8 + In(3)…(21),determinewhethertheseriesconvergesordiverges(here p isreal). (3)n =1( Š 1)n ln( n +3) (4)n =1( Š 1)nn n3+1 (5)n =1cos( n/ 2) n4 / 5(6)n =1( Š 1)nn ( n3 / 2+1)2 / 3(7)n =2( Š 1)nn (ln n )p(8)n =1( Š 1)nn3 2n(9)n =2( Š 1)n(ln n )p n (10)n =1( Š 1)nsin n (11)n =1( Š 1)nn2 n4+1 (12)n =1( Š 1)n n1+1 /nŠ n (13)n =1( Š 1)n(n n Š 1)n(14)n =1( Š 1)nnn n (15)n =1( Š 1)n n + p (16)n =1( Š 1)n np(17)n =1( Š 1)n( n2+ n +1) (2 n +3)2(18)n =1ln2n n sin( n/ 4)(19)n =1( Š 1)nsin2( n/ 2) n (20)n =2( Š 1)n n Š ( Š 1)n(21)n =1( Š 1)n n n In(22)…(24),“nd n forwhichtheapproximationbypartialsums s sniscorrectwithinanerrornotexceeding10Š N. (22)n =1( Š 1)n Š 1 4 n (23)n =1( Š 1)nn 10n(24)n =1( Š 1)nn1 / 3 n2 / 3+6

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55.RATIOANDROOTTESTS117 (25)Provethatthesumofthealternatingharmonicseriesisn =1( Š 1)n Š 1 n =ln2 Hint :Showthatapartialsumofthealternatingharmonicseriesis s2 n= h2 nŠ hn,where hn= an+ln n andthesequence { an} isde“ned inExercise52.4.16.Thenusetheresultofthelatterexercisetoprove that sn ln2as n 55.RatioandRootTests55.1.AbsolutelyConvergentSeries.Definition 8.7(AbsoluteConvergence) Aseries aniscalled absolutelyconvergent iftheseriesofabsolutevalues | an| isconvergent. Theabsoluteconvergenceisstrongerthanconvergence,meaning thatthereareconvergentseriesthatdonotconvergeabsolutely.For example,thealternatingharmonicseries an, an=( Š 1)n Š 1/n ,is convergent,butnotabsolutelyconvergentbecausetheseriesofabsolute values | an| =1 /n isnothingbuttheharmonicseries 1 /n ,whichis divergent(asa p Š serieswith p =1).Ontheotherhand,theabsolute convergenceimpliesconvergence. Theorem 8.20(ConvergenceandAbsoluteConvergence) Every absolutelyconvergentseriesisconvergent. Proof. Foranysequence { an} ,thefollowinginequalityholds; 0 an+ | an| 2 | an| because | an| iseither anor Š an.Itshowsthattheseries bn,where bn= an+ | an| ,convergesbythecomparisontestbecause 2 | an| = 2 | an| convergesif anconvergesabsolutely.Hence,theseries an= bnŠ | an| convergesasthedierenceoftwoconvergent series. Example 8.29 Testtheseries [sin n Š 2cos(2 n )] /n3 / 2forabsoluteconvergence. Solution: Makinguseoftheinequality | A + B || A | + | B | andthe propertiesthat | sin x | 1and | cos x | 1,oneinfers | an| = | sin n Š 2cos(2 n ) | n3 / 2 | sin n | +2 | cos(2 n ) | n3 / 2 3 n3 / 2. Theseriesofabsolutevalues | an| convergesbycomparisonwiththe convergent p Š series3 nŠ 3 / 2(here p =3 / 2 > 1).Sotheseriesin questionconvergesabsolutely.

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1188.SEQUENCESANDSERIES Definition 8.8(ConditionalConvergence) Aseries aniscalled conditionallyconvergent ifitisconvergentbutnotabsolutelyconvergent. Thus,allconvergentseriesareseparatedintotwoclassesofconditionallyconvergentandabsolutelyconvergentseries.Thekeydifferencebetweenpropertiesofabsolutelyconvergentandconditionally convergentseriesisstudiedinSection56.55.2.RatioTest.Theorem 8.21(RatioTest) Givenaseries an,supposethe followinglimitexists: limn an +1 an = c, where c 0 or c = € If c< 1 ,then anconvergesabsolutely. € If c> 1 ,then andiverges. € If c =1 ,thenthetestgivesnoinformation. Proof. If c< 1,thentheexistenceofthelimitmeansthat,for any > 0,thereisaninteger N suchthat Š < an +1 an Š c< = an +1 an 0 smallenoughsothatthenumber q = c + < 1.Inparticular,put n = N + k Š 1,where k 2.Applyingtheinequality | an +1| 1,thenthereisaninteger N suchthat | an +1| / | an| > 1or | an +1| > | an| 0forall n N .Hence,thenecessaryconditionfora seriestoconverge, an 0as n doesnothold;thatis,theseries andiverges.If c =1,itissucient,togiveexamplesofaconvergent anddivergentseriesforwhich c =1.Considera p -series nŠ p.One has c =limn an +1 an =limn np ( n +1)p=limn 1 (1+1 /n )p=1 forany p .Buta p -seriesconvergesif p> 1anddivergesotherwise.

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55.RATIOANDROOTTESTS119 Example 8.30 Findallvaluesof p and q forwhichtheseries n =1npqnconvergesabsolutely. Solution: Here an= npqn.Onehas c =limn an +1 an =limn ( n +1)p np| q |n +1 | q |n= | q | limn (1+1 /n )p 1 = | q | So,for | q | < 1andany p ,theseriesconvergesabsolutelybytheratio test.If q = 1,theratiotestisinconclusive,andthesecaseshavetobe studiedbydierentmeans.If | q | =1,then | an| = np= 1 /nŠ p, whichisa p -seriesthatconvergesif Š p> 1or p< Š 1.Thus,theseries convergesabsolutelyforall p if | q | < 1andfor p< Š 1if q = 1.Note that,for Š 1 p< 0and q = Š 1,theseriesconditionallyconverges (i.e.,itisconvergentbutnotabsolutelyconvergent).Inthiscase,itis aconvergentalternating p -series ( Š 1)n/nŠ p. Theratio testisadvantageoustotestconvergenceofserieswhosetermscontain productsofnumbers,e.g.,factorials n !=1 2 3 n Example 8.31 Testtheseries n =0xn/n forconvergencewhere, byde“nition, 0!=1 Solution: Bytheratiotest limn | an +1| | an| =limn | x |n +1 ( n +1)! n | x |n=limn | x | n +1 =0 Theseriesconvergesforallreal x .Lateritwillbeshownthatthesum oftheseriesistheexponentialfunction ex. 55.3.RootTest.Theorem 8.22(RootTest) Givenaseries an,supposethefollowinglimitexists: limn n | an| = c, where c 0 or c = € If c< 1 ,then anconvergesabsolutely. € If c> 1 ,then andiverges. € If c =1 ,thenthetestgivesnoinformation. Proof. If c< 1,then,asintheproofoftheratiotest,theexistence ofthelimitmeansthatthereisaninteger N andanumber q c
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1208.SEQUENCESANDSERIES Thisshowsthattheseries | an| convergesbycomparisonwiththe convergentgeometricseries qn,0 1,thenthereexistsaninteger N suchthatn | an| > 1 forall n N ,andhencethenecessaryconditionforconvergence, an 0as n ,doesnothold.Theseries andiverges.If c =1, considera p -series:n nŠ p=(n n )Š p 1Š p=1byTheorem8.14.But a p -seriesconvergesif p> 1anddivergesif p< 1.Theroottestis inconclusive. Example 8.32 Testtheconvergenceoftheseries an,where an=[(2 n2+5) / (3 n2+2)]n. Solution: Here | an| = an,andtheabsoluteconvergenceisequivalent totheconvergence.Onehas limn n | an| =limn 2 n2+5 3 n2+2 =limn 2+5 /n2 3+2 /n2= 2 3 < 1 Sotheseriesconverges. 55.4.OscillatoryBehaviorofSequencesintheRootandRatioTests.Considerasequencede“nedrecursivelyby a1=1 / 2and an +1= 1 2 sin n 2 an= cn= | an +1| | an| = 1 2 sin n 2 Anattempttotesttheabsoluteconvergenceof anbytheratiotest failsbecausethesequence cnoscillatesbetween0and1 / 2and,hence, doesnotconverge.Ontheotherhand,theabsoluteconvergenceofthe series ancaneasilybeestablishedbycomparisonwiththeconvergent geometricalseries: | an| 1 2 | an Š 1| 1 4 | an Š 2| 1 2nwheretheinequality | sin x | 1hasbeenused.Similarly,thesequence usedintheroottestmayalsoexhibitoscillatorybehaviorandbenonconvergent.Forexample, an= 1 2 sin n 2 n= cn=n | an| = 1 2 sin n 2 Theseries anconvergesabsolutelybycomparisonwiththegeometric series: | an| 2Š n. Theratioandroottests,asstatedinTheorems8.21and8.22, assumethe existence ofthelimitlimn cn= c ,whiletheaboveexamplesshowthatthereareabsolutelyconvergentseriesforwhichthis

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55.RATIOANDROOTTESTS121 hypothesisisnotful“lled.Whatcanbesaidabouttheabsoluteconvergenceofaserieswhenthislimitdoesnotexist?Toanswerthisquestion,recallthat,intheproofoftheratioorroottest,theexistenceof limn cn= c< 1hasbeenusedonlytoestablishtheboundednessof thesequence cn q< 1forall n N ,whichis sucientfortheseries antoconverge .Buttheboundednesspropertydoes not implythe convergence!Evidently,theboundednessconditionholdsintheabove examples, cn1 2< 1forall n> 1,whilethesequence { cn} doesnot converge.Similarly,theexistenceofthelimitvalue c> 1hasonly beenusedtoshowthatn | an| 1or | an| 1for in“nitelymany n to concludethatthesequence { an} cannotconvergeto0andhence andiverges.If | an +1| / | an| 1forall n N ,thenagain { an} cannot convergeto0(bytheproofoftheratiotest).Thus,thehypothesisof convergenceof { cn} intherootorratiotestcanbeweakenedtoobtain widerscopesofthetests. Theorem 8.23(RatioandRootTestsRe“ned) Givenaseries nan,put cn= | an +1| / | an| or cn=n | an| .Then cn q< 1forall n N = anconverges n | an| 1forin“nitelymany n| an +1| | an| 1forall n N = andiverges forsomeinteger N .55.5.WiderScopeoftheRootTest.Thedierenceinthescopesofthe rootandratiotestsiselucidatedbythefollowinglemmathatisproved inmoreadvancedcalculuscourses. Lemma 8.1 Ifthelimitof | an +1| / | an| exists,thensodoesthelimit ofn | an| andinthiscase (8.14)limn n | an| =limn | an +1| | an| Ifthesequencen | an| doesnotconverge,neitherdoes | an +1| / | an| Theconverseisnottrue.Forexample,put an=2Š n Š 1(3 Š ( Š 1)n) > 0.Then limn n | an| =limn n 3 Š ( Š 1)n 2n +1= 1 2 limn n 3 Š ( Š 1)n 2 = 1 2

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1228.SEQUENCESANDSERIES because(3 Š ( Š 1)n) / 2=1foreven n and(3 Š ( Š 1)n) / 2=2forodd n ,whilen p 1forany p> 0.Incontrast,thesequence | an +1| | an| = 2n +1 2n +23 Š ( Š 1)n +1 3 Š ( Š 1)n= 1 2 3+( Š 1)n 3 Š ( Š 1)n= 1 ,n even1 4,n odd oscillatesbetween1and1 / 4and,hence,doesnotconverge.Sothe convergenceofthesequence {n | an|} doesnotimplytheconvergence of {| an +1/an|} andthesequence {n | an|} mayconvergeevenifthe sequence {| an +1/an|} doesnot. Lemma8.1showsthattheratiotesthas thesamepredictingpowerastheroottestonlyif | an +1| / | an| converges TheroottestasstatedinTheorem8.23has wider scope,meaning that whenevertheratiotestshowsconvergence,theroottestdoes,too, andwhenevertheroottestisinconclusive,theratiotestis,too .The subtletytonotehereisthattheconverseofthelatterstatementisnot generallytrue. Theinconclusivenessoftheratiotestdoesnotimplythe inconclusivenessoftheroottest .Theassertioncanbeillustratedwith thefollowingexample.Consideraconvergentseriesobtainedfromthe sumoftwogeometricseriesinwhichtheorderofsummationischanged:n =1an= 1 2 + 1 3 + 1 22+ 1 32+ 1 23+ 1 33+ = 1 2k =01 2k+ 1 3k =01 3k= 1 2 1 1 Š1 2+ 1 3 1 1 Š1 3= 3 2 wherethesumofageometricserieshasbeenused(Theorem8.9).Now notethatif n =2 k iseven,then a2 k=(1 / 3)k,and a2 k Š 1=(1 / 2)kif n =2 k Š 1isodd.Takethesubsequenceofratiosforeven n =2 k c2 k= a2 k +1/a2 k=(2 / 3)k/ 9.Itconvergesto0as k .Ontheother hand,thesubsequenceofratiosforodd n =2 k Š 1diverges: c2 k Š 1= (3 / 2)k as k .Sothelimitof cndoesnotexist;moreover,the ratiotest(asinTheorem8.23)failsmiserablytodetecttheconvergence because cnisnotevenbounded.Theseriesconvergesbytheroottest. Indeed, c2 k=2 k a2 k=1 / 2 < 1and c2 k Š 1=2 k Š 1 a2 k Š 1=1 / 3 < 1. Althoughthesequence cndoesnotconverge(itoscillatesbetween1 / 3 and1 / 2),itisbounded, cn q =1 / 2 < 1forall n ,andhencethe seriesconvergesbyTheorem8.23. Theratiotestappearstobesensitive totheorderofsummation,whilethisisnotsofortheroottest .

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55.RATIOANDROOTTESTS123 55.6.WhentheRatioTestIsInconclusive.Theorem 8.24(DeMorgansTest) Let anbeaseriesinwhich | an +1| / | an| 1 as n .Theseriesconvergesabsolutelyif limn n an +1 an Š 1 = b< Š 1 Theproofofthistheoremislefttothereaderasanexercise(see Exercise55.7.26).Considertheasymptoticbehavioroftheratio cn= | an +1| / | an| as n .Thetheoremassertsthatif cnbehavesas cn 1+ b/n forlarge n (i.e.,neglectingtermsoforder1 /npwhere p> 1),thentheseries anconvergesif b< Š 1. Fora p -series,theratiotestisinconclusive(seetheproofofthe ratiotest).However,DeMorganstestresolvestheinconclusiveness. Indeed,forlarge n cn= np ( n +1)p= 1 (1+1 /n )p 1 Š p n wheretheasymptoticbehaviorhasbeenfoundfromthelinearization f ( x )=(1+ x )Š p f (0)+ f(0) x =1 Š px forsmall x =1 /n .So b = Š p andtheseriesconvergesif b< Š 1or p> 1. ThisillustratesabasictechnicaltricktoapplyingDeMorganstest. Supposethatthereisafunction f ( x )suchthat | an +1| / | an| = f (1 /n ).If f isdierentiableat x =0,thenitslinearization L ( x )= f (0)+ f(0) x isagoodapproximationinthesensethat limx 0f ( x ) Š L ( x ) x =limx 0f ( x ) Š f (0) x Š f(0)= f(0) Š f(0)=0 Inotherwords, f ( x )= f (0)+ f(0) x + x ( x )where ( x ) 0as x 0 (theerrorofthelinearapproximationdecreasestozerofasterthan x ). Therefore | an +1| | an| = f (1 /n )= f (0)+ f(0) n + (1 n) n =1+ f(0) n + (1 n) n Bytheinconclusivenessoftheratiotest, f (0)=1.Then limn n | an +1| | an| Š 1 =limn f(0)+ (1 n) = f(0) andtheseries anconvergesabsolutelyif f(0) < Š 1. Example 8.33 Investigatetheconvergenceoftheseriesn =1p ( p +1) ( p + n Š 1) n !

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1248.SEQUENCESANDSERIES Solution: Theratiosequencehastheform | an +1| | an| = | p ( p +1) ( p + n Š 1)( p + n ) | ( n +1)! n | p ( p +1) ( p + n Š 1) | = | n + p | n +1 = | 1+p n| 1+1 nSoitconvergesto1andtheratiotestisinconclusive.Considerthe function f ( x )=(1+ px ) / (1+ x ).Asaratiooflinearfunctions,itis dierentiableat x =0.Bytheratiorule f( x )= p (1+ x ) Š (1+ px ) (1+ x )2= p Š 1 (1+ x )2= f(0)= p Š 1 ByDeMorganstest,theseriesconverges(absolutely)if f(0) < Š 1or p Š 1 < Š 1or p< 0. 55.7.Exercises.In(1)…(18),usetheratioorrootteststodeterminewhethertheseries isabsolutelyconvergent(here p isreal). (1)n =1(1000)2 n n (2)n =1n2 2 3 n(3)n =1n pn,p =0 (4)n =1( n !)2 (2 n )! (5)n =22nn nn(6)n =1pnn nn(7) 100 1 + 100 101 1 3 + 100 101 102 1 3 5 + (8) 4 2 + 4 7 2 6 + 4 7 10 2 6 10 + (9)n =1( 2 Š3 2)( 2 Š5 2) ( 2 Š2 n +1 2) (10)n =1n2 (2+1 n)n(11)n =1( n !)2 2n2(12)n =1 1+ 1 n n2(13)n =1 2 n3+ n 3 n3+5 n(14)n =1 pn+ np 2n+ n2n,p> 0

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55.RATIOANDROOTTESTS125 (15)n =1np (ln n )n(16)n =12 4 (2 n ) n (17)n =1pnn 5 8 (3 n +2) (18)n =2 n Š 1 n +1 n ( n Š 1)(19)Forwhichintegers p> 0istheseries n =1( n !)2/ ( pn )!convergent? (20)Considerageometricserieswith q =1 / 2inwhichtheorderof termsischangedbyswappingtermsineachconsecutivepair: a1+ a2+ a3+ = 1 2 +1+ 1 8 + 1 4 + 1 32 + 1 16 + 1 128 + 1 64 + Testtheconvergenceofthisseriesusingtherootandratio tests. In(21)…(24),useDeMorganstesttoinvestigatetheconvergenceof theseries. (21) 1 2p+ 1 3 2 4p+ 1 3 5 2 4 6p+ (22)a b+a ( a + c ) b ( b + c )+a ( a + c )( a +2 c ) b ( b + c )( b +2 c )+ ,a> 0 ,b> 0 ,c> 0 (23) n =1 p ( p +1) ( p + n Š 1) n nq(24) n =11 3 5 (2 n Š 1) 2 4 6 (2 n )p1 nq(25)(EstimatingSums).Givenaseries anwithpositiveterms, put cn= an +1/an.Supposethat cn c< 1,thatis,theseries converges, an= s .Let snbeapartialsum.Provethat s Š sn an +1 1 Š cn +1if { cn} isadecreasingsequence,and s Š sn an +1 1 Š c if { cn} isanincreasingsequence. Hint :Usethegeometric seriesasintheproofoftheratiotesttoestimatetheremainder s Š sn= an +1+ an +2+ (26)ProveDeMorganstest. Hints :Comparetheseries | an| withtheconvergent p -series bn,where bn= A/npand p = Š (1 Š b ) / 2 > 1if b< Š 1.Showthat n ( bn +1/bnŠ 1) b as n .Next,showthat,bychoosingtheconstant A ,one canalwaysmake | an|
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1268.SEQUENCESANDSERIES 56.Rearrangements Herecharacteristicdistinctpropertiesofconditionallyconvergent andabsolutelyconvergentseriesareestablishedthroughtheconcept ofrearrangement. Definition 8.9(Rearrangement) Let { kn} n =1 2 ,... ,bean integer-valuedpositivesequenceinwhicheverypositiveintegerappears onlyonce.Givenaseries an,put a n= akn.Theseries a nis calleda rearrangement of an. Thesequence { kn} isobtainedbya permutation oftermsofthe sequenceofallintegers { 1 2 3 4 ,... } .Forexample, { 1 2 3 4 5 6 7 8 9 ,... }Š{ kn} = { 1 3 2 5 7 4 9 11 6 ,... } Arearrangementofaseriesisaccordinglyobtainedbyapermutationof termsinthesum.Fora“nitesum,arearrangement(orapermutation) ofitstermsdoesnotchangethevalueofthesum.Thisisnotgenerally soforconvergentseries. Consideranalternatingharmonicseries: (8.15)n =1an=n =1( Š 1)n Š 1 n =1 Š 1 2 + 1 3 Š 1 4 + 1 5 Š 1 6 + Theseriesisconvergentbutnotabsolutelyconvergent(itssumis s = ln2;seeExercise54.2.25).Oneofitsrearrangementsreads (8.16)n =1a n=1+ 1 3 Š 1 2 + 1 5 + 1 7 Š 1 4 + 1 9 + 1 11 Š 1 6 + inwhichtwopositivetermsarealwaysfollowedbyonenegative.Let snand s nbepartialsumsof(8.15)and(8.16),respectively.Put hn= 1+1 / 2+ +1 /n (apartialsumoftheharmonicseries).Then s2 n=1 Š 1 2 + 1 3 Š 1 4 + + 1 2 n Š 1 Š 1 2 n =1+ 1 3 + 1 5 + + 1 2 n Š 1 Š 1 2 1+ 1 2 + 1 3 + + 1 n = h2 nŠ 1 2 Š 1 4 ŠŠ 1 2 n Š 1 2 hn= h2 nŠ hn

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56.REARRANGEMENTS127 Inasimilarfashion, s 3 n=1+ 1 3 + 1 5 + 1 7 + + 1 4 n Š 1 Š 1 2 Š 1 4 ŠŠ 1 2 n = h4 nŠ 1 2 Š 1 4 ŠŠ 1 4 n Š 1 2 hn= h4 nŠ 1 2 h2 nŠ 1 2 hn=( h4 nŠ h2 n)+ 1 2 ( h2 nŠ hn)= s4 n+ 1 2 s2 n. Sincethesequence { sn} convergesto s ,itssubsequence { s4 n} and { s2 n} alsoconvergeto s .Bybasiclimitlawsforsequences,thesequence { s 3 n} convergesandbytakingthelimit n intheaboveequality, limn s 3 n=limn ( s4 n+1 2s2 n)=limn s4 n+1 2limn s2 n= s + s/ 2=3 s/ 2 Iftherearrangedseries(8.16)converges,thenthesubsequence { s 3 n} ofthesequenceofpartialsums { s n} mustconvergetothesumofthe seriesandthelattermustbe s=3 s/ 2.Thus,arearrangementofthe seriesmaychangeitssum!Thisfactisnotspeci“ctotheexample consideredbutinherentinallconditionallyconvergentseries.Terms ofaconditionallyconvergentseriesoccurwithdierentsigns(positive andnegative).Byregroupingpositiveandnegativeterms,itwillbe provedthatthesumofaconditionallyconvergentseriescanbemade anynumber or .Theanalysisbeginsbystudyingtheproperties ofsumsofpositiveandnegativetermsofaconditionallyconvergent series. Givenanumber x ,put x=( x | x | ) / 2.Thenumber x+= x if x> 0and x+=0otherwise.Similarly, xŠ= x if x< 0and xŠ=0 otherwise. Lemma 8.2 Givenaseries an,considertwoseries a+ nand aŠ n,where a n=( an| an| ) / 2 (theseriesofpositiveandnegative terms).Then (i) If anconvergesabsolutely,then a+ nand aŠ nconverge. (ii) If anisconditionallyconvergent,then a+ nand aŠ ndiverge. Proof. Let an= s< and | an| = t ,where t< if anconvergesabsolutelyand t = ifitisconditionallyconvergent.Let s nbepartialsumsof a n, snbepartialsumsof an,and tnbe partialsumsof | an| .Since sn s and tn t as n ,oneinfers that a+ nŠ aŠ n= | an| a+ n+ aŠ n= an= s+ nŠ sŠ n= tns+ n+ sŠ n= sn= s n= 1 2 ( sn tn) .

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1288.SEQUENCESANDSERIES If anconvergesabsolutely,then t< andhencelimn s n= ( s t ) / 2;thatis,bothseries a nconverge.If anisconditionallyconvergent,then t = andlimn s n= (theseries a ndiverge). Theorem 8.25(RiemannsRearrangementTheorem) Let anbeaseriesthatconverges,butnotabsolutely.Then,forany c thatisa realnumberor ,thereexistsarearrangement a nwhosesequence ofpartialsums { s n} convergesto c Proof. Let p1,p2,.... denotenonnegativetermsof anintheorderinwhichtheyoccur,andlet q1,q2,... denotenegativetermsof anintheorderinwhichtheyoccur.InthenotationofLemma8.2,the series pnand a+ naswellas qnand aŠ nmayonlydierby zeroterms(ifsome an=0).Sotheseries pnand qndiverge. Considerthefollowingrearrangement.Givenanumber c ,take“rst k1terms pn,suchthatthenumber c liesbetweenthepartialsums sk1= p1+ p2+ + pk1and sk1Š 1;thatis, k1isde“nedbythecondition sk1Š pk1c>sk1+ m1or | c Š sk1+ m1| < | qm1| .Thiscan alwaysbedonebecausepartialsumsof pncanbelargerthanany number,whilepartialsumsof qncanbesmallerthananynumber owingtothedivergenceoftheseseries.So s1 sn sk1, 1 n k1, where | c Š sk1| c ,andtake m2nextterms qn,where m2isthesmallest integerforwhich sk1+ m1+ k1+ m2
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56.REARRANGEMENTS129 onecantakeanydivergentsequence cn (or Š )andconstructa rearrangementsuchthat sk1overshoots c1and sk1+ m1undershoots c1, sk1+ m1+ k2overshoots c2and sk1+ m1+ k2+ m2undershoots c2,andsoon. Obviously,thissequenceofpartialsumdiverges. Absolutelyconvergentserieshaveadrasticallydierentproperty. Theorem 8.26(RearrangementandAbsoluteConvergence) Ifa series anconvergesabsolutely,theneveryrearrangementof anconverges,andtheyallconvergetothesamesum Proof. Let tm= | a1| + | a2| + + | am| beapartialsumofthe seriesofabsolutevalues.Thesequence { tm} increasesmonotonically convergingtoanumber t bythehypothesis;thatis,forany > 0, thereisaninteger N suchthat t Š tm< forall m>N .Therefore,nk = N +1| ak| = tmŠ tN +1= | tmŠ t + t Š tN +1| | tmŠ t | + | t Š tN +1| < 2 So,bytaking N largeenough,thesumof any numberofterms | ak| k>N ,canbemadesmallerthan any preassignedpositivenumber. Let snand s nbepartialsumsof ananditsrearrangement a n. Onecantake n>N largeenoughsuchthat s ncontains a1, a2,..., aN(i.e.,theintegers1 2 ,...,N areinthesetofintegers k1,k2,...,knin thenotationsofDe“nition8.9).Thenthedierence s nŠ sncontains onlyterms akwith k>N (theterms a1, a2,..., aNarecancelled).Let m bethelargestintegeramongst { k1,k2,...,kn} suchthat amisnot cancelledin s nŠ sn.Thenthedierencecontainssomeoftheterms aN +1,aN +2,...,amwithpossiblyreversedsings(thetermsof snthat arenotcancelledcontributeto s nŠ snwithanoppositesign).Applying theinequality | b c || b | + | c | tothesumofalltermsin s nŠ sn,itis concludedthat | s nŠ sn|mk = N +1| ak| < 2 forall n exceedingsomeinteger.If s n sand sn s as n ,then | sŠ s | < 2 ,whichshowsthat s= s because > 0isarbitrary. Thus, anabsolutelyconvergentseriesismuchlikea“nitesum.The sumdoesnotdependontheorderinwhichthesummationiscarried out.Incontrast,thesumofaconditionallyconvergentseriesdepends onthesummationorder.Thisisthecharacteristicdierencebetween thesetwoclassesofconvergentseries.

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1308.SEQUENCESANDSERIES 56.1.StrategyforTestingSeries.Itwouldnotbewisetoapplytestsfor convergenceinaspeci“corderto“ndonethat“nallyworks.Instead, aproperstrategy,aswithintegration,istoclassifytheseriesaccording toitsform.Oneshouldalsokeepinmindthataconclusionaboutthe convergenceofaseriescanbereachedindierentways. 1.Necessaryconditionforconvergence .Itisisalwayseasier tocheck“rstthecondition an 0as n thanitistoinvestigate theseries anforconvergence.Iftheconditiondoesnothold,the seriesdiverges. 2.Specialseries .Aseries ancoincideswith(orisacombinationoforisequivalentto)specialseriessuchasa p -series,alternating p -series,geometricseries,telescopicseries,andsoon.Theirconvergencepropertiesareknown. 3.Seriessimilartospecialones .Ifaseries anhasaform thatissimilartooneofthespecialseries,thenoneofthecomparison testsshouldbeconsidered.Forexample,if anisarationaloralgebraicfunction(containsrootsofpolynomials),thentheseriesshould becomparedwitha p -series. 4.Alternatingseries .If an=( Š 1)nbn, bn 0,thenthealternatingseriestestisanobviouspossibility. 5.Ratioandroottests .Absoluteconvergenceimpliesconvergence.So,iftheratioorroottestshowsconvergence,thentheseriesin questionconvergesabsolutely.Ifthesetestsshowdivergence,thenthe seriesinquestionmaystillconvergebutnotabsolutely,andafurther investigationisrequired.Theroottestisconvenientforseriesofthe form ( bn)n.Theratiotestisconvenientwhen aninvolvesthefactorial n !orsimilarproductsofintegers.Theroottesthasawiderscope, butitismorediculttouse.Theratiotestisofteninconclusiveif anisarationaloralgebraicfunction( cn= | an +1| / | an| 1).Inthiscase, theasymptoticbehaviorof cnisrathereasyto“nd, cn 1+ b/n as n ,andthenuseDeMorganstest. 6.Seriesofnonnegativeterms .If an= f ( n ) 0andthe integral 1f ( x ) dx iseasytoevaluate,thentheintegraltestiseective. Also,itcanbeusedincombinationwiththecomparisontest: an f ( n )and 1f ( x ) dx convergesandsois an,or f ( n ) anand 1f ( x ) dx divergesandsois an. Example 8.34 Testtheseries ( n +1) / ( n2+ n +1) forconvergence. Solution: Forlarge n ,theleadingtermsofthetopandbottomof theratioare n and n2,respectively.So an 1 /n asymptoticallyfor

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56.REARRANGEMENTS131 large n .Theseriesresemblestheharmonicseries,whichdiverges.Itis natural,then,totrytoprovethedivergenceoftheseriesbycomparing itwiththeharmonicseries: n +1 n2+ n +1 > n n2+ n +1 n n2+ n2+ n2= 1 3 n Thus,theseriesindeeddivergesbycomparisonwiththeharmonicseries. Example 8.35 Testtheseries 3n/ (2 4 6 (2 n )) forconvergence. Solution: Eachterm aninvolvesafactorial-likeproductofintegers, whichsuggeststheuseoftheratiotest: an +1 an= 3n +1 2 4 (2 n ) (2 n +2) 2 4 (2 n ) 3n= 3 2 n +2 0 So,theseriesconverges. Example 8.36 Testtheseries sin( n2) eŠ nforconvergence. Solution: Onehas | an| = | sin( n2) | eŠ n eŠ n. Theseries eŠ nconvergesbytheintegraltest: 1eŠ xdx =2 1ueŠ udu = Š 2 ueŠ u 1Š 2 1eŠ udu = 2 e + 2 e = 4 e < Herethechangeofvariableshasbeencarriedout“rst, x = u2, dx = 2 udu .Thentheintegrationbypartshasbeendonetoevaluatethe improperintegral.Hence,theseriesinquestionconvergesbythecomparisontest. Example 8.37 Put hn=1+1 / 2+ +1 /n (apartialsumof theharmonicseries).Investigatetheconvergenceof npeŠ qhn,where p = q Solution: Theratiotestisinconclusive: an +1 an= ( n +1)peŠ qhn +1 npeŠ qhn= ( n +1)p npeŠ q ( hn +1Š hn)= (1+1 n)p 1 eŠq n +1limn 1+ 1 n peŠq n +1=limn 1+ 1 n plimn eŠq n +1=1p e0=1 ToapplyDeMorganstest,theasymptoticbehaviorof an +1/anhas tobeinvestigated.Put f ( x )=(1+ x )pexp( Š qx/ (1+ x ))sothat

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1328.SEQUENCESANDSERIES an +1/an= f (1 /n ).Thelinearizationof f ( x )at x =0maybeobtained bycalculating f(0),whichistechnicallyinvolvedgiventheexplicit formof f ( x ).Instead,onecanlinearizethefactorsin f ( x ).Let g ( x )= (1+ x )p.Then g( x )= p (1+ x )p Š 1and,hence, g ( x ) g (0)+ g(0) x = 1+ px near x =0.Forsmall u eŠ qu 1 Š qu .Setting u = x (1+ x )Š 1andusingthelinearization(1+ x )Š 1 1 Š x ,one“nds eŠ qu 1 Š qu 1 Š qx (1 Š x ) 1 Š qx bykeepingonlytermslinearin x .Therefore f ( x ) (1+ px )(1 Š qx ) 1+( p Š q ) x = an +1 an 1+ p Š q n Thus,theseriesconvergesif p Š q< Š 1or p 0(12)n =1(n p2Š 1)n(13)n =1n enp(14)n =1npn Š 1 pnŠ (1 Š 1 /n )n, | p | < 1(15)n =1(n p Š 1) ,p 0 (16)n =1( Š 1)n p + n (17) n =1( Š 1)nn Š 1 n +1 1 100 n (18)n =1( a1 /nŠ b1 /n) ,a> 0 ,b> 0(19)n =1n n n

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57.POWERSERIES133 In(20)and(21),usethesumofthealternatingharmonicseries n =1( Š 1)n Š 1/n =ln2to“ndthesumsofitsrearrangement. (20)1+ 1 3 Š 1 2 + 1 5 + 1 7 Š 1 4 + (21)1 Š 1 2 Š 1 4 + 1 3 Š 1 6 Š 1 8 + In(22)and(23),“ndtworearrangementsoftheconditionallyconvergentseriesthatconvergeto+ andto Š (22)n =1( Š 1)n Š 1 n (23)n =1( Š 1)n Š 1 n (24)Letallpositivetermsofthealternatingharmonicseriesbedividedintogroupsof p termsintheorderinwhichtheyappear intheseriesandletallnegativetermsbedividedintogroups of q termsintheorderinwhichtheyappear.Considerthe rearrangementinwhichthe“rstgroupofpositive p termsis followedbythe“rstgroupofnegative q terms,which,inturn, isfollowedbythesecondgroupof p positivetermsandthe latterisfollowedbythesecondgroupof q negativeterms,and soon.Ifthesumofthealternatingharmonicseriesisln2, provethatthesumofthisrearrangementisln2+1 2ln( p/q ). (25)Let { an} 1bethesequenceofpositiverootsoftheequation tan x = x intheincreasingorder.Testtheconvergenceofthe series n =1( an)Š 2. (26)If anisconditionallyconvergent,showthat limn s+ n sŠ n=1 ,s n= 1 2nk =1( | ak| ak) 57.PowerSeries Definition 8.10(PowerSeries) Givenasequence { cn} ,theseriesn =0cnxn= c0+ c1x + c2x2+ c3x3+ iscalleda powerseries inthevariable x .Thenumbers cnarecalled the coecients oftheseries. Ingeneral,theserieswillconvergeordiverge,dependingonthe choiceof x .Thepowerseriesalwaysconvergesfor x =0tothenumber c0.

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1348.SEQUENCESANDSERIES Example 8.38 Forwhatvaluesof x doesthepowerseries n =1xn/n converge? Solution: Bytheroottest,n | xn| n = | x | n n | x | as n Sotheseriesconvergesforall Š 1 1 or x< Š 1.Theroottestisinconclusivefor x = 1.Thesevalues havetobeinvestigatedbydierentmeans.For x =1,thepowerseries becomestheharmonicseries 1 /n ,whichisdivergent.For x = Š 1, thepowerseriesbecomesthealternatingharmonicseries ( Š 1)n/n whichisconvergent.Thus,thepowerseriesconvergesif x [ Š 1 1) anddivergesotherwise. Givenanumber a ,considerapowerseriesinthevariable y = x Š a :n =0cnyn=n =0cn( x Š a )n. Itisalsocalleda powerseriescenteredat a ora powerseriesabout a Let S bethesetofallvaluesof x forwhichapowerseriesin x convergesandlet Sabethesetofallvaluesof x forwhichthecorresponding powerseriesin( x Š a )converges.Whatistherelationbetween S and Sa?Sincetheseriesareobtainedfromoneanotherbymerelyshifting thevalueofthevariablebyanumber a x x Š a ,theset Sais thereforeobtainedbyaddingthenumber a toeveryelementof S : (8.18) x Sa x Š a S = Sa= { x | x Š a S } Forexample,theseries n =0( x Š 2)n/n convergesif x Š 2 [ Š 1 1) or x [1 3)anddivergesotherwisebyExample8.38.Thus,theproblemof“ndingtheset Saisequivalenttotheproblemof“ndingthe set S .57.1.PowerSeriesasaFunction.Supposethatapowerseriesin x convergesonaset S .Thenitde“nesafunctionon S : f ( x )= n =0cnxn,x S. Theset S iscalledthe domain ofsuchafunction.Functionsde“ned bypowerseriesaremostcommoninapplications.Inwhatfollows,it willbeshownthatfamiliarelementaryfunctionssuchassin x ,cos x andexp x ,etccanalsoberepresentedaspowerseries.Therearemany other(special)functionsthatarede“nedbypowerseries.

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57.POWERSERIES135 Example 8.39 FindthedomainoftheBesselfunctionoforder 0 thatisde“nedbythepowerseries J0( x )=n =0( Š 1)n 22 n( n !)2x2 n, where,bycommonconvention, 0!=1 Solution: Since an= cnx2 ncontainsthefactorial,theratiotestis moreconvenient: | an +1| | an| = x2| cn +1| | cn| = x222 n( n !)2 22( n +1)(( n +1)!)2= x2 22( n +1)2 0 as n .Sotheseriesconvergesforall x Valuesofafunctionde“nedbyapowerseriescanbeestimatedby partialsumsthatarepolynomialsinthevariable x : f ( x ) fn( x )=nk =0ckxk= c0+ c1x + c2x2+ + cnxn. Thus,partialsumsde“nea sequenceofpolynomialsthatconvergesto thefunction on S fn( x ) f ( x )forall x S .Theaccuracyof theapproximationisdeterminedbytheremainder Rn( x )= f ( x ) Š fn( x ).TheaccuracyassessmentisdiscussedinSection8.59.Sincethe remainder Rn( x )isafunctionon S ,theerroroftheapproximationis not generallyuniform;thatis,itdependson x .Inthisregard,recall alsoausefulresultgiveninExercise55.7.2557.2.RadiusofConvergence.Theset S onwhichapowerseriesis convergentisanimportantcharacteristicanditspropertieshavetobe studied. Lemma 8.3(PropertiesofaPowerSeries) (i) .Ifapowerseries cnxnconvergeswhen x = b =0 ,thenitconvergeswhenever | x | < | b | (ii) .Ifapowerseries cnxndivergeswhen x = d =0 ,thenitdiverges whenever | x | > | d | Proof. If cnbnconverges,then,bythenecessaryconditionfor convergence, cnbn 0as n .Thismeans,inparticular,that,for =1,thereexistsaninteger N suchthat | cnbn| < =1forall n>N Thus,for n>N | cnxn| = cnbnxn bn = | cnbn| x b n< x b n. whichshowsthattheseries cnxnconvergesbycomparisonwiththe geometricseries qn,where q = x/b and | x/b | < 1or | x | < | b | .

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1368.SEQUENCESANDSERIES Supposethat cndndiverges.If x isanynumbersuchthat | x | > | d | then cnxncannotconvergebecause,bypart(i)ofthelemma,the convergenceof cnxnimpliestheconvergenceof cndn.Therefore, cnxndiverges. Thislemmaallowsustoestablishthefollowingdescriptionofthe set S Theorem 8.27(ConvergencePropertiesofaPowerSeries) Fora powerseries cnxn,thereareonlythreepossibilities: (i) Theseriesconvergesonlywhen x =0 (ii) Theseriesconvergesforall x (iii) Thereisapositivenumber R suchthattheseriesconvergesif | x | R Proof. Supposethatneithercase1norcase2istrue.Thenthere arenumbers b =0and d =0suchthatthepowerseriesconvergesfor x = b anddivergesfor x = d .ByLemma8.3,thesetofconvergence S liesintheinterval | x || d | forall x S .Thisshowsthat | d | isan upperboundfortheset S .Bythecompletenessaxiom, S hasaleast upperbound R =sup S .If | x | >R ,then x S ,and cnxndiverges. If | x | | x | .Since b S cnxnconvergesby Lemma8.3. Theorem8.27showsthatapowerseriesconvergesina single open interval( Š R,R )anddivergesoutsidethisinterval.Theset S mayor maynotincludethepoints x = R .Thisquestionrequiresaspecial investigationjustlikeinExample8.38.Sothenumber R ischaracteristicforconvergencepropertiesofapowerseries. Definition 8.11(RadiusofConvergence) The radiusofconvergence ofapowerseries cnxnisapositivenumber R> 0 suchthatthe seriesconvergesintheopeninterval ( Š R,R ) anddivergesoutsideit. Apowerseriesissaidtohavea zeroradiusofconvergence R =0 ,ifit convergesonlywhen x =0 .Apowerseriesissaidtohavean in“nite radiusofconvergence R = ,ifitconvergesforallvaluesof x Theratioorroottestcanbeusedtodeterminetheradiusofconvergence.

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57.POWERSERIES137 Corollary 8.3(RadiusofConvergenceofaPowerSeries) Given apowerseries cnxn, iflimn | cn +1| | cn| = = R = 1 iflimn n | cn| = = R = 1 where R =0 if = and R = if =0 Proof. Put an= cnxnintheratiotest(Theorem8.21).Then | an +1| / | an| = | x || cn +1| / | cn|| x | .Theseriesconvergesif | x | < 1, whichshowsthat R =1 / .Similarly,usingtheroottest(Theorem 8.22),n | an| = | x |n | cn|| x | < 1,whichshowsthat R =1 / Remark. IfthesequencesinCorollary8.3donotconverge,then Theorem8.23fromSection55.4shouldbeappliedto an= cnxnto determinetheradiusofconvergence.Also,oneshouldkeepinmind thattheroottesthaswiderscope(recallLemma8.1inSection55.5) Oncetheradiusofconvergencehasbeenfoundand0 0 Solution: | cn +1| | cn| = qn +1 n +2 n +1 qn= q n +1 n +2 = q 1+1 /n 1+2 /n q = Therefore, R =1 / =1 /q .If x = Š 1 /q ,then cnxn=( Š 1)n/ n +1= ( Š 1)nbn.Thesequence bnconverges monotonically to0sothat ( Š 1)nbnconvergesbythealternatingseriestest.If x = Š 1 /q ,then cnxn= 1 / n +1 > 1 / 2 n n 1.The p -series 1 /n1 / 2diverges( p =1 / 2 < 1)sothat 1 / n +1divergesbythecomparisontest.Thus,the intervalofconvergenceis S =[ Š 1 /q, 1 /q ). Example 8.41 Findtheradiusofconvergenceandtheintervalof convergenceofthepowerseries n2( x +1)n/qn,where q> 0 Solution: Put y = x +1.If S istheintervalofconvergenceof cnyn, where cn= n2/qn,thentheintervalofconvergenceinquestionisobtainedbyadding Š 1toallnumbersin S accordingtotherule(8.18).

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1388.SEQUENCESANDSERIES ByCorollary8.3, limn n | cn| =limn 1 qn n2= 1 q (limn n n )2= 1 q = So R =1 / = q .If y = q ,then cnyn= n2,andtheseries n2diverges ( an= n2doesnotconvergeto0).If y = Š q ,then cnxn=( Š 1)nn2, andtheseriesdivergesbecause an=( Š 1)nn2,doesnotconvergeto0. Theseriesconvergesonlyif | y | = | x +1 | 0 (6)n =0 n ( x +1)n(7)n =1( Š 1)nn2+1 n3+3 ( x Š 1)n(8)n =1(4 x +1)n n2(9)n =1xn 1 3 5 (2 n Š 1) (10)n =1n2x2 n 2 4 (2 n ) (11)n =0( n !)k ( kn )! xn,k> 0(integer) (12)n =04n n ( x +3)n(13)n =1 1+ 1 n n2xn(14)n =1p ( p Š 1) ( p Š n +1) n xn(15)n =1xn a2 n+ b2 n(16)n =1xn p n,p> 0 (17)n =12 4 (2 n ) 3 5 (2 n +1) xn(18)n =1( Š 1)n n n e nxn(19)n =1[3+( Š 1)n]n n xn(20)Let p
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58.REPRESENTATIONOFFUNCTIONSASPOWERSERIES139 (21)The Airyfunction isde“nedbythepowerseries A ( x )=1+ x3 2 3 + x6 2 3 5 6 + x9 2 3 5 6 8 9 + Finditsdomain. (22)Afunction f isde“nedbythepowerseries f ( x )= p + qx + px2+ qx3+ px4+ qx5+ ; thatis,itscoecients c2 k= p and c2 k Š 1= q ,where p and q arereal.Findthedomainof f andanexplicitexpressionof f ( x )(thesumoftheseries). (23)If f ( x )= cnxn,where cn +4= cnforall n 0,“ndthe domainof f andaformulafor f ( x ). (24)Thepowerseries cnxnand bnxnhavetheradiiofconvergence R1and R2,respectively.Whatistheradiusofconvergenceof ( cn+ bn) xn? (25)Supposethattheradiusofconvergenceof cnxnis R .What istheradiusofconvergenceof cnxkn,where k> 0isan integer? 58.RepresentationofFunctionsasPowerSeries Considerapowerseries 1 Š x2+ x4Š x6+ x8+ =n =0( Š 1)nx2 n. Itisageometricserieswith q = Š x2,andthereforeitconvergesfor all | q | = x2< 1or x ( Š 1 1).Usingtheformulaforthesumofa geometricseries,oneinfersthat 1 1+ x2=1 Š x2+ x4+ =n =0( Š 1)nx2 nforall Š 1 0 ,anddeterminetheintervalofitsvalidity.

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1408.SEQUENCESANDSERIES Solution: Put y = x Š a .Thefunctioncanberewritteninaform thatresemblesthesumofageometricseries: 1 x = 1 a (1+ y/a ) = 1 an =0 Š y a n=n =0( Š 1)n an +1( x Š a )n,x (0 2 a ) Thegeometricseriesconvergesif | q | = |Š y/a | = | y | /a< 1,andhence thisrepresentationisvalidonlyif Š a 0 ,thenthefunction f de“nedby f ( x )= c0+ c1( x Š a )+ c2( x Š a )2+ =n =0cn( x Š a )n

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58.REPRESENTATIONOFFUNCTIONSASPOWERSERIES141 isdierentiable(andthereforecontinuous)ontheinterval ( a Š R,a + R ) and f( x )= c1+2 c2( x Š a )+3 c3( x Š a )2+ =n =1ncn( x Š a )n Š 1, f ( x ) dx = C + c0( x Š a )+ c1( x Š a )2 2 + = C +n =0cn( x Š a )n +1 n +1 Theradiiofconvergenceofthesepowerseriesareboth R Thus,for powerseries ,thedierentiationorintegrationandthe summationcanbecarriedoutinanyorder: d dx cn( x Š a )n= d dx [ cn( x Š a )n] cn( x Š a )n dx = [ cn( x Š a )n] dx. Remark. Theorem8.28statesthattheradiusofconvergenceof apowerseriesdoesnotchangeafterdierentiationorintegrationof theseries.Thisdoesnotmeanthatthe intervalofconvergence does notchange.Itmayhappenthattheoriginalseriesconvergesatan endpoint,whereasthedierentiatedseriesdivergesthere. Example 8.43 Findtheintervalsofconvergencefor f f,and fif f ( x )= n =1xn/n2. Solution: Here cn=1 /n2andhencen | cn| =1 /n n2=(1 /n n )2 1= .Sotheradiusofconvergenceis R =1 / =1.For x = 1, theseriesisa p -series 1 /n2thatconverges( p =2 > 1).Thus, f ( x ) isde“nedontheclosedinterval x [ Š 1 1].ByTheorem8.28,the derivatives f( x )= n =1xn Š 1/n and f( x )= n =2( n Š 1) xn Š 2/n have thesameradiusofconvergence R =1.For x = Š 1,theseries f( Š 1)= ( Š 1)n Š 1/n isthealternatingharmonicseriesthatconverges,whereas theseries f( Š 1)= ( Š 1)n( n Š 1) /n divergesbecausethesequenceof itstermsdoesnotconvergeto0: | ( Š 1)n( n Š 1) /n | =1 Š 1 /n 1 =0. For x =1,theseries f(1)= 1 /n istheharmonicseriesandhence diverges.Theseries f(1)= ( n Š 1) /n alsodiverges(( n Š 1) /n does notconvergeto0).Thus,theintervalsofconvergencefor f f,and fare,respectively,[ Š 1 1],[ Š 1 1),and( Š 1 1). Theterm-by-termintegrationofapowerseriescanbeusedtoobtainapowerseriesrepresentationofantiderivatives. Example 8.44 Findapowerseriesrepresentationfor tanŠ 1x .

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1428.SEQUENCESANDSERIES Solution: tanŠ 1x = dx 1+ x2= n =0( Š x2)n dx = C +n =0( Š 1)nx2 n +1 2 n +1 SincetanŠ 10=0,theintegrationconstant C satis“esthecondition 0= C +0or C =0.Thegeometricserieswith q = Š x2convergesif | q | < 1or x2< 1or | x | < 1.Hence,theradiusofconvergenceofthe seriesfortanŠ 1x is R =1. Inparticular,thenumber1 / 3islessthantheradiusofconvergenceofthepowerseriesfortanŠ 1x .SothenumbertanŠ 1(1 / 3)= / 6canbewrittenasthenumericalseriesbysubstituting x =1 / 3 intothepowerseriesfortanŠ 1x .Thisleadstothefollowingrepresentationofthenumber : =2 3n =0( Š 1)n (2 n +1)3n.58.2.Continuityatendpointsoftheintervalofconvergence.Consider thegeometricserieswith q = Š xn =0( Š 1)nxn= 1 1+ x Byintegratingtheseriesterm-by-term,thepowerseriesforthelogarithmfunctionisdeduced: dx 1+ x =ln(1+ x )= C +n =0( Š 1)nxn +1 n +1 whichholdsfor | x | < 1byTheorem8.28.Theconstant C isdetermined bysetting x =0,whichyields C =ln1=0.Thus, (8.19)ln(1+ x )=n =1( Š 1)n Š 1 n xnThisseriesdivergesat x = Š 1asaharmonicseries,butitconvergesat theotherendpoint x =1asthealternatingharmonicseries.InExercise 54.2.25ithasbeenshownthatthesumofthealternatingharmonic seriesisln2.Itfollowsthenthatthepowerseriesrepresentation(8.19) alsoholdsat x =1 limx 1Šn =1( Š 1)n Š 1 n xn=ln2

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58.REPRESENTATIONOFFUNCTIONSASPOWERSERIES143 Here x 1Šmeanstheleftlimit( x approaches1fromtheleft).This observationisnotspeci“ctotheexampleconsideredbutratherisof generalnature. Theorem 8.29(Abelstheorem) Ifthepowerseries f ( x )= n =0cnxn, | x |
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1448.SEQUENCESANDSERIES Thecondition f (0)= e0=1determinestheconstant c0=1.Thus, theexponentialfunctionhasthefollowingpowerseriesrepresentation: (8.20) ex=1+ x 1! + x2 2! + x3 3! + =n =0xn n Theseriesconvergesontheentirerealline.Inparticular,thenumber e hasthefollowingseriesrepresentation: e =1+ 1 1! + 1 2! + 1 3! + =n =01 n .58.4.ApproximationofDeniteIntegrals.Ifaninde“niteintegralof f ( x )isdiculttoobtain,thentheevaluationoftheintegral b af ( x ) dx posesaproblem.Apowerseriesrepresentationoersasimplewayto approximatethevalueoftheintegral.Supposethat f ( x )= cnxnfor Š R
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58.REPRESENTATIONOFFUNCTIONSASPOWERSERIES145 1 / ( n !(2 n +1)): 1 0eŠ x2dx Šnk =0( Š 1)k k !(2 k +1) bn +1= 1 ( n +1)!(2 n +3) < 10Š 5. Adirectcalculationshowsthat b7 1 32 10Š 5and b8 1 46 10Š 6. So n =7issucienttoapproximatetheintegralwiththerequired accuracy. 58.5.Exercises.In(1)…(3),“ndapowerseriesrepresentationforthefunctionanddeterminetheintervalofconvergence. (1) f ( x )= 1 1 Š x4(2) f ( x )= x 3 x2+2 (3) f ( x )= x10 x Š 1 In(4)…(6),usedierentiationto“ndapowerseriesrepresentationfor thefunctionanddeterminetheintervalofconvergence. (4) f ( x )= 1 (1+ x )2(5) f ( x )= x3 (1 Š 4 x2)2(6) f ( x )= 1 (1+ x4)3In(7)…(9),useterm-by-termdierentiationto“ndthesumoftheseries: (7) x + x3 3 + x5 5 + (8)1+ x2 2! + x4 4! + (9)1+ 1 2 x + 1 3 2 4 x2+ 1 3 5 2 4 6 x3+ Hint :In(9),multiplythederivativeby1 Š x In(10)…(12),useintegrationto“ndapowerseriesrepresentationfor thefunctionanddeterminetheradiusofconvergence. (10) f ( x )=ln(1 Š x )x(11) f ( x )=ln 1+ x2 1 Š x2 (12) f ( x )=tanŠ 1(3 x ) In(13)…(15),useterm-by-termintegrationto“ndthesumoftheseries. (13) x +2 x2+3 x3+ (14) x Š 4 x2+9 x3Š 16 x4+ (15)1 2 x +2 3 x2+3 4 x3+

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1468.SEQUENCESANDSERIES In(16)…(18),“ndapowerseriesrepresentationfortheinde“niteintegralanddeterminetheintervalofconvergence. (16) ln(1 Š x ) x dx (17) exŠ 1 Š x x2dx (18) tanŠ 1( x2) dx (19)Findapowerseriesrepresentationforsin x andcos x using thedierentialequation f+ f =0.Determinetheintervalof convergence. Hint :Show“rstthat a sin x + b cos x satis“estheequationfor arbitrarynumbers a and b (20)ShowthattheBesselfunctionoforder0de“nedinExample 8.34satis“esthedierentialequation: x2J 0( x )+ xJ 0( x )+ x2J0( x )=0 In(21)…(23),usedierentiationorintegrationto“ndthesumofthe series. (21)n =1nxn Š 1(22)n =2n ( n Š 1) xn Š 2(23)n =0( Š 1)nxn +1 2n( n +1) In(24)…(26),howmanytermsdoesoneneedinapowerseriesapproximationtoevaluatetheintegralwiththeabsoluteerrornotexceeding 10Š 6? (24) 1 0dx 1+ x8(25) 1 0eŠ xŠ 1 x dx (26) 1 / 2 0ln(1+ x4) dx (27)Findtheradiusofconvergenceofthehypergeometricseries: 1+ ab 1! c x + a ( a +1) b ( b +1) 2! c ( c +1) x2+ a ( a +1)( a +2) b ( b +1)( b +2) 3! c ( c +1)( c +2) x3+ where a b ,and c arereals.UseDeMorganstesttodetermine theintervalofconvergence. In(28)…(30),usethefractiondecompositionto“ndthepowerseries representationofthefunctionanddeterminetheintervalofconvergence. (28) 12 Š 5 x 6 Š 5 x Š x2(29) x (1 Š x )(1+ x2) (30) 1 1+ x + x2+ x3(31)Findthesumoftheseriesn =1( Š 1)n Š 1 2 n Š 1 Hint :UseAbelstheoremandExample8.44.

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59.TAYLORSERIES147 (32)Showthattheseries y = n =0xn/ ( n !)2satis“esthedierentialequation xy+ yŠ y =0. 59.TaylorSeries59.1.RealAnalyticFunctions.Supposeafunction f isrepresentedby apowerseries( R> 0): f ( x )= c0+ c1( x Š a )+ c2( x Š a )2+ =n =0cn( x Š a )n. where | x Š a | 0 ,thenitscoecientsare cn= f( n )( a ) n Definition 8.12(RealAnalyticFunctions) Afunction f onan openinterval I issaidtobeanalyticif,forany a I ,ithasapower seriesrepresentation f ( x )= cn( x Š a )nthatconvergesinsomeopen interval ( a Š ,a + ) I ,where > 0 Theclassofanalyticfunctionsplaysasigni“cantroleinapplications.Theirpropertiesarediscussednext.

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1488.SEQUENCESANDSERIES Theorem 8.31(PowerSeriesRepresentationofAnalyticFunctions) Afunction f thatisanalyticonanopeninterval I hasthe powerseriesrepresentation (8.21) f ( x )=n =0f( n )( a ) n ( x Š a )nforany a I thatconvergesinanopensubintervalof I thatincludes a ThistheoremfollowsfromDe“nition8.12andTheorem8.30. InExample8.42itwasfoundthat (8.22) 1 x =n =0( Š 1)n an +1( x Š a )n,x (0 2 a ) Thisshowsthatthefunction f ( x )=1 /x isanalyticforall x> 0 because a canbeanypositivenumber;thatis,thefunctionhasa powerseriesrepresentationthatconvergesinanopensubintervalof (0 )containingany a> 0.Similarly,theanalyticityof f ( x )=1 /x canbeestablishedforall x< 0. Itisimportanttoemphasizethatapowerseriesforananalytic functiondoes not necessarilyconvergeontheentiredomainofthe function.Butananalyticfunctioncan always berepresentedbya convergentpowerseriesinaneighborhoodofeverypointofitsdomain. Equation(8.22)illustratesthepoint. Theorem 8.32(PropertiesofAnalyticFunctions) (i) Thesumsandproductsofanalyticfunctionsareanalytic. (ii) Thereciprocal 1 /f ofananalyticfunction f isanalyticif f is nowherezero. (iii) Thecomposition f ( g ( x )) ofanalyticfunctions f and g isanalytic. (iv) .Analyticfunctionsaredierentiablein“nitelymanytimes. Aproofofproperties(i)…(iii)isgiveninmoreadvancedcalculus courses.Property(iv)followsfromTheorem8.28. Itsconverseisnot generallytrue.Therearefunctionsthataredierentiablein“nitely manytimesatapoint,buttheycannotberepresentedbyapowerseriesthatconvergesinanopenintervalthatincludesthispoint .Asan example,considerthefunction f ( x )= eŠ 1 /x2if x =0and f (0)=0 .

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59.TAYLORSERIES149 Thefunctioniscontinuousat x =0becauselimx 0eŠ 1 /x2=limu eŠ u=0= f (0).Itisdierentiableat x =0because f(0)=limx 0f ( x ) Š f (0) x =limx 0eŠ 1 /x2 x =0 The“rstequalityisthede“nitionof f(0).Thelastlimitisestablished byinvestigatingtheleftandrightlimits x 0withthehelpofthe substitution x =1 /u as x 0;theleftandrightlimits coincidebecause ueŠ u2 0as u (theexponentialfunction decreasesfasterthananypowerfunction).Inasimilarfashion,itcan beprovedthat f( n )(0)=0forall n (seeExercise59.5.24).Thus, f ( x ) hasnopowerseriesrepresentation cnxninaneighborhoodof x =0 because,ifitdid,then,byTheorem8.31thefunctionshouldhavebeen identically0insomeinterval( Š ), > 0,(as f( n )(0)=0forall n ), whichisnottrue( f ( x ) =0forall x =0).Hence,thefunctionisnot analyticat x =0.59.2.TaylorandMaclaurinSeries.Definition 8.13(TaylorandMaclaurinSeries) Theseriesin (8.21)iscalledthe Taylorseries ofafunction f at a (orabout a orcenteredat a ).ThespecialcaseoftheTaylorserieswhen a =0 is calledthe Maclaurinseries ofafunction f ItisimportanttoknowMaclaurinseriesofelementaryfunctions. Inparticular,Maclaurinseriesforthepowerfunction(1+ x )p,where p isarealnumber,iscalledthe binomialseries .Thenumbers p n = p ( p Š 1) ( p Š n +1) n arecalled binomialcoecients Theorem 8.33(Maclaurinseriesofelementaryfunctions) Forany real x ,thefollowingpowerseriesrepresentationshold ex=1+ x + x2 2! + x3 3! + =n =1xn n sin x = x Š x3 3! + x5 5! Š x7 7! + =n =0( Š 1)nx2 n +1 (2 n +1)! cos x =1 Š x2 2! + x4 4! Š x6 6! + =n =0( Š 1)nx2 n (2 n )!

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1508.SEQUENCESANDSERIES For | x | < 1 ,thefollowingpowerseriesrepresentationshold ln(1+ x )= x Š x2 2 + x3 3 Š x4 4 + =n =1( Š 1)n Š 1 n xn(1+ x )p==1+ p 1! x + p ( p Š 1) 2! x2+ =n =0 p n xnProof. TheMaclaurinseriesoftheexponentialfunction exand thelogarithmfunctionln(1+ x )havebeenalreadyestablishedinSections58.3and58.2,respectively.Let f ( x )=sin x .Onehas f( x )= (sin x )=cos x and f( x )=(cos x )= Š sin x .Hence, f(2 n )( x )=( Š 1)nsin x,f(2 n +1)( x )=( Š 1)ncos x, and f(2 n )(0)=0, f(2 n +1)(0)=( Š 1)n Š 1.ByTheorem8.30,thenon-zero coecientsoftheMaclaurinseriesare c2 n +1= f(2 n +1)(0) (2 n +1)! = ( Š 1)n (2 n +1)! n =0 1 2 ,... Bytheratiotest, limn | c2 n +3x2 n +3| | c2 n +1x2 n +1| =limn x2(2 n +1)! (2 n +3)! =limn x2 (2 n +2)(2 n +3) =0 thepowerseriesconvergesforany x .TheMaclaurinseriesfor f ( x )= cos x =(sin x )isobtainedbydierentiatingtheseriesforsin x termby-term.ByTheorem8.28,italsoconvergesontheentirerealline. Finally,put f ( x )=(1+ x )p.Thederivativesare f( x )= p (1+ x )p Š 1, f( x )= p ( p Š 1)(1+ x )p Š 2,and,ingeneral, f( n )( x )= p ( p Š 1) ( p Š n +1)(1+ x )p Š n. Sothecoecientsofthepowerseriescoincidewiththebinomialcoef“cientsintroducedearlier: cn= f( n )(0) n = p ( p Š 1) ( p Š n +1) n = p n Theysatisfytherecurrencerelation cn +1= cn( p Š n ) / ( n +1).Therefore, byCorollary8.3,theradiusofconvergence R isdeterminedby limn | cn +1| | cn| =limn | p Š n | n +1 =limn | 1 Šp n| 1+1 n=1= R =1

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59.TAYLORSERIES151 59.3.TaylorSeriesofAnalyticFunctions.Everyanalyticfunctionina neighborhoodofapointisrepresentedbytheTaylorseriesaboutthat point.IftheTaylorseriesconvergesontheentirerealline,thenthe functionisanalyticeverywhere.Inparticular,theexponential exand trigonometricfunctionssin x andcos x areanalyticeverywhere.Moreover,thepropertiesofanalyticfunctionsstatedinTheorem8.32allows ustoadd,multiply,andmakeacompositionoftheTaylorseries(on thecommonintervalsoftheirconvergence)justlikeordinarysumsto obtaintheTaylorseriesrepresentationofthesums,products,andcompositionsofanalyticfunctions.Theseareextremelyusefulproperties inapplications. Example 8.47 Find“rstfourtermsoftheTaylorseriesforthe function f ( x )=exp(tanŠ 1x ) about x =0 Solution: Calculationofthederivativesofsuchafunctionisrather tedious.Instead,notethat exandtanŠ 1x arebothanalyticinaneighborhoodof x =0.SothecompositionoftheirTaylorseries(see(8.20) andExample8.44)givesthesought-afterTaylorseries.Onlymonomials1, x x2,and x3havetoberetainedwhencalculatingthecomposition.Thisimpliesthatitissucienttoretaintwoleadingtermsin theTaylorseriestanŠ 1x = x Š x3/ 3+ andfourleadingtermsin theTaylorseries(8.20)oftheexponentialfunction: etanŠ 1x=1+tanŠ 1x + 1 2 tanŠ 1x 2+ 1 6 tanŠ 1x 3+ =1+ x Š x3 3 + + 1 2 x Š x3 3 + 2+ 1 6 x Š x3 3 + 3+ =1+ x Š x3 3 + 1 2 x2+ 1 6 x3+ =1+ x + 1 2 x2Š 1 6 x3+ 59.4.ApproximationsbyTaylorPolynomials.Ananalyticfunction f can beapproximatedbyapartialsumoftheTaylorseries: f ( x ) nk =0f( k )( a ) k ( x Š a )k= Tn( x ) Thepolynomial Tn( x )iscalleda Taylorpolynomial ofdegree n about a .TheconvergenceoftheTaylorseriesguaranteesthattheremainder convergesto0: Rn( x )= f ( x ) Š Tn( x ) 0as n for | x Š a |
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1528.SEQUENCESANDSERIES where R istheradiusofconvergenceoftheTaylorseries.Theaccuracy oftheTaylorpolynomialapproximationforafunctionisassessedin TaylorstheoremdiscussedinCalculusI.Hereitisrestatedinaslightly dierentform. Theorem 8.34(TaylorsTheorem) Supposeafunction f isanalyticnear a andlet Tn( x ) beitsTaylorpolynomialsabout a .Then,for every n andany | x Š a |
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59.TAYLORSERIES153 Figure8.9. Anillustrationofanapproximationof f ( x )=sin x (thedashedredcurve)byitsTaylorpolynomialsat x =0(thesolidbluecurve).As n increases, Tn( x )approaches f ( x )=sin x .Theapproximationbecomesbetterinalargerintervalforalarger n inaccordancewiththeanalysisofExample8.50. Example 8.48 Approximatethefunction ln(1+sin x ) bytheTaylor polynomialofdegree3abouttheorigin. Solution: Firstnotethatthefunctioninquestionisacomposition ofanalyticfunctions.Sinceln(1+ u )= u Š u2/ 2+ u3/ 3+ ,the polynomial T3( x )maybeobtainedbyretainingonlytermsatmost cubic in x ,whenexpanding u =sin x x Š x3/ 6.Hence,neglecting allmonomialsofdegree4andhigher ln(1+sin x ) x Š x3 6 Š 1 2 x Š x3 6 2+ 1 3 x Š x3 6 3 x Š 1 2 x2+ 1 6 x3= T3( x )

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1548.SEQUENCESANDSERIES Example 8.49 Approximatethefunction xpbytheTaylorpolynomialofdegree 2 about x =1 Solution: If f ( x )= xp,then f( x )= pxp Š 1and f( x )= p ( p Š 1) xp Š 2. Therefore f (1)=1, f(1)= p ,and f(1)= p ( p Š 1).Theneeded Taylorpolynomialis T2( x )= f (1)+ f(1)( x Š 1)+ f(1) 2 ( x Š 1)2=1+ p ( x Š 1) Š p ( p Š 1) 2 ( x Š 1)2Alternatively,onecouldusetheidentity f ( x )= xp=(1+( x Š 1))p= (1+ u )pandthebinomialseriesin u Example 8.50 FindanupperboundontheerroroftheTaylor polynomialapproximationabout x =0 forthefunction f ( x )=sin x Inparticular,giveanupperboundfortheerroroftheapproximation sin x T7( x ) intheinterval [ Š 1 1] Solution: TheMaclaurinseriesforsin x containsonlyoddpowers of x andsoaretheTaylorpolynomials( f(2 n +2)( x )=( Š 1)n +1sin x and,hence, f(2 n +2)(0)=0).Therefore T2 n +1( x )= T2 n +2( x )andthe accuracyoftheapproximationofsin x by T2 n +1( x )isdeterminedby | f(2 n +3)( x ) | = | ( Š 1)n +1cos x | 1= M uniformlyforall x andall n ByCorollary8.4, sin x Š T2 n +1( x ) = sin x Š T2 n +2( x ) = | x |2 n +3 (2 n +3)! Theerroroftheapproximation sin x x Š x3 3! + x5 5! Š x7 7! | x | 1 islessthan1 / 9! 0 000003. Corollary8.4alsoshowsthat,fora“xed n ,theerrormaygrow withincreasing | x | (asintheaboveexample).ThisimpliesthatTaylor polynomialsofhigherdegreesareneededtoachievethesameaccuracy forlarge | x | asforsmaller | x | (seeFig.8.9).Toavoidusinghigh-degree Taylorpolynomialstoapproximatethefunctionatlarge | x | ,onecan useTaylorpolynomialsaboutsome a closetotherangeof x inwhich theapproximationisneeded.

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59.TAYLORSERIES155 59.5.Exercises.In(1)…(5),“ndtheMaclaurinseriesforthefunctionandtheradiusof convergence. (1)ln(1+ x )(2)tan x (3)sinh x (4)cosh x (5) x6+2 x5Š x3+ x Š 3 In(6)…(9),“ndtheTaylorseriesforthefunctionabout a andtheradius ofconvergence. (6)cos x,a = (7)1 / x,a =4 (8)sin x,a = / 2(9)(1+ x )2 / 3,a =7 In(10)…(13),usetheMaclaurinseriesforbasicfunctionsto“ndthe Maclaurinseriesforthefunction. (10) x cos( x2/ 2)(11) x 3 1+ x4(12) x Š sin x x3(13) x2tanŠ 1( x2) In(14)…(17),usetheproductsandcompositionoftheMaclaurinseries forbasicfunctionsto“ndthe“rstthreenon-vanishingtermsofthe Maclaurinseriesforthefunction. (14)sin( (cos x ))(15) esin x(16)tanŠ 1x ln(1+ x )(17)ln(cos x ) (18)Findthe“rst“venonvanishingtermsoftheMaclaurinseries for f ( x )= ex/ cos x Hint :Put f ( x )= c0+ c1x + + c4x4+ ,thenusetheproductoftheMaclaurinseriesto“ndthe coecientsfrom ex= f ( x )cos x In(19)…(21),“ndthedegreeofaTaylorpolynomialtoapproximate theintegrandsothattheerrorofapproximatingtheintegraldoesnot exceed10Š 3. (19) 1 / 2 0tanŠ 1( x2) dx (20) 1 0exŠ 1 x dx (21) 1 / 2 0(1+ x4)1 / 4dx In(22)…(24),“ndthesumoftheseries. (22)n =0( Š 1)n2 n 62 n(2 n )! (23)n =0( Š 1)n3n 2nn (24)1 Š ln2+ (ln2)2 2! Š (ln2)3 3! + (25)(i)Forthefunction f ( x )= eŠ 1 /x2if x =0and f (0)=0,show that f( n )(0)=0forall n andhence f cannotberepresented asapowerseriesnear0.(ii)Let f ( x )= eŠ 1 /xif x> 0and f ( x )=0if x 0.Isthisfunctionanalyticeverywhere? (26)Usetheidentity / 6=sinŠ 1(1 / 2)tocalculatethenumber withintheerrornotexceeding10Š 4.

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1568.SEQUENCESANDSERIES (27)Findcos(1)withintheerrornotexceeding10Š 6. (28)Usethe“rstthreetermsofthebinomialseriestoapproximate3 9andestimatetheerror. In(29)…(32),usevariousmethodsto“ndtheMaclaurinseriesofthe functionandinvestigateitsconvergenceradius. (29) x 0eŠ t2dt (30) x 0dt 1 Š t4(31) x 0sin t t dt (32) x 0tanŠ 1x t dt (33)Let f ( x )= n =0cnxn,where | n cn|
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CHAPTER9 FurtherApplicationsofIntegration 60.ArcLength60.1.TheLengthofaCurve.Wehaveseenvariousapplicationsofintegrationtothecomputationoftheareaofadomainandtothecomputationofthevolumeofasolid.Itisperhapsmoresurprisingthat wecanalsouseintegrationtocomputethelengthofacurvebetween twogivenpoints.Thismaysoundcounterintuitiveat“rst,sincein theapplicationswehaveseensofar,integrationwasusedtocompute someparameterofanobjectthatexistedina higherdimension than thefunctionthatwasbeingintegrated. Let f beafunctionsothat,ontheinterval[ a,b ],thederivative fof f existsandisacontinuousfunction.Wewouldliketoknowthe lengthofthecurveof f ,startingatthepoint A =( a,f ( a ))andending atthepoint B =( b,f ( b )). Intuitively,wecanimaginethatwelayaropeoverthegraphof f betweenthetwoendpoints,mark A and B ontherope,thenstraighten thatropeout,andmeasurethedistancebetweenthem. Amoreformalde“nition,whichisusefulintheactualcomputation ofthelengthofthecurve,isthefollowing.Cuttheinterval[ a,b ] into n equalparts,usingpoints a = x0
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1589.FURTHERAPPLICATIONSOFINTEGRATION Figure9.1. Arclengthasalimit. Letusnowreturnto(9.1)inordertocomputelimn Kn.Let ( b Š a ) /n =( xiŠ xi Š 1) /n = x .Notethatthen | Pi Š 1Pi| = ( xiŠ xi Š 1)2+( yiŠ yi Š 1)2= ( x )2+( x )2( yiŠ yi Š 1)2 ( xiŠ xi Š 1)2= x 1+ ( yiŠ yi Š 1)2 ( xiŠ xi Š 1)2. Nowobservethatsince fiscontinuous,theMeanValueTheorem (Theorem6.9)impliesthatthereisarealnumber x i [ xi Š 1,xi]such thatyiŠ yi Š 1 xiŠ xi Š 1= f( x i).Hence,thepreviouschainofequalitiesyields | Pi Š 1Pi| = x 1+ f( x i)2. Summingoverall i ,weget Kn=ni =1 x 1+ f( x i)2. As n goestoin“nity,theleft-handside,byde“nition,convergestothe lengthofthecurveof f between A and B ,whiletheright-handside, beingaRiemannsum,convergesto b a 1+ f( x )2dx .Hence,wehave provedthefollowingtheorem. Theorem 9.1 If fisacontinuousfunctionontheinterval [ a,b ] thenthelengthofthegraphof f ( x ) fromthepoint ( a,f ( a )) tothepoint

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60.ARCLENGTH159 Figure9.2. Thecurveof f ( x )=2 3x3 / 2. ( b,f ( b )) isequalto L = b a 1+ f( x )2dx. Example 9.1 Findthelengthofthecurveof f ( x )=2 3x3 / 2from (0 0) to (1 2 / 3) .SeeFigure9.2foranillustration. Solution: Wehave f( x )= x ,so fisacontinuousfunctionon[0 1], andthereforeTheorem9.1applies.Usingthattheorem,weobtain L = 1 0 1+( x )2dx = 1 0 1+ xdx = 2 3 (1+ x )3 / 2 1 0= 4 2 3 Š 2 3 Notethattheresultisremarkablyclosetothelengthofthe straight line thatconnectsthetwopointsinquestion,whichis 13 / 3. Wecanuseournewtechniquetoverifyaclassicformula. Example 9.2 UseTheorem9.1tocomputethecircumferenceofa circleofradius1. Solution: Letusplacethecenteroftheunitcircleattheorigin.Then theboundaryofthecircleisthesetofpointssatisfying x2+ y2=1.

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1609.FURTHERAPPLICATIONSOFINTEGRATION Figure9.3. Onequarteroftheunitcircle. WewanttouseTheorem9.1,soweneedapartofthecirclewherethat satis“estheverticallinetest(so y isafunctionof f )andwherethe tangentlinetothecircleisneververtical(sothat f( x )exists).For instance,wecanchoosethequarterofthecirclethatstartsinthepoint x = Š 2 2 andendsinthepoint x = 2 2 .SeeFigure9.3foran illustration.Onthatpartofthecurve, f( x )= Šx 1 Š x2iscontinuous, soTheorem9.1implies L = 2 / 2 Š 2 / 2 1+ f( x )2dx = 2 / 2 Š 2 / 2 1+ x2 1 Š x2dx = 2 / 2 Š 2 / 21 1 Š x2dx =sinŠ 1x 2 / 2 Š 2 / 2= / 2 Thisimpliesthatthecircumferenceofthefullcircleisfourtimesthis much,thatis,2

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60.ARCLENGTH161 60.2.Remarks.Recallthatinthe“rstparagraphofthissection,we discussedwhyitmayseemcounterintuitivethatintegrationplaysarole inthecomputationofarclengths.Nowwecanseethatthepurported contradictionexplainedthereisresolvedbythefactthattheintegrand inTheorem9.1contains f,not f Comparedtootherformulaswelearnedinourearlierstudiesof integration,itisrelativelyrarethattheformulagivenbyTheorem9.1 canbeexplicitlycomputed,since b a 1+ f( x )2dx isoftendicult tohandle.Therefore,wemustoftenresorttoapproximateintegration whilecomputingarclengths.60.3.Exercises.(1)Findthelengthofthecurve f ( x )= x2/ 2betweenthepoints givenby x =0and x =1. (2)Findthelengthofthecurve f ( x )=3 x2/ 2betweenthepoints givenby x =0and x =1. (3)Findthelengthofthecurve f ( x )=x2 4Šln x 2betweenthepoints givenby x =2and x =4. (4)Findthelengthofthecurve f ( x )=ln(cos x )betweenthe pointsgivenby x =0and x = / 4. (5)Findthelengthofthecurve f ( x )=ln(sin x )betweenthe points a = / 4and b =3 / 4. (6)Findthelengthofthecurve f ( x )=ln x Šx2 8from x =4to x =5. (7)VerifythatTheorem9.1providesthecorrectvalueforthearc lengthof f when f isa linearfunction (8)Let f =1 4( x2+2 x )3 / 2Š1 2ln( x +1+ x2+2 x ).Findthelength ofthecurveof f from x =1to x =2. (9)Let f =1 4(2 x +4) x2+4 x +3 Š1 2ln( x +2+ x2+4 x +3). Findthelengthofthecurveof f from x =1to x =2. (10)Let f ( x )=1 2x x2Š 1 Š1 2ln( x + x2Š 1).Findthelength ofthecurveof f from x =1to x =2. (11)Let f ( x )=1 2x x2+1+1 2ln( x + x2+1).Findthelength ofthecurveof f from x =1to x =2. (12)Let f and g betwofunctionsthathavecontinuousderivatives ontheinterval[ a,b ]andassumethat f( x )= Š g( x )forall x [ a,b ].Provethatthecurvesof f and g havethesame lengthbetween x = a and a = b .Tryto“ndtwodierent arguments. (13)Useamethodofapproximateintegrationtoestimatethelength ofthecurve f ( x )= ex2asfrom(0 1)to(1 ,e ).

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1629.FURTHERAPPLICATIONSOFINTEGRATION (14)Useamethodofapproximateintegrationtoestimatethelength ofthecurve f ( x )= exasfrom(0 1)to(1 ,e ).Comparethe resultwiththatofthepreviousexercise. (15)Computetheexactarclengthinthepreviousexercisebyeither usingasoftwarepackageoranappropriatesubstitutionto“nd theneededinde“niteintegral. (16)Useamethodofapproximateintegrationtoestimatethelength ofthecurve f ( x )=sin x from(0 0)to( 0). (17)Useamethodofapproximateintegrationtoestimatethelength of f ( x )= x3from(0 0)to(1 1). (18)Usetheresultofthepreviousexercisetoestimatethelength of f ( x )=cos x from x = Š to x = (19)Findthelengthofthecurve f ( x )=ln x from x =1to x = 2.Youcanuseasoftwarepackagetocomputetheneeded inde“niteintegral. (20)Useamethodofapproximateintegrationtoestimatethelength of f ( x )=tan x from x =0to x = / 4. 61.SurfaceArea61.1.TheDenitionofSurfaceArea.Inthelastsection,wede“nedthe lengthofacurve,anddeducedaformulaforthecomputationofthat length.Letusnowtakeacurve,sayofafunction f ( x )= y ,where x [ a,b ]and fiscontinuouson[ a,b ].Letus rotate thiscurvearound thehorizontalaxis,asshowninFigure9.4.Whatistheareaofthe obtained surfaceofrevolution ? Thede“nitionoftheareainquestion,anditscomputation,willbe quitesimilartowhatwehavediscussedintheprevioussectionforthe arclength. Cuttheinterval[ a,b ]into n equalparts,usingpoints a = x0
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61.SURFACEAREA163 Figure9.4. Asurfaceobtainedbyrotatingacurve. Figure9.5. Approximatingasurfaceofrevolution. Infact,itcanbeprovedthatthelimit (9.3) A ( S )=limn ni =1Sn,i=limn ni =1 ( yi Š 1+ yi) liexists.Wede“ne thislimit tobetheareaofthesurfaceofrevolution obtainedwhenthecurveof f isrotatedaroundthehorizontalaxis, with x [ a,b ]. Inordertocomputethissurfacearea,recallfromthelastsection thatthereexistsarealnumber x i [ xi Š 1,xi]suchthat li= | Pi Š 1Pi| =

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1649.FURTHERAPPLICATIONSOFINTEGRATION x 1+ f( x i)2.Alsonotethat,since f iscontinuous,smallchanges in x leadtosmallchangesin f ( x )= y ,soif n islargeenough,then f ( x i) f ( xi)= yiand f ( x i) f ( xi Š 1)= yi Š 1.Therefore,(9.3) implies (9.4) A ( S )=limn ni =12 f ( x i) x 1+ f( x i)2, where x =( b Š a ) /n .Nownoticethatthelastexpressionobtained for S ( A )isthelimitofaRiemannsum,thatis,anintegral(ofthe function2 f ( x ) 1+ f( x )2).Thismeansthatweprovedthefollowing theorem. Theorem 9.2 Let f beafunctionthat fiscontinuousonthe interval [ a,b ] .Let S bethesurfaceobtainedbyrotatingthecurve y = f ( x ) ,where x [ a,b ] ,aroundthehorizontalaxis. Thentheareaof S is A ( S )= b a2 f ( x ) 1+ f( x )2dx. Example 9.3 Computethesurfaceareaofasphereofradius r Solution: Suchaspherecanbeobtainedbyrotatingthesemicircle givenbytheequation f ( x )= r2Š x2aroundthehorizontalaxis.See Figure9.6foranillustration. Theorem9.2thenyields A ( S )= r Š r2 r2Š x2 1+ x2 r2Š x2dx = r Š r2 r2Š x2 r2 r2Š x2dx = r Š r2 rdx =2 rx r Š r=4 r2. 61.2.Variations.Ifwerotateourcurveaboutthe verticalaxis instead ofthehorizontalaxis,thenmostofthepreviousargumentremains valid.Theonlydierenceisthatwhenthepoint Pi=( xi,yi)isrotated, itisrotatedinacircleofradius xi,not yi.Thisleadstothefollowing theorem.

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61.SURFACEAREA165 Figure9.6. Asphereasasurfaceofrevolution. Theorem 9.3 Let f beafunctionsuchthat fiscontinuouson theinterval [ a,b ] .Let S bethesurfaceobtainedbyrotatingthecurve y = f ( x ) ,where x [ a,b ] ,abouttheverticalaxis. Thentheareaof S is A ( S )= b a2 x 1+ f( x )2dx. Notethatthe f ( x )termintheintegrandofTheorem9.2isreplaced by x Example 9.4 Rotatethecurvegivenby y = f ( x )= x2/ 2 ,with x [0 1] ,aroundtheverticalaxis.Findtheareaoftheobtainedsurface. Figure9.7. Thecurveof y = x2/ 2andthesurface obtainedbyitsrotation.

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1669.FURTHERAPPLICATIONSOFINTEGRATION Solution: Theorem9.3implies A ( S )= 1 02 x 1+ x2dx =2 1 3 ( x2+1)3 / 2 1 0=2 2 2 Š 1 3 0 6095 NotethatanotherwayofwritingtheresultofTheorem9.2is A ( S )= b a2 f ( x ) 1+ dy dx 2dx. Byinterchangingtherolesof x and y ,thisimpliesthat,forcurvesgiven byanequation g ( y )= x ,thefollowingholds. Theorem 9.4 Let g beafunctionsuchthat giscontinuouson theinterval [ a,b ] .Let S bethesurfaceobtainedbyrotatingthecurve x = g ( y ) ,where y [ a,b ] ,aroundtheverticalaxis. Thentheareaof S isgivenby A ( S )= b a2 g ( y ) 1+ dx dy 2dy. Whilethesurfaceareaofaconecanbecomputedbyelementary methods,itiselucidatingtocomputeitwithournewmethodandsee thattheresultiswhatweexpectittobe. Example 9.5 Findthesurfaceareaofarightconeofbaseradius R andheight h Solution: Thebasecircleoftheconehasarea R2 .Inordertocomputethelateralsurface,notethatthelateralsurfacecanbeobtainedby rotatingthelinesegmentgivenbytheequation x = ŠRy h+ R = g ( y ), where y [0 ,h ],aroundtheverticalaxis.Therefore,Theorem9.4 applies,andforthelateralsurfacearea,ityields

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61.SURFACEAREA167 A ( S )= h 02 Š Ry h + R 1+ R2 h2dy =2 R R2+ h2 h2Š y2 2 h + y h 0= R R2+ h2= Rs, where s = R2+ h2isthe slantheight ofthecone. Sothetotalsurfaceareaoftheconeisthesumoftheareaofits baseplustheareaofitslateralsurface,thatis, R2 + Rs = R ( R + s ). Theorem9.4alsohasaversionthatappliestocurvesgivenasfunctionsof y thatarerotatedaroundthehorizontalaxis. Theorem 9.5 Let g beafunctionsuchthat giscontinuouson theinterval [ a,b ] .Let S bethesurfaceobtainedbyrotatingthecurve x = g ( y ) ,where y [ a,b ] ,aboutthehorizontalaxis. Thentheareaof S isgivenby A ( S )= b a2 y 1+ dx dy 2dy. Notethatforthecomputationofsomesurfaceareas,wewillhave achoiceoftwotheoremsdiscussedinthissection.Thereaderisencouragedtodescribethecurveswhoserotationsleadtosuchsurfaces.61.3.Exercises.(1)Rotatethecurveof y = f ( x )= x2+5,where x [2 3],about the y axis.Findthesurfacearea. (2)Rotatethecurveof y = f ( x )=9 x2+1,where x [2 3],about the y axis.Findthesurfacearea. (3)Rotatethecurveof y = f ( x )=4 3x3 / 2,where x [0 1],about the y axis.Findthesurfacearea. (4)Computethesurfaceareaobtainedbyrotatingthecurve y = ex,for x [0 1],aboutthe x axis. (5)Rotatethecurveof y = f ( x )= x ,where x [0 1],about the x axis.Findthesurfacearea. (6a)Rotatethecurveof y = f ( x )=3 x ,where x [3 4],about the y axis.Findthesurfacearea.

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1689.FURTHERAPPLICATIONSOFINTEGRATION (6b)Rotatethecurveof y = f ( x )= 9 x +1,where x [3 4], aboutthe x axis.Findthesurfacearea. (7)Rotatethecurveof y = f ( x )= x3,where x [0 1],aboutthe x axis.Findthesurfacearea. (8)Rotatethecurveof y = f ( x )= x3,where x [0 1],aboutthe y axis.Findthesurfacearea. (9)Rotatethecurveof x = g ( y )=2 3( y +1)3 / 2,where y [2 3], aboutthe x axis.Findthesurfacearea. (10)Rotatethecurveof x = g ( y )=1 2y2,where y [2 3],about the x axis.Findthesurfacearea. (11)SolveExample9.5usingTheorem9.3. (12)Let f beafunctionsuchthat fiscontinuouson[ a,b ].Prove thattheareaofthesurfaceobtainedbyrotatingthegraphof f from x = a to x = b aboutthe y axisisthesameasthe areaofthesurfaceobtainedbyrotatingthesamesegmentof thegraphof f1( x )= f ( x )+ C aboutthe y axis.Here C isan arbitraryconstant.Tryto“ndtwodierentarguments. (13)Let f beafunctionsuchthat fiscontinuouson[ a,b ].Prove thattheareaofthesurfaceobtainedbyrotatingthegraphof f from x = a to x = b aboutthe x axisisthesameastheareaof thesurfaceobtainedbyrotatingthegraphof f2( x )= f ( x + C ) from x = a Š C to x = b Š C aboutthe x axis.Here C isan arbitraryconstant.Tryto“ndtwodierentarguments. (14)Rotatethecurveof y = f ( x )= x2,where x [2 3],aboutthe x axis.Findthesurfacearea.( Hint :Youmaywanttousethe factthat 1 / 1+ x2dx =sinhŠ 1( x ).) Intheremainingexercisesofthissection,useamethodofapproximateintegrationtocomputetheareaofthegivensurface ofrevolution. (15) f ( x )=sin x x [0 ],rotatedaboutthe x axis. (16) f ( x )=sin x x [0 ,/ 2],rotatedaboutthe y axis. (17) f ( x )=ln x x [1 2],rotatedaboutthe x axis. (18) f ( x )=ln x x [1 2],rotatedaboutthe y axis. (19) g ( y )= eŠ y2, y [0 1],rotatedaboutthe x axis. (20) g ( y )= eŠ y2, y [0 1],rotatedaboutthe y axis. 62.ApplicationstoPhysicsandEngineering62.1.CenterofMass. 62.1.1.One-DimensionalSystems.Letusassumethatwehavetwo objectsofmass m1and m2placedonthelineofrealnumbers,at

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62.APPLICATIONSTOPHYSICSANDENGINEERING169 points x1and x2,respectively.Wewantto“ndthepoint xgsuchthat ifweplaceafulcrumundertheinterval[ x1,x2]at xg,theobjectsatthe endpointsoftheintervalwillbalance.SeeFigure9.8foranillustration. Weassumethattheinterval[ x1,x2],orthestickrepresentingit,has negligiblemass. If m1= m2,thenweclearlyhave xg=( x1+ x2) / 2.Otherwise, wemakeuseofthewell-knownfactofphysicsthattheintervalwill balanceifthe moments onthetwosidesofthefulcrumareequal,that is,when (9.5) m1( xgŠ x1)= m2( x2Š xg) holds.Solving(9.5)for xg,weget (9.6) xg= m1x1+ m2x2 m1+ m2. Thepoint xgofthereallineiscalledthe centerofmass or centerof gravity ofthesystemdescribedabove,thatis,thesystemofanobjectof mass m1at x1andanobjectofmass m2at x2.The moment ofanobject withrespecttoapoint P isthemassoftheobjecttimesthedistanceof theobjectfrom P .Inparticular,intheabovesystem,thetwoobjects hadmoments m1x1and m2x2withrespecttotheorigin.Sothetotal systemhadmoment m1x1+ m2x2.Notethatifwereplacethetwoobjectsbyasimpleobjectofmass m1+ m2placedat xg,thenthemoment ofthesystemabouttheorigindoesnotchange.Thisisanimportant propertythatonlythecenterofmasshas,andthereforewerepeatit. Ifweconcentratethetotalmassofthesystematthecenterofmass, themomentofthesystemwithrespecttotheoriginwillnotchange. Ifweconsiderasystemof k distinctobjectsofmass m1,m2,...,mkplacedatpoints x1,x2,...,xkalongthehorizontalaxis,thenwecan useananalogousargumenttoshowthatthecenterofmassofthe systemisat Figure9.8. Centerofmass.

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1709.FURTHERAPPLICATIONSOFINTEGRATION (9.7) xg= k i =1mixi k i =1mi.62.2.Two-DimensionalSystems. 62.2.1.DiscreteTwo-DimensionalSystems.Letusnowconsiderthe moregeneralcasewhenthe k objectsofmass m1,m2,...,mkareplaced inpoints( x1,y1) ( x2,y2) ,..., ( xk,yk)ofthe plane .Wewouldliketo “ndthecenterofmass( xg,yg)ofthissystem.Inotherwords,we assumethataplateofnegligiblemassisplacedunderoursystem,and wewantto“ndthepoint( xg,yg)withthepropertythatifweplacea fulcrumundertheplateatthatpoint,theplatewillbalance. Usingmethodssimilartotheone-dimensionalcase,itcanbeproved thattheplatewillbalanceifthefulcrumisplacedat( xg,yg)with (9.8) xg= k i =1mixi k i =1miand yg= k i =1miyi k i =1mi. Thiscorrespondstotheintuitivelyappropriateconceptthatthe platewillbalanceifitbalancesbothhorizontallyŽandvertically.Ž Thesum Mx= k i =1miyiiscalledthe momentofthesystemwith respecttothe x axis .Thisnameisduetothefactthatifwetriedto balancethesystemonthe x axis,thelargerthenumber Mx,themore wouldtheweightsofthesystemrotatetheplate.Similarly,thesum My= k i =1mixiiscalledthe momentofthesystemwithrespecttothe y axis .62.2.2.SymmetryLines.Nowletusconsiderthe continuous versionof theproblem.Let P beaplateandletustryto“ndthecenterofmass of P .(Wenolongerassumethatthemassoftheplateisnegligible; infact,thatmassistheobjectofourstudynow.)Letusassume,for therestofthischapter,thatthemassof P isuniformlydistributed over P .Letusalsoassumethatthe density ofthematerialofwhich P ismadeis1.Thatis,themassofaunitsquarewithin P is1.Once theseassumptionsaremade,itisclearthatthecenterofmassof P is determinedbythepurely geometric characteristicsof P .Inorderto emphasizethis,iftheassumptionsofthisparagraphhold,wewillcall thecenterofmassof P the centroid of P Sometimeswecan“ndthecentroidof P withoutcomputation.A symmetryline of P isastraightline t suchthattheimageof P when re”ectedthrough t is P itself.Thatimpliesthatthetwopartsinto which t cuts P arecongruent,andtheplatebalancesontheline t Consequently,thecentroid C of P mustbeon t ,sinceifweconcentrate theentiremassof P in C ,itstillhastobalanceontheline P .

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62.APPLICATIONSTOPHYSICSANDENGINEERING171 Theargumentofthepreviousparagraphshowsthatthecentroid of P mustbeon every symmetrylineof P .Soif P hasmorethan onesymmetryline,thenthesesymmetrylinesmustallintersectinone point,namely,inthecentroidof P .Inthiscase,weobtainthecenter ofmassof P astheintersectionofanytwosymmetrylinesof P .For example,wecan“ndthecenterofmassofacircle,ellipse,rectangle, orrhombusinthisway.62.2.3.AFormulaforContinuousTwo-DimensionalSystems.Letuskeep theconditionsfromtheprevioussectionandletusimposethenewconditionthat P isadomainunderacurveŽ;thatis,thebordersof P are theverticallines x = a and x = b ,thehorizontalaxis,andthegraph ofthecontinuousfunction f ( x )= y Wewouldliketouseformula(9.8)to“ndtheapproximatelocation ofthecentroidof P .Letuscuttheinterval[ a,b ]into n equalparts, usingtheintermediatepoints a = x0
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1729.FURTHERAPPLICATIONSOFINTEGRATION centerofmassistheuniquepoint( xg,yg)withthepropertythatif theentiremass A = b af ( x ) dx of P isplacedinthatpoint,thenthe momentsofthisone-objectsystemareidenticaltothemomentsof P Inotherwords, Mx= ygA and My= xgA .Therefore, xg= My/A and yg= Mx/A ,whichmeansthatformulas(9.9)and(9.10)implythe followingtheorem. Theorem 9.6 Let f beafunctionsuchthat f ( x ) 0 if x [ a,b ] Let D beadomainwhosebordersaretheverticallines x = a and x = b thehorizontalaxis,andthecurveofthefunction f ( x )= y .Let A ( D ) denotetheareaof D Let xgand ygbethecoordinatesofthecentroidof D .Thenwehave xg= b axf ( x ) dx A ( D ) and yg= b a 1 2[ f ( x )]2dx A ( D ) Example 9.6 Findthecoordinatesofthecentroidofthequarter oftheunitcirclethatisinthenortheasternquadrant. Notethatifweaskedthesamequestionforthe entire unitcircle, theanswerwouldobviouslybethat xg= yg=0,sincethecentroid ofanydomainmustbeonallsymmetrylines.Ifweaskedthesame questionforthehalfoftheunitcirclethatisinthenorthernhalf-plane, then xg=0wouldclearlyhold,sincetheverticalaxisisasymmetry lineofthatsemicircle. Solution: (ofExample9.6):Notethatthedomain D inquestionhas asymmetryline,namely,thelinedeterminedbytheequation x = y Sothecentroidof D isonthatline,thatis, xg= yg.Therefore,it sucestocomputeoneof xgand yg.Wehave f ( x )= 1 Š x2= y ,so ygissomewhateasiertocompute.Theorem9.6yields yg= 1 0 1 2 (1 Š x2) dx / 4 = 2 x Š x3 3 1 0= 2 2 3 = 4 3 Sothecenterofgravityofthequarteroftheunitcircleinthenortheasternquadrantisat(4 3 ,4 3 ).SeeFigure9.9foranillustration.

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62.APPLICATIONSTOPHYSICSANDENGINEERING173 Notethat4 / (3 ) 0 424.Thismakesperfectsensesincethisshows thatthecentroidof D isclosertothehorizontalaxis(thebottomof thequartercircle)thantothe y =1line(thetopofthequartercircle). Thatisreasonable,sincethebottomof D iswiderthanthetopof D soitconstitutesalargerportionofthetotalweightof D thanthetop of D Example 9.7 Findthecentroidofthedomain D whoseborders aretheverticallines x =0 and x =1 ,thehorizontalaxis,andthe graphofthefunction f ( x )= x2= y Solution: Thedomain D inquestiondoesnothaveasymmetryline, sowemustuseTheorem9.6tocomputebothof xgand yg.Wehave xg= b axf ( x ) dx A ( D ) = 1 0x x2dx 1 0x2dx = x4/ 4 1 0 x3/ 3 1 0= 3 4 Figure9.9. Thecentroidofaquarteroftheunitcircle.

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1749.FURTHERAPPLICATIONSOFINTEGRATION and yg= b a 1 2[ f ( x )]2dx A ( D ) = 1 0( x4/ 2) dx 1 / 3 = x5/ 10 1 0 1 / 3 = 3 10 Sothecentroidof D isat(0 75 0 3).Thisagreeswithourintuition, sincethebottomof D islargerthanitstop,andtheleft-handsideof D issmallerthantheright-handsideof D .SeeFigure9.10foran illustration. 62.3.Exercises.(1)Findthecentroidoftheunitsemicirclethatliesinthenorthern half-plane. (2)Findthecentroidoftheunitsemicirclethatliesinthewestern half-plane. (3)Findthecentroidoftheplatewhosebordersarethelines x =0 and x = / 2,thegraphofthefunction f ( x )=sin x ,andthe horizontalaxis. (4)Findthecentroidoftheplatewhosebordersarethevertical lines x =1and x =2,thehorizontalaxis,andthegraphof thefunction f ( x )=ln x Figure9.10. Centerofmassoftheareade“nedinExample9.7.

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62.APPLICATIONSTOPHYSICSANDENGINEERING175 (5)Findthecentroidoftheplatewhosebordersarethevertical lines x =1and x =2,thehorizontalaxis,andthegraphof thefunction f ( x )= ex. (6)Findthecentroidoftheplatewhosebordersarethevertical lines x =0and x =1,thehorizontalaxis,andthegraphof thefunction f ( x )= x3. (7)Findthecentroidoftheplatewhosebordersarethevertical lines x =0and x =1,thehorizontalaxis,andthegraphof thefunction f ( x )= x (8)Findthecentroidoftheplatewhosebordersarethevertical lines x =0and x =1,thehorizontalaxis,andthegraphof thefunction f ( x )= x (9)Findthecentroidoftheplatewhoseverticesareat(0 0), (10 10),and(20 0). (10)Findthecentroidoftheplatewhoseverticesareat(0 0), (10 10),and(50 0). (11)Findthecentroidoftheplatewhoseverticesareat(0 0), (15 0),(1 1),and(8 1). (12)Let ABC beanytriangle.Letusassignobjectsofequalmass to A B ,and C .Let A1denotethemidpointofthesegment BC ,let B1denotethemidpointofthesegment AC ,andlet C1denotethemidpointofthesegment AB .Provethatthe lines AA1, BB1,and CC1intersectinonepoint,centroidof thesystem ABC (13)Keepthenotationoftheprecedingexercise.Provethatthe centerofmassof ABC isthesameasthecentroidof A1B1C1, ifweassignobjectsofequalweightto A1, B1,and C1. (14)Anobjectconsistsoftwosquares.The“rstisthesquare withvertices(0 0),(0 2),(2 0),and(2 2),andtheotheris thesquarewithvertices(0 2),(1 2),(0 3),and(1 3).The densityofthematerialofthesmallsquareistwicethedensity ofthematerialofthelargesquare.Whereisthecentroidof thisobject? (15)Let n> 1beapositiveinteger.Findthecentroid M ( n )ofthe platewhosebordersaretheverticallines x =1and x = n thehorizontalaxis,andthegraphofthefunction f ( x )= xŠ 3. (16)Keepthenotationoftheprecedingexercise.Describethebehaviorofthepoint M ( n )as n goestoin“nity. (17)Howdoestheanswertothetwoprecedingexerciseschangeif f isreplacedby g ,where g ( x )= xŠ 2?

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1769.FURTHERAPPLICATIONSOFINTEGRATION Figure9.11. Thedemandfunction p ( x )ofacommodity. (18)Solveexercise15with f unchanged,butwiththetwovertical lineschangedto x =1 /n and x =1.Tryto“ndananswer withoutadditionalintegration. (19)Findthecentroidoftheplatewhosebordersarethehorizontal lines y =0and y =1,theverticalaxis,andthegraphofthe function x = y (20)Findthecentroidoftheplatewhosebordersarethehorizontal lines y =0and y =1,theverticalaxis,andthegraphofthe function x = y +1. 63.ApplicationstoEconomicsandtheLifeSciences63.1.ConsumerSurplus.Letusconsidertheproblemofpricingsome merchandisewhosevalueishighlysubjective;thatis,itisworthmore tosomecustomersthantoothers.Examplesofthiscouldbetickets forvarioussportingevents,airlineticketstovacationdestinations,or popularbooks. Let p ( x )bethe demand functionofthiscommodity.Thatis, p ( x ) isthepricethatwillresultinselling x unitsofthecommodity.Lower pricesusuallyleadtohighersales,therefore p ( x )isusuallya decreasing function asillustratedinFigure9.11. Theareaunderthegraphof p representsthetotalrevenuethe companycouldpossiblyhave,ifitmanagedtochargeeachcustomer themaximumpricethatthatcustomeriswillingtopay.Indeed,ifthe highestamountanyoneiswillingtopayforoneunitis p ( x1),and x1customersarewillingtopaythatprice,thentherevenuecomingfrom thesemostenthusedcustomersis x1p ( x1),whichistheareaofthe domainunderthegraphof p thatisbetweentheverticallines x =0

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63.APPLICATIONSTOECONOMICSANDTHELIFESCIENCES177 and y = x1.Wecouldcontinueinthisway,notingthatifthesecond highestpricethatsomecustomersarewillingtopayis p ( x2),andthere are x2Š x1peoplewhoarewillingtopaythisprice(notincludingthose whoarewillingtopayeven p ( x1)),thentherevenuefromthemwill be( x2Š x1) p ( x2).Thisistheareaofthedomainunderthegraphof p thatisbetweenthelines x = x1and x = x2,andsoon. Ifthesellerdecidestosetone“xedprice p ( z ),thenthesellerwill sell z items,foratotalrevenueof zp ( z )(theareaoftherectangle R borderedbythetwocoordinateaxesandthelines x = z and y = p ( z )). Thismeansthatthecustomerswhowouldhavepaidanevenhigher priceforthesegoodshavesavedmoney.Besideslosingthatpotential revenue,theselleralsolosesrevenuebynotgettinganypurchasesfrom customerswhowerewillingtopaysomeamount,butnot p ( z ),forone unit. Let xnbethenumberofitemsthatthesellercansellatthelowest priceatwhichthesellerisstillwillingtoselltheseitems.Itisadirect consequenceoftheabovediscussionthatthetotalamountsavedbyall customerswhoboughttheitemat p ( z )dollarsisthearea underthe curveof p butabovetherectangle R ,thatis, (9.11)ni =1( p ( xi) Š p ( z ))( xiŠ xi Š 1) Ifthenumber n ofpricesatwhichvariouscustomersarewillingto buygoestoin“nity,thentheRiemannsumin(9.11)approachesthe de“niteintegral (9.12) CS = z 0( p ( x ) Š p ( z )) dx. Ineconomics, CS iscalledthe customersurplus forthegivencommodity. Similarly,theintegral zp ( x ) dx istheamountofmissedrevenue,thatis,themoneythecompanycould havereceivedfrombuyerswhofoundtheproducttooexpensive.Note thatthisistheareaofthedomainunderthegraphof p ,butonthe rightof R Example 9.8 Ticketsforacertain”ightarenormallypricedat $ 300,andinanaveragemonth,500ticketsaresold.Researchshows that,forevery $ 10thatthepriceisreduced,thenumberofticketssold

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1789.FURTHERAPPLICATIONSOFINTEGRATION goesupby20.Findthedemandcurveandcomputetheconsumer surplusfortheseticketsifthepriceissetat $ 240. Solution: Iftheairlinewantstosell x tickets,thenthepricethatthe airlineneedstochargeis p ( x )=300 Š 10 x Š 500 20 =300 Š x Š 500 2 Indeed,inordertosell x Š 500extratickets,theairlineneedstodecrease itspriceby$10foreach20-packofextratickets. Ifthepriceissetat p ( z )=$240,thenthelastdisplayedequation showsthat z =60.Nowwecanapplyformula(9.12)tocomputethat thecustomersurplusis CS = 60 0( p ( x ) Š 240) dx = 60 0 60 Š x Š 500 2 dx = 60 0 310 Š x 2 dx =17 700 Socustomerswouldsaveatotalof$17,700inanaveragemonthifthe priceoftheticketsweresetat$240. 63.2.SurvivalandRenewal.Example 9.9 Letusassumethattherearecurrently30,000people intheUnitedStateswhohaveacertainillness.Letusalsoassume thatweknowthatthefractionofthatpopulationwhowillstillhavethe illness t monthsfromnowisgivenbythefunction f ( t )= eŠ 0 05 t.We alsoknowthateverymonth1000newpatientswillgettheillness.How manypeopleintheUnitedStateswillhavetheillnessin20months? Solution: Clearly, f (20)= eŠ 1=0 368ofthepeoplewhocurrently havetheillnesswillstillhaveit20monthsfromnow.Nowwehave tocomputethenumberofpeoplewhowillgettheillness between now ( t =0)and20monthsfromnow( t =20)andwillstillbeill20months fromnow. Subdividetheinterval[0 20]into n equalsubintervalsusingthe points 0= t0
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63.APPLICATIONSTOECONOMICSANDTHELIFESCIENCES179 fromnow,inotherwords,approximately20 Š ti Š 1monthsfromgetting theillness,thefractionofthemwhowillstillhavetheillnesswillbe f (20 Š ti Š 1).Sotheirnumberwillbeabout1000 t f (20 Š ti Š 1). Summingoverallallowedvaluesof i ,wegetthatthetotalnumberof peopleintheUnitedStateswhowillhavetheillness20monthsfrom nowis 30 000 f (20)+ni =11000 t f (20 Š ti Š 1) WerecognizethattheabovesumisaRiemannsum,so,as n goesto in“nity,theaboveexpressionconvergesto D =30 000 f (20)+ 20 01000 f (20 Š t ) dt =30 000 eŠ 1+1000 20 0e0 05 t Š 1dt 11 036 38+12 642 41 23 679 So20monthsfromnow,23,679peopleintheUnitedStateswill stillhavetheillness. Notethattheresultofthepreviousexampleshowsthatthenumber ofpeopleintheUnitedStateswhohavetheillnesswill decrease during thenext20months.Tryto“ndanintuitiveexplanationforthatfact thatdoesnotinvolveintegration.63.3.Exercises.(1)Howmuchmoneywouldcustomerssaveintotalifthe”ights discussedinExample9.8weresoldfor$210each? (2)Considerthe”ightsdiscussedinExample9.8andassumethat theticketsaresoldattheiroriginalpriceof$300.Compute themissedrevenuefortheairline. (3)Considerthe”ightsdiscussedinExample9.8andassumethat theticketsaresoldatareducedpriceof$270.Computethe missedrevenuefortheairline. (4)Ademandcurveisgivenby p ( x )=600 / ( x +10).Findthe consumersurpluswhenthesellingpriceis$20. (5)Ademandcurveisgivenby p ( x )=1000 / ( x +40).Findthe missedrevenuewhenthesellingpriceis$25. (6)Wedeposit P dollarsintoabankaccountatanannualinterest rateof r %.Letusassumethattheinterestiscompounded n timesayear.Findaformulaforouraccountbalanceafter t

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1809.FURTHERAPPLICATIONSOFINTEGRATION years.Thenusethatresulttocomputeouraccountbalance after t yearsiftheinterestiscompounded continuously (7)Wedeposit$100,000intoabankaccount,whereitwillearn anannualinterestof5%.Theinterestiscompoundedcontinuously.Eachyear,wedeposit$2000intothissameaccountin a continuous manner.Whatwillouraccountbalancebein15 years? (8)Considerthebankaccountoftheprecedingexercise,withthe soledierencethat,after10years,weincreaseourannual depositto$3000.Whatwillouraccountbalancebein15 years? (9)Wedeposit$100,000intoabankaccount,whereitwillearn anannualinterestof5%.Theinterestiscompoundedcontinuously.Eachyear,wewithdrawatotalof$4000inacontinuous manner.Whatwillouraccountbalancebein20years? (10)Acountrycurrentlyhasapopulationof80millionpeopleand anaturalgrowthrateof1.5%.Thenaturalgrowth g ofthe populationinagivenyeariscomputedasthedierencebetweenthenumberofbirthsandthenumberofdeathsinthat year,whilethenaturalgrowthrateforthatyearis g divided bythesizeofthepopulationatthebeginningofthatyear. Letusassumethateachyear1.1millionpeopleemigrate fromthiscountry.Ifthecurrenttrendscontinue,howlarge willthepopulationofthiscountrybein20years? (11)Acountrycurrentlyhasapopulationof100millionpeople andanaturalgrowthrateof1%.Seetheprecedingexercise forthede“nitionofnaturalgrowthrate. Letusassumethateachyear0.5millionpeopleemigrate fromthiscountry.Ifthecurrenttrendscontinue,howlarge willthepopulationofthiscountrybein25years? (12)Acountrycurrentlyhasapopulationof80millionpeopleand anaturalgrowthrateof Š 0 5%.Letusassumethateachyear 0.35millionpeopleimmigratetothiscountry.Ifthecurrent trendscontinue,howlargewillthepopulationofthiscountry bein20years?

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63.APPLICATIONSTOECONOMICSANDTHELIFESCIENCES181 (13)Ticketstoacertainsectionofthearenaforabasketballgame usuallycost$50.Thisresultsinthesalesof1000tickets.For everydollarthatthepriceisdropped,thenumberoftickets soldgoesup1%.Findthedemandfunctionforthesetickets. (14)Computetheconsumersurplusfortheticketsdiscussedinthe previousexerciseiftheticketsaresoldat$40. (15)Let S ( x )bethesupplyfunctionforacertaincommodity.That is, S ( x )isthepricethatoneunitofthecommodityhastocost inordertoattractenoughsellerstoprovide x unitsforsale. Notethat S ( x )isanincreasingfunction,sinceahigher priceisneededtoattractmoresellers. Letusassumethattheunitsaresoldata“xedprice T = S ( t ).Thatmeansthatthesellerswhowouldbewillingto sellatalowerpricearemakingapro“t.Thetotalamountof thepro“tmadebyallsellersissometimescalledthe producer surplus .Provethattheproducersurplusforthiscommodity canbecomputedbytheformula t 0( T Š S ( x )) dx. (16)Letusassumethatthesupplyfunctionforacertaincommodityisgivenby S ( x )= x2+2 x .Computetheproducersurplus (whichwede“nedintheprecedingexercise)ifthesellingprice issetat$24. (17)Howdoestheproducersurpluschangeinthesituationdescribedinthepreviousexerciseifthesellingpriceisraisedto $35? (18)Letusassumethatthesupplyfunctionforticketsto”ights betweentwogivencitiesisgivenby S ( x )= x2+10 x +5. Computetheproducersurplusifthesellingpriceissetat $205. (19)Thegatesofafootballstadiumopenat3p.m.After t hours, fansenterthestadiumatarateof f ( t )= Š t3+20 t2+30 t + 64peopleperhour,where t =0.Howmanyfansenterthe stadiumbetween5p.m.and6p.m.? (20)Therearecurrently50,000peopleintheUnitedStateswho takeacertainmedication.Thefractionofthatpopulation whowillstilltakethatmedication t monthsfromnowisgiven bythefunction f ( t )= eŠ 0 01 t.Everymonth,themedication willbetakenby500newpatients.Howmanypeopleinthe Unitedstateswilltakethemedicationayearfromnow?

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1829.FURTHERAPPLICATIONSOFINTEGRATION 64.Probability ThewordprobabilityŽisoftenusedininformalconversations,even ifitissometimesnotclearwhatthespeakermeansbythatword.It turnsoutthattherearetwodistinctconceptsofprobability.Thesetwo conceptscomplementeachotherinthattheyareapplicableindierent circumstances,anduseverydierentmethods.64.1.DiscreteProbability.Letussaythatwearetossingafaircoin fourtimes.Whatistheprobabilitythatwewillgetatleastthree heads? Thisisasituationinwhichthe event thatwestudy,thatis,the sequenceoffourcointosses,hasonlya “nitenumber ofoutcomes. Indeed,thereare24possibleoutcomes,sinceeachcointosshastwo outcomes(headsortails),andthecoinistossedfourtimes. Amongthese16possibleoutcomes,“veare favorable outcomes, namely, HHHH HHHT HHTH HTHH ,and THHH .Furthermore,eachsingleoutcome(favorableornot)isequallylikelytooccur, sincethecoinisfair,andtheresultofeachcointossisequallylikely tobeheadsortails. Inthissituation,thatis,whenthenumberofallpossibleoutcomes is“nite,andeachoutcomeisequallylikelytooccur,de“ne (9.13)Probabilityofevent= Numberoffavorableoutcomes Numberofalloutcomes which,inourexample,showsthattheprobabilityofgettingatleast threeheadsis5 / 16. Probabilitiesde“nedbyformula(9.13)arecalled discreteprobabilities .Theformulaisapplicableonlywhenthenumberofpossible outcomesis“nite.Ifwewanttoapplythisformulaincomplicated situations,weneedadvancedtechniquestocountthenumberofall outcomesandthenumberoffavorableoutcomes.Thefascinatingdisciplinestudyingthosetechniquesiscalled enumerativecombinatorics andwillnotbediscussedinthisbook.64.2.ContinuousProbability.LetussaywewanttoknowtheprobabilitythatduringthenextcalendaryearthecityofGainesville,Florida, willhavemorethan40inchesofprecipitation,orwewanttoknowthe probabilitythatGainesvillewillhavelessthan50inchesofprecipitation,orthatGainesvillewillhaveatleast42butatmost48inchesof precipitation. Inthiscase,formula(9.13)isnotapplicable,sinceboththenumber offavorableoutcomesandalloutcomesisin“nite.Indeed,theamount

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64.PROBABILITY183 ofprecipitationinGainesvillenextyearcanbeanynonnegativereal number.Furthermore,notalloutcomesareequallylikely.Receiving verylittleorverymuchprecipitationisfarlesslikelythanreceiving closetotheusualamount.Weneedatotallydierentapproach.Our approach,whiledierentfromtheoneintheprevioussection,shares someofthemostimportantfeaturesofthatapproach.Forinstance, theprobabilityofaneventthatiscertaintohappenwillbe1,while theprobabilityofaneventthatneverhappenswillbe0. Similarsituationsoccurwhenwewanttoknowtheprobabilitythat acertaindevicewillworkformorethan t years,orthatarandomly selectedpersonweighsmorethan p poundsbutlessthan q pounds,or thatthebloodpressureofarandomlyselectedpersonisbelowagiven value.Thequantitiesmentionedherearecalled randomvariables Wewouldliketode“netheprobability F ( a )thattheamount X ofprecipitationinGainesvillenextyearwillbeatmost a inches.This probabilitywillsometimesbedenotedby P ( X a ). Whilewedonotyetknowhowtocompute F ( a ),weknowthatthe function F hastosatisfythefollowingrequirements: (1)Wewillhave F ( a1) F ( a2)if a1
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1849.FURTHERAPPLICATIONSOFINTEGRATION Figure9.12. Theprobability P ( a X b )asanarea. holds.Inotherwords, P ( a X b )isequaltotheareaofthedomain betweenthegraphofthedensityfunction f ,thehorizontalaxis,and theverticallines x = a and x = b .SeeFigure9.12foranillustration. Indeed,wehave P ( a X b )= P ( X b ) Š P ( X a ) = F ( b ) Š F ( a ) = b Šf ( x ) dx Š a Šf ( x ) dx = b af ( x ) dx. Example 9.10 Letthecontinuousrandomvariable X havedensity function f ( x )= 0 if x, 6 x (1 Š x ) if 0 x 1 0 if x> 1 Verifythat f isindeedadensityfunctionandcomputetheprobability P (0 3 X 0 6) Solution: Inordertoseethat f isindeedadensityfunction,we mustverifythatitsde“niteintegral,takenovertheentirelineofreal numbers,isequalto1.Thisisnotdicult,since f ( x )=0outsidethe interval[0 1].Thisleadsto Šf ( x ) dx =6 1 0 x Š x2 dx =6 x2 2 Š x3 3 1 0=6 1 6 =1 So f isindeedavaliddensityfunction.

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64.PROBABILITY185 Figure9.13. Thegraphofthefunction f inExample9.10. Wecanuseformula(9.14)tocomputetherequestedprobability. Weget P (0 3 X 0 6)= 0 6 0 3f ( x ) dx =6 0 6 0 3( x Š x2) dx =6 x2 2 Š x3 3 0 6 0 3=6(0 108 Š 0 036) =0 432 64.2.1.ExponentialDistribution.Considerthefollowingdensityfunction.Let beapositiverealnumberandlet f ( x )= 0if x< 0 eŠ xif0 x. (9.15) Weseethat f isadecreasingfunctionontheinterval[0 ).Figure 9.14showshowthespeedatwhich f decreasesdependsontheparameter Itturnsoutthatthisdensityfunctionisaveryfrequentlyoccurring one.Therefore,ithasaname.Itiscalledthe exponentialdensityfunction withparameter .Usingtherightconstant andundertheright circumstances,itcanbeusedinmanyscenarios,typicallyconnected to waitingtimes .Forinstance,itcouldbeusedtomeasuretheprobabilitythat,givenastartingmoment,agivencellphonewillringinless than t minutes,orthat,atagivenlocation,itwillstartraininginless

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1869.FURTHERAPPLICATIONSOFINTEGRATION Figure9.14. =1(red), =1 5(blue), =2(orange). than h hours,orthat,givenarandomstore,acustomerwillenterin s seconds.Theexponentialdensityfunctionwillgiveagoodapproximationtocomputetheseprobabilitiesifthementionedprocessestake placeataroughlyconstantrate.Thatis,weshouldchooseapartof thedaywhenthatgivencellphonereceivescallsatroughlyconstant frequency,aseasonwhenitrainsatthatlocationatroughlyconstant timeperiods,oratimeofdaywhencustomersenterthatstoreata roughlyconstantrate. Example 9.11 Theprobabilitythatacertainkindofnewrefrigeratorwillneedamajorrepairin x yearsisgivenbytheexponential densityfunctionwithparameter =1 / 9 .Whatistheprobabilitythat anewrefrigeratorwill not needamajorrepairfor10years? Solution: First,wecomputetheprobabilitythattherefrigerator will needamajorrepairin10years.Let X denotethenumberofyears passingbeforethe“rstmajorrepairisneeded.Thenthatprobabilityis P ( X< 10)= 10 Šf ( x ) dx = 10 01 9 eŠ x/ 9dx = Š eŠ x/ 9 10 0=1 Š eŠ 10 / 9=0 671 .

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64.PROBABILITY187 Therefore,theprobabilitythattherefrigeratorwill not needamajor repairin10yearsis P ( X 10)=1 Š 0 671=0 329. 64.2.2.Mean.Ifwewanttocomputetheaverageweightofapersonselectedfromagivenpopulationof n people,wecansimplytake theweights a1,a2,...,anofthosepeopleandcomputetheirarithmetic mean,oraverage,thatis,therealnumber A = a1+ a2+ + an n Thiscouldtakeaverylongtimeif n isaverylargenumber.Ifthedata aregiveninamoreorganizedform,wemaybeabletosavesometime. Inparticular,ifweknowthatthereare b1peopleinthepopulation whoseweightis x1,thereare b2peoplewhoseweightis x2,andsoon, thenwecancomputetheaverageweightofthepopulationas (9.16) A = b1x1+ b2x2+ + bkxk b1+ b2+ + bk, sincethisfractionisthetotalweightofthepopulationdividedbythe numberofpeopleinthepopulation. Nownotethat pi= bi/ ( b1+ b2+ + bk)isjusttheprobabilitythat arandomlyselectedpersonofthispopulationhasweight xi.Therefore, (9.16)canbewrittenas (9.17) p1x1+ p2x2+ + pkxk. Theoretically,theweightofapersoncantakein“nitelymanyvaluessincethemeasuringscalecanbealwaysbemoreprecise.Itisnot diculttoprovethat,as k goestoin“nity,thesumin(9.17)willturn intoaRiemannsum,andtheweights xiwillbemeasuredbyacontinuousrandomvariable X ,andtheprobabilities piwillbeexpressibleby thede“niteintegralsofadensityfunction.Thisleadstothefollowing de“nition. Definition 9.1 Let f bethedensityfunctionofthecontinuous randomvariable X .Thenthevalueof ( X )= Štf ( t ) dt iscalledthe averagevalue or mean or expectedvalue of X Example 9.12 Let X bethecontinuousrandomvariablewhose densityfunctionistheexponentialdensityfunctionwithparameter thatwede“nedin(9.15).Then ( X )=1 / .

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1889.FURTHERAPPLICATIONSOFINTEGRATION Solution: UsingDe“nition9.1,wehave ( X )= Štf ( t ) dt = 0teŠ tdt = Š teŠ tŠ 1 eŠ t 0= 1 Inotherwords,theparameterandthemeanofanexponentialdistributionarereciprocalsofeachother.Inviewofthis,wecanreformulatetheresultofExample9.11asfollows.Iftheaveragetimebeforea newrefrigeratorneedsamajorrepairis9years,thentheprobability thatarefrigeratorwillnotneedamajorrepairfor10yearsis0.329.64.2.3.NormalDistribution.Let bearealnumberandlet bepositiverealnumber.Considerthedensityfunction f ( x )= 1 2 exp Š ( x Š )2 2 2 Thedistributionde“nedbythisdensityfunctioniscalledthe normaldistribution withparameters and .Thisdistributionisdenoted by N ( , ).Inparticular,if =0and =1,thentheobtained distribution N (0 1)iscalledthe standardnormaldistribution Plottingthegraphof f forvariousvaluesof and ,weseethat thegraphhasabellcurve;itshighestpointisreachedwhen x = anditincreasesontheleftofthatanddecreasesontherightofthat. Thesmallerthevalueof ,thesteeperistheriseandfallofthegraph of f .SeeFigure9.15foranillustration. Itcanbeprovedthat ispreciselythemeanof N ( , ).The constant iscalledthe standarddeviation of N ( , ).Itmeasureshow spreadoutthevaluesofourvariable X are.(Theprecisede“nitionis that isthesquarerootofthemeanof( X Š )2. ) Manyscenariosaremodeledbyanormaldistribution,suchastest scores,athleticresults,orannualsnowfallatagivenlocation. Example 9.13 Inanaverageyear,Northtowngets10feetofsnow, withastandarddeviationof2feet.Whatistheprobabilitythat,ina randomyear,Northtowngetsbetween9and12feetofsnowifsnowfall ismodeledbyanormaldistribution?

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64.PROBABILITY189 Figure9.15. =1(red), =1 5(orange), =2 (green),and =2 5(blue). Solution: Let X denotethesnowfallinarandomyearinNorthtown. Weneedto“ndtheprobability P (9 X 12).Assnowfallismodeled byanormaldistribution,thegivenparametersimplythatthatmust bethedistribution N (10 2).Therefore,byformula(9.14),wehave P (9 X 12)= 12 91 2 2 exp Š ( x Š 10)2 8 dx =0 5328 wherethede“niteintegralhastobecomputedbysomeapproximation method(orasoftwarepackage)since eŠ x2hasnoantiderivativeamong elementaryfunctions. 64.3.Exercises.(1)Forwhichvalueof c will f ( x )= cx4beadensityfunctionon [0 1]? (2)Forwhichvalueof c will f ( x )=c 1+4 x2beadensityfunction on[ Š ]? (3)Considerthevalueof c thatmakesthefunction f oftheprecedingexerciseadensityfunction.Usingthatvalue c ,“nd P ( Š 2
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1909.FURTHERAPPLICATIONSOFINTEGRATION (6)Letussaythatthelifetimeofabicycletire(measuredin months)hasanexponentialdistributionwith =7.Whatis theprobabilitythatatirewilllastbetween5and8months? (7)Whatistheprobabilitythatthetireoftheprecedingexercise willlastmorethan14months? (8)Findthenumber a ofmonthssuchthatthereisexactly1 / 2 probabilitythatthetireofthepreviousexercisewilllastat least a months. (9)Letusassumethatthelifetimeofagivenproduct,measured inmonths,hasanexponentialdistributionwithparameter Whatistheprobabilitythattheproductwilllastformore than twice itsexpectedlifetime? (10)Consideringtheproductoftheprecedingexercise,whatisthe probabilitythatitwilllastatleast k timesitsexpectedlifetime,where k isanypositiverealnumber? (11)Consideringtheproductoftheprecedingexercise,whatis morelikely,thatitslifetimewillbeatmost Š 1months orthatitwillbeatleast +1months? (12)Theaveragescoreonanexamis100points.Inordertopass, astudentcannotbemorethan2standarddeviationsbelow theaverage.Ifthescoreshaveanormaldistributionwitha standarddeviationof6,howlargeafractionofthestudents willpasstheexam? (13)UsingtheconditionsofExample9.13,whatistheprobability thatNorthtownwillgetlessthan5feetofsnowinagiven year? (14)Useasoftwarepackagetoprovethatthestandardnormal distributionfunctionisindeedadistributionfunction. (15)TheheightoftheadultmalepopulationoftheNetherlands hasanormaldistributionwithameanof73inchesanda standarddeviationof3inches.Whatpercentageoftheadult malepopulationismorethan75inchestall? (16)ConsidertheadultmalepopulationoftheNetherlandsgiven intheprecedingexercise.Whatpercentageofthatpopulation isbetween70and75inchestall? (17)Provethatthemeanofthestandardnormaldistributionis indeed0. (18)Let X betherandomvariablethatcountsthegoalsscored byanoensivesoccerplayerofacertaineliteleagueduring anentireseason.Anoensiveplayerisconsideredexceptional ifthenumberofgoalshescoresexceedstheaverageofall

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64.PROBABILITY191 oensiveplayersbyatleast3standarddeviations.Letussay that X hasdistribution N (33 3).Whatpercentageofoensive playersisconsideredexceptional? (19)Forthe standard normaldistribution,“ndtheprobabilitythat thevalueoftherandomvariableiswithin3standarddeviationsofthemean. (20)For any normaldistribution,“ndtheprobabilitythatthevalue oftherandomvariableiswithin3standarddeviationsofthe mean.

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CHAPTER10 PlanarCurves 65.ParametricCurves Everypointinaplanecanbede“nedasanorderedpairofrealnumbers( x,y )calledthe rectangularorCartesiancoordinates .Agraphof afunction f isthesetpointsinaplanewhosecoordinatessatisfy thecondition y = f ( x ).Thegraphgivesasimpleexampleofaplanar curve.Moregenerally,aplanarcurvecanbede“nedasthesetofpoints whosecoordinatessatisfythecondition F ( x,y )=0calledthe Cartesianequation ofacurve.Inmanyinstances,anequation F ( x,y )=0 hasmultiplesolutionsforeverygiven x .Forexample,considerthe circleofunitradius: (10.1) x2+ y2=1= y = 1 Š x2,x [ Š 1 1] Thetwosolutionsrepresenttwosemicircles.Thegraph y = 1 Š x2isthesemicircleabovethe x axis,whilethegraph y = Š 1 Š x2is thesemicirclebelowthe x axis.Theunionofthetwographsisthefull circle.Thisexampleshowsade“ciencyindescribingplanarcurvesby thegraphofafunctionbecausethecurvescannotalwaysberepresented asthegraphofasinglefunction. Ontheotherhand,(10.1)admitsadierentsolution: (10.2) x2+ y2=1= x =cos t,y =sin t,t [0 2 ] whichimmediatelyfollowsfromthetrigonometricidentitycos2t + sin2t =1forallvaluesof t Thisrepresentationmeansthatapointofthecoordinateplaneis assignedtoeveryvalueof t [0 2 ]bytherule( x,y )=(cos t, sin t ). Thecoordinatesofpointsofthecirclearefunctionsofathirdvariable calleda parameter .As t changes,thepoint(cos t, sin t )tracesoutthe circleofunitradiuscenteredattheoriginintheplane.Theparameter t hasasimplegeometricalinterpretation.Itistheanglecounted counterclockwisefromthepositive x axistoarayfromtheoriginon whichthepoint(cos t, sin t )lies.Thisobservationadmitsanatural generalization. Definition 10.1(Parametriccurves) Let x ( t ) and y ( t ) becontinuousfunctionson [ a,b ] .A parametriccurve inthecoordinateplaneis193

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19410.PLANARCURVES Figure10.1. Circle: x ( t )=cos t y ( t )=sin t Figure10.2. Parametriccurve.As t increasesfrom a to b ,thepoint( x ( t ) ,y ( t ))tracesoutacurveinthe xy plane. thesetofpointssatisfyingtheconditions,calledthe parametricequations x = x ( t ) ,y = y ( t ) ,t [ a,b ] Thepoints ( x ( a ) ,y ( a )) and ( x ( b ) ,y ( b )) arecalledthe initial and terminal pointsofthecurve,respectively. Thegraphofafunction f isaparticularexampleofaparametric curve: x = t y = f ( t ).

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65.PARAMETRICCURVES195 Figure10.3. Thespiral x = t cos t y = t sin t t [0 2 ].Thedistancefromtheorigin R = x2+ y2= t increaseslinearlyastheangle t ,countedcounterclockwisefromthepositive x axis,increasesfrom0to2 Parametriccurvesarecommonineverydaylife.Thepositionofa particleinaplaneisde“nedbyitsrectangularcoordinates( x,y )inthe plane.Whentheparticlemoves,itscoordinatesbecomefunctionsof time t sothattheparametriccurve x = x ( t ), y = y ( t )isthetrajectory oftheparticle. Example 10.1 Sketchthecurvewiththeparametricequations x = t cos t y = t sin t t [0 2 ] Solution: Abasicapproachtovisualizetheshapeofaparametric curveistoplotpoints( x ( tk) ,y ( tk)), k =1 2 ,...,n ,correspondingto successivevaluesof t : t1
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19610.PLANARCURVES rotatesabouttheorigin,thedistance R betweenitandtheorigin increaseslinearlywiththerotationangle.Suchamotionoccursalong anunwindingspiral.Intheinterval t [0 2 ],thespiralmakesone fullturnfromtheinitialpoint(0 0)totheterminalpoint(2 0). 65.1.ParametricCurvesandCurvesasPointSets.Ifacurveisde“ned asapointsetinthecoordinateplane,forexample,bytheCartesian equation F ( x,y )=0,thentherearemanyparametricequationsthat describeit.Forexample,thecircle(10.1)mayalsobedescribedbythe followingparametricequations: (10.3) x2+ y2=1= x =cos(3 ) ,y = Š sin(3 ) [0 2 ] Whatisthedierencebetween(10.2)and(10.3)?First,notethat, astheparameter t in(10.2)increases,thepoint(cos t, sin t )tracesout thecircle counterclockwise (theinitialpoint(1 0)movesupwardas y = sin t> 0for0
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65.PARAMETRICCURVES197 particularcaralongtheroad.Naturallydierentcarsmovesdierently alongtheverysameroad. Example 10.2 Supposeacurve C isdescribedbytheparametric equations x = x ( t ) y = y ( t ) if t [ a,b ] .Findtheparametricequations of C suchthatthecurveistracedoutbackward,thatis,fromthepoint ( x ( b ) ,y ( b )) to ( x ( a ) ,y ( a )) (theinitialandterminalpointsareswapped). Solution: Onehasto“ndanewparameter t = g ( ),suchthat g ( b )= a and g ( a )= b .When increasesfrom a to b ,theparameter t decreasesfrom b to a ,andthesought-afterparametricequationsare obtainedbythecomposition X ( )= x ( g ( ))and Y ( )=y ( g ( )).The simplestpossibilityistolookforalinearrelationbetween t and g ( )= c + d .Thecoecients c and d are“xedbytheconditions g ( a )= b or b = c + da and g ( b )= a or a = c + db .Therefore,by subtractingtheseequations, b Š a =( c + da ) Š ( c + db )= Š ( b Š a ) d or d = Š 1.Byaddingtheseequationswith d = Š 1, b + a =( c Š a )+( c Š b ) or c = a + b .Hence, t =( a +b ) Š ,sothattheparametricequations of C withreversedorientationare x = x ( a + b Š ) ,y = y ( a + b Š ) [ a,b ] Forexample,if C isthecircleorientedcounterclockwiseasin(10.2), thenthesamecircleorientedclockwiseisdescribedby x =cos(2 Š )=cos y =sin(2 Š )= Š sin [0 2 ]. 65.2.TheCycloid.Thecurvetracedbya“xedpointonthecircumferenceofacircleasthecirclerollsalongastraightlineiscalleda cycloid (seeFigure10.4).To“nditsparametricequations,suppose thatthecirclehasaradius R anditrollsalongthe xaxis .Letthe “xedpoint P onthecircumferencebeinitiallyattheoriginsothatthe centerofthecircleispositionedatthepoint(0 ,R )(onthe y axis).Let CP denotethestraightlinesegmentbetweenthecenterofthecircle C and P .Initially, CP isperpendiculartothe x axis.Asthecirclerolls, thesegment CP rotatesaboutthecenterofthecircle.Therefore,itis naturaltochoosetheangleofrotation asaparameter.Thecoordinatesof P arefunctionsof tobefound.Ifthecirclerollsadistance D sothatitscenterisat( D,R ),thenthearclength R ofthepart ofthecirclebetween P andthetouchpoint T hastobeequalto D thatis, D = R .Let Q beapointonthesegment CT suchthat PQ and CT areperpendicular.Considertheright-angledtriangle CPQ Itshypotenuse CP haslength | CP | = R ,andthelengthsofitscatheti are | CQ | = | CP | cos = R cos and | PQ | = | CP | sin = R sin .Let

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19810.PLANARCURVES Figure10.4. De“nitionofacycloid.Adiskofradius R isrollingalongthe x axis.Acurvetracedoutbya “xedpointonitsedgeiscalledacycloid. Figure10.5. Overallshapeofacycloid. ( x,y )becoordinatesofthepoint P .Theparametricequationsofthe cycloidare x = D Š| PQ | = R Š R sin = R ( Š sin ) y = R Š| CQ | = R Š R cos = R (1 Š cos ) Itlookslikeanupwardarcovertheinterval0 x 2 R ,withmaximalheight ymax=2 R ( = / 2),andthearcrepeatsitselfoverthe nextintervalofthelengthofcircumference2 R andsoon. Remark. In1696,theSwissmathematicianJohannBernoulli posedthe brachistochroneproblem :Findthecurvealongwhichaparticlewillslide(withoutfriction)intheshortesttime(underthein”uence ofgravity)fromapoint A toalowerpoint B notdirectlybeneath A Theparticlewilltaketheleasttimeslidingfrom A to B ifthecurveis apartofaninvertedarchofacycloid.

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65.PARAMETRICCURVES199 65.3.FamiliesofCurves.Dierentvaluesof R de“nedierentcycloids. Ingeneral,iftheparametricequationscontainanumericalparameter, thentheparametricequationsde“nea family ofcurves;eachfamily membercorrespondstoaparticularvalueofthenumericalparameter. Example 10.3 Investigatethefamilyofcurveswiththeparametric equations x = a cos t y = b sin(2 t ) t [0 2 ] ,where a and b are positivenumbers. Solution: Consider“rstthesimplestcase a = b =1.Thefunction x ( t )=cos t hasaperiodof2 ,and y ( t )=sin(2 t )hasaperiodof .Theinitialpointis( x (0) ,y (0))=(1 0).As t increases,thepoint movesupwardsothat x ( t )decreases(becomeslessthan1),while y ( t ) increases,reachingitsmaximumvalue1at t = / 4.Afterthat, y ( t ) beginstodecrease,while x ( t )continuestodecrease.At t = / 2,the pointarrivesattheoriginandpassesthroughitintothethirdquadrant sothat x ( t)and y ( t )continuetodecrease.When t =3 / 4, y ( t )attains theminimumvalue Š 1andbeginstoincreasefor t> 3 / 4,while x ( t )=cos t isstilldecreasingtowarditsminimalvalue Š 1,whichis reachedat t = ,andthecurvecrossesthe x axismovingintothe secondquadrant.Inthesecondquadrant 1)orcompresses( a< 1)anygeometricalsethorizontallyinthe coordinateplane.Thetransformation y by doesthesamebutinthe verticaldirection.Sothefamilyofcurvesconsistsofcurvesofthe shapestretchedto“tintotherectangleboundedbythelines x = a and y = b 65.4.Exercises.In(1)…(9),sketchthecurvebyplottingitspoints.Includethearrow showingtheorientationofthecurve.Eliminatetheparameterto“nd aCartesianequationofthecurve.

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20010.PLANARCURVES (1) x =1+2 t,y =3 Š t (2) x = t,y =2 Š t (3) x = t2,y = t3(4) x =1+2 et,y =3 Š et(5) x =cosh t,y =sinh t (6) x =2sin t,y =3cos t (7) x =2 Š 3cos t,y = Š 1+sin t (8) x =cos t,y =sin(4 t )(9) x = t2sin t,y = t2cos t In(10)…(12),sketchthefamilyofparametriccurves (10) x = a2Š b2 a cos3t,y = a2Š b2 b sin3t (11) x = a cosh2t,y = b sinh2t (12) x = a (sinh t Š t ) ,y = a (cosh t Š 1) In(13)…(16),“ndparametricequationsofthecurvede“nedbyaCartesianequationandsketchit.Investigatethedependenceoftheshape ofthecurveontheparameter a (13) x2 / 3+ y2 / 3= a2 / 3(14) x2+ y2= a tanŠ 1( y/x ) (15) x3+ y3=3 axy (16)( x2+ y2)2= a2( x2Š y2) Hint :Put y = tx in(15)and y = x tan t in(16). (17)Thecurves x = a sin( nt ), y = b cos t ,where n isapositiveinteger,arecalled Lissajous“gures .Investigatehowthesecurves dependon a b ,and n (18)Consideradiskofradius R .Let P beapointonthediskat adistance b fromitscenter.Findtheparametricequationsof thecurvetracedoutbythepoint P asthediskrollsalonga straightline.Thecurveiscalleda trochoid .Aretheequations wellde“nedif b>R .Sketchthecurvefor bR (19)The swallowtailcatastrophecurves arede“nedbytheparametricequations x =2 ct Š 4 t3, y = Š ct2+3 t4.Sketchthese curvesforafewvaluesof c .Whatfeaturesdothecurveshave incommon?Howdotheychangewhen c increases? (20)Ahypocycloidisthecurvetracedoutbyapointonthecircle thatisrollingalonga“xedcirclesothatitremainswithinthe latter.Findparametricequationsofthehypocycloidifthe radiusofthemovingcircleis a>b ,where b istheradiusof the“xedcircle.Sketchthecurveiftheratio b/a isaninteger n =2and n =4. (21)Anepicycloidisthecurvetracedoutbyapointonthecircle thatisrollingalonga“xedcirclesothatitremainsoutof

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66.CALCULUSWITHPARAMETRICCURVES201 thelatter.Findparametricequationsoftheepicycloidifthe radiusofthemovingcircleis a and b istheradiusofthe“xed circle.Sketchthecurveiftheratio a = b 66.CalculuswithParametricCurves66.1.TangentLinetoaParametricCurve.Consideraparametriccurve x = x ( t ), y = y ( t ),wherethefunctions x ( t )and y ( t )arecontinuouslydierentiableandthederivatives x( t )and y( t )donotvanish simultaneouslyforany t .Suchparametriccurvesarecalled smooth Theorem 10.1(TangentLinetoaSmoothCurve) Asmoothparametriccurve x = x ( t ) y = y ( t ) hasatangentlineatanypoint ( x0,y0) anditsequationis (10.4) x( t0)( y Š y0) Š y( t0)( x Š x0)=0 where ( x0,y0)=( x ( t0) ,y ( t0)) Proof. Takeapointofthecurve( x0,y0)=( x ( t0) ,y ( t0))correspondingtoaparticularvalue t = t0.Supposethat x( t0) =0.Then, bythecontinuityof x( t ),thereisaneighborhood I=( t0Š ,t0+ ) forsome > 0suchthat x( t ) =0forall t I;thatis,thederivativeiseitherpositiveornegativein I.Bytheinversefunction theorem(studiedinCalculusI),thereisaninversefunction t = f ( x ) thatisdierentiableinsomeopenintervalthatcontains x0.Substituting t = f ( x )intothesecondparametricequation y = y ( t ),one obtainsthatnearthepoint( x0,y0)thecurvecanberepresentedasa partofthegraph y = F ( x )suchthat y0= F ( x0).Thefunction F is dierentiableasthecompositionoftwodierentiablefunctions.The derivative F( x0)determinestheslopeofthetangenttothegraph,and theequationofthetangentlinereads (10.5) y = y0+ F( x0)( x Š x0) Byconstruction, y = F ( x )= y ( t )= F ( x ( t ))forall t I. Dierentiationofthisequationwithrespectto t bymeansofthechain ruleyields y( t )= F( x ( t )) x( t )= F( x ( t ))= y( t ) x( t ) = F( x0)= y( t0) x( t0) Substitutingthisequationinto(10.5),thelattercanbewrittenin theform(10.4).If x( t0)=0,then y( t0) =0bythede“nitionof asmoothcurvesothatthereisadierentiableinverse t = g ( y )and

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20210.PLANARCURVES hence x = G ( y )= x ( g ( y )).Similarargumentsleadtotheconclusion thatthetangentlinetothegraph x = G ( y )hastheform(10.4).The detailsarelefttothereaderasanexercise. Theruleforcalculatingtheslopeofthetangentlinecanalsobe obtainedbymeansoftheconceptofthedierential.Recallthatthe dierentialsoftworelatedquantities y = F ( x )areproportional: dy = F( x ) dx .Ontheotherhand, x = x ( t ), y = y ( t )andtherefore dx = x( t ) dt and dy = y( t ) dt .Hence, F( x )= dy dx =dy dt dx dt= y( t ) x( t ) Thesemanipulationswithdierentialsarebasedonatacit assumption that,forasmoothcurve x = x ( t ), y = y ( t ),thereexistsa dierentiable function F suchthat y = F ( x ).Intheproofofthetangentlinetheorem,thishasbeenshowntobetrueasaconsequenceoftheinverse functiontheorem.Theuseofthedierentialsestablishesthefollowing helpfulrulestocalculatethederivatives: d dx =d dt dx dt= 1 x( t ) d dt and d dy =d dt dy dt= 1 y( t ) d dt .66.2.ConcavityofaParametricCurve.Theconcavityofagraph y = F ( x )isdeterminedbythesignofthesecondderivative F( x ).If F( x ) > 0,thegraphisconcaveupward,anditisconcavedownwardif F( x ) < 0.If y ( t )and x ( t )aretwicedierentiable,thentheconcavity ofthecurvecanbedeterminedbythesignofthesecondderivative d2y d2x = d dx dy dx = 1 xd dt dy dx = y x x= yxŠ xy ( x)3. Example 10.4 Acurve C isde“nedbytheparametricequations x = t2, y = t3Š 3 t (i) Showthat C hastwotangentlinesatthepoint (3 0) (ii) Findthepointson C wherethetangentlineishorizontalorvertical. (iii) Determinewherethecurveisconcaveupwardordownward. Solution: (i)Notethat y ( t )= t ( t2Š 3)=0hasthreesolutions t =0 and t = 3.Butthecurvehasonlytwopointsofintersectionwith the x axis,(0 0)and(3 0),because x ( 3)=3;thatis,thecurveis self-intersectingatthepoint(3 0).Thisexplainswhythecurvemay havetwotangentlines.Onehas x( t )=2 t and y( t )=3 t2Š 3sothat

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66.CALCULUSWITHPARAMETRICCURVES203 x( 3)= 2 3and y( 3)=6.Sotheslopesofthetangentlines are( y/x)( 3)= 3,andtheequationsofthelinesread y = 3( x Š 3)and y = Š 3( x Š 3) (ii)Thetangentlinebecomeshorizontalwhen y( t )=3 t2Š 3=0 (seeEq.(10.4)).Thishappenswhen t = 1.Thus,thetangent lineishorizontalatthepoints(1 2).Thetangentlineisverticalif x( t )=2 t =0or t =0.Sothetangentlineisverticalattheorigin (0 0). (iii)Thesecondderivativeis d2y d2x = 1 xd dt dy dx = 1 2 t d dt 3 t2Š 3 2 t = 3 4 t d dt t Š 1 t = 3 4 t 1+ 1 t2 Thisequationshowsthatthecurveisconcaveupwardif t> 0(the secondderivativeispositive)andthecurveisconcavedownwardif t< 0(thesecondderivativeisnegative). 66.3.CuspsofPlanarCurves.Consideracurvede“nedbytheCartesianequation x2Š y3=0.Thisequationcanbesolvedfor y y = x2 / 3, suchthat dy/dx =2 3xŠ 1 / 3.For x> 0,theslopeofthetangentline diverges, y( x ) as x 0+(as x approachesto0fromtheright). For x< 0,italsodiverges, y( x ) Š as x 0Š(as x approaches0 fromtheleft).Thetwobranchesofthecurve( x> 0and x< 0)are joinedat x =0andhavea common tangentline,whichisthevertical line x =0(the y axis)inthiscase,buttheslopesuersajumpdiscontinuity(from Š to ).Sothecurveisnotsmoothat x =0and exhibitsahornlikeshapenear x =0.Suchapointofaplanarcurve iscalleda cusp Aparametriccurve x = x ( t ), y = y ( t )mayhavecuspseventhough bothderivatives x( t )and y( t )arecontinuousforall t .Forexample, considertheparametriccurve x = t3, y = t2.Forallvaluesof t x2Š y3=0.Sothiscurvecoincideswiththatdiscussedaboveandhas acuspattheorigin( t =0).Thederivatives x( t )=3 t2and y( t )=2 t arecontinuouseverywhere,and,inparticular, x(0)= y(0)=0atthe cusppoint.Despitethecontinuityofthederivatives,theslopeofthe curveisnotde“nedsince dy/dx = y/xisanundeterminedform0 0. Acloserinvestigationshowsthattheslope y( t ) /x( t )=2 3tŠ 1suers ajumpdiscontinuity(from Š to+ as t changesfromnegative topositive).Thede“nitionofasmoothparametriccurverequires thatthederivatives x( t )and y( t )arecontinuousand donot vanish simultaneouslyatany t .Thisconditioneliminatespossiblecuspsthat mayoccuratpointswherebothderivativesvanish.

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20410.PLANARCURVES Figure10.6. Plotof y = x2 / 3.Thecurvehasacuspat theorigin. Furthermore,considerthecurve x = t2, y = t3.Theslope dy/dx = y/x=3 2t iscontinuouseverywhereand,inparticular,at t =0,where x(0)= y(0)=0.Nevertheless,thecurvehasacuspattheorigin.To seethis,letusinvestigatetheCartesianequationofthiscurve x3Š y2= 0,whichcanbesolvedfor x x = y2 / 3.Therefore,thederivative dx/dy =2 3yŠ 1 / 3exhibitsajumpdiscontinuityfrom Š to as y changesfromnegativetopositive.Thetwobranchesofthecurve( y> 0 and y< 0)haveacommontangentline(thehorizontalline y =0), butattheirjoiningpointacuspisformed.Notealsothattherate dx/dy = x/y=2 3tŠ 1suersafamiliarin“nitejumpdiscontinuity,thus indicatingacusp.Thisexampleshowsthatbothrates dy/dx = y/xand dx/dy = x/ymustbestudiedtodeterminewhetherthereisa cuspatthepointwhere y= x=0. Example 10.5 Findthetangentlinetotheastroidde“nedbythe parametricequation x = a cos3t y = a sin3t t [0 2 ] atthepoints t = / 4 .Determinethepointswherethetangentlineishorizontal andvertical.Isthecurvesmooth?Specifytheregionsofupwardand downwardconcavity.Usetheresultstosketchthecurve. Solution: Theslopeofthetangentlineatagenericpointis dy dx = 1 xd dt y = Š 1 3 a cos2t sin t 3 a sin2t cos t = Š tan t. Thevalue t = / 4correspondstothepoint x = a/ 23 / 2, y = a/ 23 / 2becausesin( / 4)=cos( / 4)=1 / 2,andtheslopeatthispointis

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66.CALCULUSWITHPARAMETRICCURVES205 Figure10.7. Thecurve x = t2, y = t3.Thederivatives x(0)= y(0)=0vanishat t =0.Thecurvehasacusp atthepoint( x (0) ,y (0))=(0 0). Š 1.Sothetangentlineis y = a 2 2 Š x Š a 2 2 or y = a 2 Š x. Theslope dy/dt = Š tan t vanishesat t =0and t = sothetangent lineishorizontal( y =0)atthepoints( a, 0).However,thederivatives x( t )= Š 3 a cos2t sin t and y( t )=3 a sin2t cos t vanishsimultaneously at t =0and t = .Theinverseslope dx/dy =1 / ( dy/dx )= Š cot t exhibitsanin“nitejumpdiscontinuityat t =0and t = ,sothe curvehascuspsat( a, 0)andhenceitisnotsmoothatthesepoints. Theslope dy/dx isin“niteat t = / 2and t =3 / 2.Therefore,the curvehasaverticaltangentline( x =0)at(0 a ).However,theslope dy/dx = Š tan t hasanin“nitejumpdiscontinuityat t = / 2and t =3 / 2.Sothecurvehascuspsandisnotsmoothat(0 a ).Note alsothatbothderivatives xand yvanishatthesepoints.Thus,the curveconsistsoffoursmoothpieces,andthecurvehascuspsatthe joiningpointsofitssmoothpieces.Thesecondderivative d2y d2x = 1 xd dt dy dx = 1 3 a cos2t sin t (tan t )= 1 3 a sin t cos4t

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20610.PLANARCURVES ispositiveifsin t> 0(or y> 0)andnegativeifsin t< 0(or y< 0). Sothetwobranchesofthecurveabovethe x axisareconcaveupward, whilethetwobranchesbelowitareconcavedownward.Thecurves looklikeasquarewithvertices( a, 0),(0 a )whosesidesarebent inwardtowardtheorigin. 66.4.Exercises.In(1)…(4),“ndanequationofthetangentline(s)tothecurveatthe givenpoint.Sketchthecurveandthetangent(s). (1) x = t2+ t,y =4sin t, (0 0) (2) x =sin t +sin(2 t ) ,y =cos t +cos(2 t ) (1 Š 1) (3) x2+2 y2=3 x, (2 1) (2 Š 1) (4) x2 4 + y2=1 ( 2 1 / 2) ( Š 2 1 / 2) In(5)…(8),investigatetheconcavityofthecurve. (5) x = t3Š 12 t,y = t2Š 1(6) x =sin(2 t ) ,y =cos t (7) x = t2Š ln t,y = t2+ln t (8) x =3sin t3,y =2cos t3(9)Investigatetheslopeofthetrochoid x = R Š b sin y = R Š b cos intermsof .Findtheconditionontheparameters R and b suchthatthetrochoidhasverticaltangentlines. (10)Atwhatpointsonthecurve x =2 t3, y =1+4 t Š t2doesthe tangentlinehaveslope1? (11)Findequationsofthetangentstothecurve x =2 t3+1, y = 3 t2+1thatpassthroughthepoint(3 4). (12)Findpointsonthecurve x2+4 y2=2atwhichithasatangent lineparalleltotheline y Š x =1. In(13)…(18),investigatewhetherthecurvehascuspsornot.Ifitdoes, “ndtheirposition.Sketchthecurve. (13) x = t3,y = t3(14) x = t5,y = t2(15) x =( t2Š 1)3,y =( t3Š 1)2(16) x a n+ y b n=1 ,a> 0 ,b> 0 ,n> 0 (17)( x2+ y2)2= a2( y2Š x2)(18) x3+ y3=3 axy In(19)and(20),“ndthepointsofintersectionoftwocurves C1and C2.Ateachpointofintersection,“ndthetangentlinesto C1and C2anddeterminetheanglebetweentheselines. (19) C1: x =cos t,y =sin t ; C2: x =1+cos(2 t ) ,y =sin(2 t )

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67.POLARCOORDINATES207 (20) C1: x2+ y2= a2; C2: x2+ y2=2 ay 67.PolarCoordinates Apointonaplaneisdescribedbyanorderedpairofnumbers ( x0,y0)intherectangularcoordinatesystem.Thisdescriptionimplies ageometricalproceduretoobtainthepointastheintersectionoftwo mutuallyperpendicularlines x = x0and y = y0.Thesetofvertical andhorizontallinesformarectangulargridinaplane.Thereareother possibilitiestolabelpointsonaplanebyorderedpairofnumbers.Here thepolarcoordinatesystemisintroduced,whichismoreconvenientfor manypurposes. Fixapoint O onaplane.Ahorizontalrayfrom O iscalledthe polar axis ,andthepoint O iscalledthe origin or pole .Considerageneric point P ontheplane.Let betheanglebetweenthepolaraxisandthe ray OP from O through P .Theangle iscountedcounterclockwise fromthepolaraxis.Thepositionofthepoint P ontheray OP is uniquelydeterminedbythedistance r = | OP | .Thus,anypoint P on aplaneisuniquelyassociatedwiththeorderedpair( r, ),and r are calledthe polarcoordinates of P .Thecoordinate r iscalledthe radial variable ,and iscalledthe polarangle Allpointsontheplanethathavethesamevalueoftheradialvariableformacircleofradius r centeredattheorigin(allpointsthathave Figure10.8. De“nitionofthepolarcoordinatesina plane. r isthedistance | OP | ,and istheanglecounted counterclockwisefromthehorizontalrayoutgoingfrom O totheright.Therectangularcoordinatesofapoint P arerelatedtothepolaronesas x = r cos y = r sin .

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20810.PLANARCURVES thesamedistancefromtheorigin).Allpointsontheplanethathave thesamevalueofthepolarangleformaray(ahalf-lineboundedbythe origin).Soapoint P withpolarcoordinates( r, )istheintersection ofthecircleofradius r andtheraythatmakestheangle withthe polaraxis.Concentriccirclesandraysoriginatingfromthecenterof thecirclesformapolargridinaplane(seeFigure10.9). Torepresentallpointsofaplane,theradialvariablehastorange overtheinterval r [0 ),whilethepolarangletakesitsvaluesin theinterval[0 2 )becauseanyrayfromtheorigindoesnotchange afterrotationabouttheoriginthroughtheangle2 .Itisconvenient, though,tolet rangeoverthewholerealline.Positivevaluesof correspondtorotationanglescountedcounterclockwise,whilenegative valuesof areassociatedwithrotationanglescountedclockwise.All pairs( r, )witha“xedvalueof r andvaluesof dierentbyinteger multiplesof2 representthesamepointsoftheplane.Forexample, theorderedpairs( r, )=(1 Š )and(1 )correspondtothesame point.Indeed,bothpointsareonthecircleofunitradius.Theray = isobtainedfromthepolaraxisbycounterclockwiserotation ofthelatterthroughtheangle .Butthesamerayisobtainedby rotatingthepolaraxisthroughtheangle clockwise;thatis,therays = and = Š coincide. Furthermore,themeaningoftheradialvariable r canbeextended tothecaseinwhich r isnegativebyagreeingthatthepairs( Š r, )and ( r, + ), r> 0,representthesamepoint.Geometrically,thepoints ( r, )lieonalinethroughtheoriginatthesamedistance | r | fromthe originbutontheoppositesidesoftheorigin.Withthisagreementon extendingthemeaningofthepolarcoordinates,eachpointonaplane mayberepresentedbycountablymanypairs:(10.6)( r, ) ( r, +2 n )or( Š r, +(2 n +1) ) where n isaninteger.67.1.RectangularandPolarCoordinates.Supposethatthepolaraxis issetsothatitcoincideswiththepositive x axisoftherectangular coordinatesystem.Everypointontheplaneiseitherdescribedby therectangularcoordinates( x,y )orthepolarcoordinates( r, ).Itis easyto“ndtherelationbetweenthepolarandrectangularcoordinates ofapoint P byexaminingtherectanglewiththediagonal OP .Its horizontalandverticalsideshavelengths x and y ,respectively.The lengthofthediagonalis r .Theanglebetweenthehorizontalsideand

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67.POLARCOORDINATES209 thediagonalis .Therefore,cos = x/r andsin = y/r ,or x = r cos ,y = r sin r2= x2+ y2, tan = y x Theserelationsallowustoconvertthepolarcoordinatesofapointto rectangularcoordinatesandviceversa(seeFig.10.8). Example 10.6 Findtherectangularcoordinatesofapointwhose polarcoordinatesare (2 ,/ 6) .Findthepolarcoordinatesofapoint withrectangularcoordinates ( Š 1 1) Solution: For r =2and = / 6,onehas x =2cos( / 6)=2 3 / 2= 3and y =2sin( / 6)=2 / 2=1,so( x,y )=( 3 1).For x = Š 1and y =1,onehas r2=2or r = 2andtan = Š 1.Thepoint( Š 1 1) liesinthesecondquadrant,thatis, / 2 .Therefore, =3 / 4. Alternatively,onecantake =3 / 4 Š 2 = Š 5 / 4. 67.2.PolarGraphs.A polargraph isacurvede“nedbytheequation r = f ( )or,moregenerally, F ( r, )=0.Itconsistsofallpointsthat haveatleastonepolarrepresentation( r, )thatsatis“estheequation. Herepolarcoordinatesareunderstoodintheextendedsenseof(10.6) whentheyareallowedtotakeanyvalue. Figure10.9. Polargrid.Coordinatecurvesofthepolarcoordinates.Thecurvesofconstantvaluesof r are concentriccircles.Thecurvesofconstantvaluesof are raysoutgoingfromtheorigin.

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21010.PLANARCURVES Thesimplestpolargraphisde“nedbyaconstantfunction r = a where a isreal.Since r representsthedistancefromtheorigin,the pairs( | a | )formacircleofradius | a | centeredattheorigin.Similarly, thegraph = b ,where b isreal,isthesetofallpoints( r,b ),where r rangesovertherealaxis,whichisthelinethroughtheoriginthat makesanangle b radianswiththepolaraxis.Noticethatthepoints ( r,b ), r> 0,and( r,b ), r< 0,lieintheoppositequadrantsrelativeto theoriginasthepairs( r,b )and( Š r,b + )representthesamepoint. Ingeneral,theshapeofapolargraphcanbedeterminedbyplotting points( f ( k) ,k), k =1 2 ,...,n ,forasetofsuccessivelyincreasing valuesof 1<2<
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67.POLARCOORDINATES211 Figure10.10. Polarcurve r = .Itisaspiralbecause thedistancefromtheorigin r increaseswiththeangle asthepointrotatesabouttheoriginthroughtheangle .67.3.SymmetryofPolarGraphs.Whensketchingpolargraphs,ititis helpfultotakeadvantageofsymmetry,justlikewhenplottinggraphs y = f ( x )forsymmetric( f ( Š x )= f ( x ))orskew-symmetric( f ( Š x )= Š f ( x ))functions. (i) Ifapolarequationisunchangedwhen isreplacedby Š thecurveissymmetricaboutthepolaraxis.Notethatthe transformation( r, ) ( r, Š )meansthat( x,y ) ( x, Š y ), whichisthere”ectionaboutthe x axis(orthepolaraxis). (ii) Ifapolarequationisunchangedwhen( r, )isreplacedby ( Š r, )orby( r, + ),thecurveissymmetricabouttheorigin.Again,thesetransformationsareequivalentto( x,y ) ( Š x, Š y ),whichisthere”ectionabouttheorigin. (iii) Iftheequationisunchangedunderthetransformation( r, ) ( r, Š ),thenthecurveissymmetricabouttheverticalline = / 2.Intherectangularcoordinates,thistransformation is( x,y ) ( x, Š y ),whichisthere”ectionaboutthe yaxis. Example 10.9 Describethe cardioid r =1+sin Solution: Theequationisunchangedunder Š sothecurve issymmetricabouttheverticalaxis(the y axis).Itissucientto investigatethecurveintheinterval [ Š / 2 ,/ 2](inthefourthand “rstquadrants).Consideraraythatrotatescounterclockwisefrom = Š / 2to = / 2(fromthenegative y axistothepositive y

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21210.PLANARCURVES Figure10.11. Thecardioid r =1+sin axis).When = Š / 2, r =0.As increasesfrom Š / 2to0(the fourthquadrant),thedistancefromtheorigin r =1+sin increases monotonicallyfrom0to1( r =1onthepolaraxis).Intheinterval [0 ,/ 2](the“rstquadrant),thedistancefromtheorigin r continuesto increasemonotonicallyandreachesitsmaximalvalue2onthevertical axis.ThecardioidisshowninFig.10.11. 67.4.TangenttoaPolarGraph.To“ndatangentlinetoapolargraph r = f ( ),thepolarangleisviewedasaparametersothattheparametricequationsofthegraphare x = r cos = f ( )cos ,y = r sin = f ( )sin Bytheproductruleforthederivative, dy dx =dy d dx d= f( )sin + f ( )cos f( )cos Š f ( )sin Inparticular,ifthecurvepassesthroughtheorigin, r =0,theequation fortheslopeattheoriginissimpli“ed dy dx =tan if dr d = f( ) =0 Notethatif f( )=0,thentheslopeisanundeterminedform0 0because x( )= y( )=0foranyvalueof suchthat f( )= f ( )=0. Thismeansthatthecurvemayhaveacuspattheoriginandhenceis notsmooth.

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67.POLARCOORDINATES213 Example 10.10 Findtheslopeofthecardioid r =1+sin in termsof .Investigatethebehaviorofthecardioidneartheorigin. Solution: Here f ( )=1+sin and f( )=cos .Thisleadstothe slope dy dx = cos sin +(1+sin )cos cos2 Š (1+sin )sin = cos (1+2sin ) (1+sin )(1 Š 2sin ) wheretheidentitycos2 =1 Š sin2 hasbeenusedtotransformthedenominator.Thecardioidpassesthroughtheoriginas passesthrough Š / 2.Theslope dy/dx isundeterminedbecausethenumeratorand denominatoroftheratiovanishat = Š / 2(bothderivatives dx/d and dy/d vanish).Theleftandrightlimitshavetobeinvestigatedto seeiftheslopehasajumpdiscontinuitythusindicatingacusp.The numeratorvanishesbecauseofthefactorcos ,whilethedenominator vanishesbecauseofthefactor(1+sin ).Hence, lim ( Š / 2)dy dx = Š 1 3 lim ( Š / 2)cos 1+sin = Š 1 3 lim ( Š / 2)Š sin cos = wherelHospitalsrulehasbeenusedtoresolvetheundeterminedform0 0andthepropertythattan as ( Š / 2)hasbeeninvoked to“ndthelimit.Thecardioidhasaverticaltangentlineattheorigin. Theslopehasanin“nitejumpdiscontinuity,meaningthatthecardioid hasacuspattheorigin(seeFigure10.11). 67.5.Exercises.(1)Findrectangularcoordinatesofapointwhosepolarcoordinatesaregiven: ( r, )=(1 Š / 3) ( Š 1 5 / 6) (4 10 / 3) (2)Findpolarcoordinatesofapointwhoserectangularcoordinatesaregiven: ( x,y )=(1 1) (1 Š 1) (3 4) ( Š 2 0) (0 Š 2) In(3)…(5),convertthepolargraphequationtoaCartesianequation andsketchthecurve. (3) r =4sin (4) r =tan sec (5) r =2sin Š 4cos

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21410.PLANARCURVES In(6)…(14),sketchthecurvewiththegivenpolarequation. (6) r = 0(7) r =ln 1(8) r2Š 3 r +2=0 (9) r =4cos(6 )(10) r2=9sin(2 )(11) r =1+2cos(2 ) (12) r =2+sin(3 )(13) r =1+2sin(3 )(14) r2 =1 (15)Sketchthecurve( x2+ y2)2=4 x2y2. Hint :Usepolarcoordinates. (16)Investigatethedependenceoftheshapeofthecurve r = cos( n )astheinteger n increases.Whathappensif n isnot aninteger? (17)Showthatthecurve r =1+ a sin hasaninnerloopwhen | a | > 1and“ndtherangeof thatcorrespondstotheinner loop. (18)Forwhatvaluesof a isthecurve r =1+ a sin smooth? In(19)and(20),“ndtheslopeofthetangentlinetothegivencurve atthepointspeci“edbythevalueof andgiveanequationofthe tangentline. (19) r =2sin = / 3(20) r =1 Š 2cos = / 6 (21)Showthatthecurves r = a sin and r = a cos intersectat rightangles. In(22)…(24),investigatetheconcavityofthepolargraph: (22) r = a> 0(23) r = a (24) r = a (1+cos ) Hint :Usedierentialstoexpressthederivative d2y/dx2inpolarcoordinatessimilarlytothecalculationoftheslopeinSection67.4. 68.ParametricCurves:TheArcLengthandSurfaceArea68.1.ArcLengthofaSmoothCurve.Let C beasmoothcurvede“ned bytheparametricequations x = x ( t ), y = y ( t ),where t [ a,b ]. Supposethat C istraversedexactlyonceas t increasesfrom a to b andconsiderapartitionoftheinterval[ a,b ]suchthat t0= a and tk= t0+ k t k =0 1 2 ,...,n ,aretheendpointsofthepartition intervalsofwidth t =( b Š a ) /n .Thenthepoints Pkwithcoordinates ( x ( tk) ,y ( tk))lieonthecurvesothat P0and Pnaretheinitialand terminalpoints,respectively.Thecurve C canbeapproximatedbya polygonalpathwithvertices Pk.Byde“nition,thelength L of C is thelimitofthelengthsoftheseapproximatingpolygonsas n : (10.7) L =limn nk =1| Pk Š 1Pk| providedthelimitexists.

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68.PARAMETRICCURVES:THEARCLENGTHANDSURFACEAREA215 Bythemeanvaluetheorem,whenappliedtothefunctions x ( t )and y ( t )ontheinterval[ tk Š 1,tk],therearenumbers t kand t kin( tk Š 1,tk) suchthat xk= x ( tk) Š x ( tk Š 1)= x( t k) t, yk= y ( tk) Š y ( tk Š 1)= y( t k) t. Therefore, | Pk Š 1Pk| = ( xk)2+( yk)2= ( x( t k))2+( y( t k))2 t. Thesumin(10.7)resemblesaRiemannsumforthefunction F ( t )= ( x( t ))2+( y( t ))2.ItisnotexactlyaRiemannsumbecause t k = t kingeneral.However,if x( t )and y( t )arecontinuous,itcanbeshown thatthelimit(10.7)isthesameasif t kand t kwereequal,namely, L istheintegralof F ( t )over[ a,b ]. Theorem 10.2(ArcLengthofaCurve) Ifacurve C isdescribed bytheparametricequations x = x ( t ) y = y ( t ) t [ a,b ] ,where x( t ) and y( t ) arecontinuouson [ a,b ] and C istraversedexactlyonceas t increasesfrom a to b ,thenthelengthof C is L = b a dx dt 2+ dy dt 2dt. Figure10.12. Thearclengthofasmoothparametric curveisapproximatedbythelengthof n straightline segmentsconnectingpointsonthecurve.Thearclength isde“nedin(10.7)asthelimit n .

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21610.PLANARCURVES If C isagraph y = f ( x ),then x = t y = f ( t ),and dx = dt ,and thelengthisgivenbythefamiliarexpression L = b a 1+ dy dx 2dt. Itisconvenienttointroducethearclengthofanin“nitesimalsegment ofacurve(the dierentialofthearclength ) ds = ( dx )2+( dy )2= L = Cds = b ads dt dt. Thesymbol Cmeansthesummationoverin“nitesimalsegmentsofthe curve S (theintegralalongacurve C )andexpressesasimplefactthat thetotallengthisthesumofthelengthsofits(in“nitesimal)pieces.68.2.IndependenceofParameterization.Byitsveryde“nition,thearc lengthis independent oftheparameterizationofthecurve.Ifacurve C isde“nedasapointset,then any parametricequationscanbeused toevaluatethearclength.Let C betracedoutonlyonceby x = x ( t ), y = y ( t ),where t [ a,b ],andby x = X ( ), y = Y ( ),where [ ]. Asnoted,thereisarelationbetweentheparameters t and = g ( t ), suchthat g ( t )increasesfrom to as t increasesfrom a to b ,that is,d dt= g( t ) 0,suchthat x ( t )= X ( g ( t ))and y ( t )= Y ( g ( t )). Therefore,theintegrals(10.7)correspondingtodierentparametric equationsofthesamecurvearerelatedbyachangeoftheintegration variable: L = b a dx dt 2+ dy dt 2dt = b a dx d d dt 2+ dy d d dt 2dt = b a dx d 2+ dy d 2d dt dt = dx d 2+ dy d 2d. Thus, thearclengthisindependentofthecurveparameterizationand canbecomputedinanysuitableparameterizationofthecurve Acircleofradius R isdescribedbytheparametricequations x = R cos t y = R sin t t [0 2 ].Then dx = Š R sin tdt and dy = R cos tdt .Hence, ds2=( R sin tdt )2+( R cos tdt )2= R2(sin2t +cos2t ) dt2= R2dt2,or ds = Rdt ,and L = Cds = 2 0Rdt = R 2 0dt =2 R. Example 10.11 Findthelengthofonearchofthecycloid x = R ( Š sin ) y = R (1 Š cos ) .

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68.PARAMETRICCURVES:THEARCLENGTHANDSURFACEAREA217 Solution: Accordingtothedescriptionofthecycloid,onearchcorrespondstotheinterval [0 2 ].Thearclengthdierential ds is foundasfollows: dx = R (1 Š cos ) d,dy = R sin d, ds2= dx2+ dy2=[(1 Š cos )2+sin2 ] R2d2=[1 Š 2cos +cos2 +sin2 ] R2d2=(2 Š 2cos ) R2d2. ds = 2(1 Š cos ) Rd. Toevaluatetheintegralof 2(1 Š cos ),thedouble-angleidentityis invoked,sin2( / 2)=(1 Š cos ) / 2.Since0 / 2 when [0 2 ], thesinusisnonnegative,sin( / 2) 0,intheintegrationinterval,and hence,aftertakingthesquareroot( u2= | u | ),theabsolutevaluecan beomitted.Thus, L = R 2 0 4sin2( / 2) d =2 R 2 0sin( / 2) d =2 R [ Š 2cos( / 2)] 2 0=8 R. 68.3.AreaofaPlanarRegion.Theareaunderthecurve y = f ( x ) andabovetheinterval x [ a,b ]isgivenby A = b af ( x ) dx ,where f ( x ) 0.Supposethatthecurveisalsodescribedbyparametric equations x = x ( t ), y = y ( t ),sothatthefunction x ( t )isone-to-one. Then,bychangingtheintegrationvariable, dx = x( t ) dt and A = b aydx = y ( t ) x( t ) dt. Thenewintegrationlimitsarefoundasusual.When x = a t iseither or ,andwhen x = b t istheremainingvalue. Example 10.12 Findtheareaunderonearchofthecycloid x = R ( Š sin ) y = R (1 Š cos ) Solution: When [0 2 ], x [0 2 R ]foronearchofthecycloid, and y ( ) 0.Usingthedierential dx foundinthepreviousexample, A = 2 R 0ydx = R22 0(1 Š cos )2d = R22 0(1 Š 2cos +cos2 ) d = R22 0[1+1 2(1+cos(2 ))] d = R22 0(1+1 2) d =3 R2,

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21810.PLANARCURVES where 2 0cos d =0and 2 0cos(2 ) d =0bythe2 periodicityof thecosinefunction. 68.4.SurfaceAreaofAxiallySymmetricSurfaces.Anaxiallysymmetric surfaceisasurfacesymmetricrelativetorotationsaboutaline.Sucha lineiscalledthe symmetryaxis .Forexample,acylinderissymmetric relativetorotationsaboutitsaxis,asphereissymmetricrelativeto rotationsaboutitsdiameter,andsoon.Anaxiallysymmetricsurface issweptbyaplanarcurvewhenthelatterisrotatedaboutaline.A cylinderofradius R andheight h isobtainedbyrevolvingastraightline segmentoflength h aboutalineparalleltothesegmentatadistance R .Asphereofradius R isobtainedbyrevolvingacircleofradius R aboutitsdiameter. Figure10.13. Asurfaceisobtainedbyrotationofa smoothcurveaboutaverticalline.If ds isthearclength ofanin“nitesimalsegmentofthecurveatapoint P and R isthedistanceofthepoint P fromtherotation axis,thenthesurfaceareasweptbythecurvesegment is dA =2 Rds (thesurfaceareaofacylinderofradius R andheight ds ).

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68.PARAMETRICCURVES:THEARCLENGTHANDSURFACEAREA219 Let ds bethearclengthofanin“nitesimalsegmentofasmooth curve C positionedatapoint( x,y ).Ifthedistancebetweenthepoint ( x,y )andthesymmetryaxisis R ( x,y ),thenthearea dA ofthepart ofthesurfacesweptbythecurvesegmentwhenthelatterisrotated aboutthesymmetryaxisistheareaofacylinderofradius R ( x,y )and height ds : dA =2 R ( x,y ) ds. Thetotalsurfaceareaisthesumofareasofallsuchpartsofthesurface (10.8) A =2 CR ( x,y ) ds =2 b aR ( x ( t ) ,y ( t )) ds dt dt, where x = x ( t ), y = y ( t ), a t b areparametricequationsof C Hereitisagainassumedthatthepoint( x ( t ) ,y ( t ))tracesoutthecurve C onlyonceas t increasesfrom a to b .Inparticular,ifthesymmetry axiscoincideswiththe x axis,then R ( x,y )= | y | (thedistanceofthe point( x,y )tothe x axis)and A =2 C| y | ds =2 b a| y ( t ) | dx dt 2+ dy dt 2dt. Example 10.13 Findtheareaofthesurfaceobtainedbyrevolving onearchofthecycloid x = R ( Š sin ) y = R (1 Š cos ) aboutthe x axis. Solution: Thedierentialofthearclengthofthecycloidhasbeen computedinExample10.11.Since y ( t ) 0here,theabsolutevalue maybeomittedand A =2 Cyds =2 2 0R (1 Š cos ) 2(1 Š cos ) Rd =8 R22 0sin3( / 2) d =16 R2 0sin3udu =16 R2 0(1 Š cos2u )sin udu =16 R21 Š 1(1 Š z2) dz =16 R2( z Š z3/ 3) 1 Š 1= 64 R2 3 wherethedouble-angleidentityhasbeenusedagain,sin2( / 2)=(1 Š cos ) / 2,andthentwosuccessivechangesoftheintegrationvariable havebeendonetoevaluatetheintegral, u = / 2 [0 ]and z = cos u [ Š 1 1].

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22010.PLANARCURVES 68.5.Exercises.In(1)…(6),“ndthearclengthofthecurve. (1) x =2+3 t2, y =1 Š 2 t3betweenthepoints(2 1)and(5 Š 1). (2) x =3sin t Š sin(3 t ), y =3cos t Š cos(3 t ),0 t (3) x = t/ (1+ t ), y =ln(1+ t )betweenthepoints(0 0)and (2 / 3 ln3). (4) x =c2 acos3t y =c2 bsin3t c2= a2Š b2. (5) x =cos4t y =sin4t (6) x = a (sinh t Š t ), y = a (cosh t Š 1),0 t T (7)Findasuitableparameterizationofthecurve x2 / 3+ y2 / 3= a2 / 3anduseittocalculatethearclengthofthecurve. In(8)…(13),“ndtheareaoftheregionenclosedbythecurve(s).If necessary,“ndparametricequationsofthecurve“rst. (8) x = a cos3t y = a sin3t (theastroid). (9) x = a cos t y = b sin t (anellipse). (10) x = t2Š 2 t y = t ,andthe y axis. (11) x =2 t Š t2, y =2 t2Š t3. (12) x =c2 acos3t y =c2 bsin3t c2= a2Š b2. (13) x = a cos t y =a sin2t 2+sin t. In(14)…(22),“ndtheareaofasurfacegeneratedbyrotatingthegiven curveaboutthespeci“edaxis.Sketchthesurface. (14) x = a cos3t y = a sin3t (aboutthe x axis). (15) x = a cos3t y = a sin3t (abouttheaxis y = x ). (16) x = t3, y = t2,0 t 1(aboutthe x axis). (17) x = etŠ t y =4 et/ 2,0 t 1(aboutthe y axis). (18) x2+ y2= a2(aboutthe y axis). (19) x2 / 3+ y2 / 3= a2 / 3(20) x2+( y Š b )2= a2, b a (aboutthe x axis) (21)( x/a )2+( y/b )2=1,0
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69.AREASANDARCLENGTHSINPOLARCOORDINATES221 69.AreasandArcLengthsinPolarCoordinates69.1.AreaofaPlanarRegion.Theorem 10.3(AreaofaPlanarRegioninPolarCoordinates) Letaplanarregion D beboundedbytworaysfromtheorigin = a = b andapolargraph r = f ( ) ,where f ( ) 0 ,thatis, D = { ( r, ) | 0 r f ( ) ,a b } Thentheareaof D is A = 1 2 b a[ f ( )]2d. Proof. Considerapartitionoftheinterval[ a,b ]bypoints k= a + k k =0 1 ,...,n ,where =( b Š a ) /n .Let mkand Mkbetheminimumandmaximumvaluesof f on[ k Š 1,k].Recallthata continuousfunction f alwaysattainsitsmaximumandminimumvalues onaclosedinterval.Thearea Akoftheplanarregionboundedby therays = k Š 1, = kandthepolargraph r = f ( )isnotlessthan theareaofthedisksectorwithradius r = mkandangle andisnot greaterthantheareaofthedisksectorwithradius r = Mkandangle .Theareaofasectorofadiskwithradius R andangle radians is A =1 2R2 .Therefore, 1 2 m2 k Ak 1 2 M2 k andthetotalarea A oftheplanarregioninquestionsatis“estheinequality AL n= 1 2nk =1m2 k A 1 2nk =1M2 k = AU n,A =nk =1 Ak, whichistrueforany n .Let F ( )=1 2[ f ( )]2.Thefunction F iscontinuouson[ a,b ].Then1 2m2 kand1 2M2 karetheminimumandmaximum valuesof F onthepartitioninterval[ k Š 1,k].Thisshowsthatthe lowerandupperbounds, AL nand AU n,are loweranduppersums for thefunction F on[ a,b ].Bythede“nitionofthede“niteintegraland integrabilityofacontinuousfunction(seeCalculusI),theupperand lowersumsconvergetotheintegralof F over[ a,b ], AL n b aFd and AU n b aFd as n .Theconclusionofthetheorem, A = b aFd followsfromthesqueezeprinciple. Example 10.14 Findtheareaenclosedbyoneloopofthefour-leaf rose r =cos(2 ) .

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22210.PLANARCURVES Solution: Notethat r =1when =0,whichisthemaximalvalueof r .Thefunctioncos(2 )hastworoots = / 4thatarethenearest to =0.Hence,oneloopcorrespondstotheinterval [ Š / 4 ,/ 4]. Theareais A = 1 2 / 4 Š / 4cos2(2 ) d = 1 4 / 4 Š / 4[1+cos(4 )] d = 1 4 [ +1 4sin(4 )] / 4 Š / 4= 8 Let D beaplanarregionthatliesbetweentwopolargraphs r = f ( )and r = g ( )suchthat f ( ) g ( ) 0if [ a,b ]and0 < b Š a 2 ;thatis, D isthesetofpointswhosepolarcoordinates satisfytheinequalities: D = { ( r, ) | 0 g ( ) r f ( ) ,a b } Thentheareaof D isgivenby A = 1 2 b a[ f ( )]2d Š 1 2 b a[ g ( )]2d = 1 2 b a [ f ( )]2Š [ g ( )]2 d. Example 10.15 Findtheareaofaregion D boundedbythecardioid r =1+sin andthecircle r =3 / 2 thatliesabovethepolaraxis (inthe“rstandsecondquadrants). Solution: Thepolargraphs r =1+sin = f ( )and r =3 / 2= g ( ) areintersectingwhen f ( )= g ( )or1+sin =3 / 2orsin =1 / 2. Sincetheregion D liesinthe“rsttwoquadrants,thatis,0 thevaluesof forthepointsofintersectionhavetobechosenas = / 6= a and = Š / 6= b .Therefore, D = { ( r, ) | 3 / 2 r 1+sin ,/ 6 5 / 6} andhencetheareaof D is A = 1 2 b a[(1+sin )2Š9 4] d = 1 2 b a[ Š5 4+2sin +sin2 ] d = 1 2 b a[ Š5 4+2sin +1 2(1 Š cos(2 ))] d = 1 2 b a[ Š3 4+2sin Š1 2cos(2 )] d =1 2[ Š3 4 Š 2cos Š1 4sin(2 )] 5 / 6 / 6= 9 3 Š 2 8

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69.AREASANDARCLENGTHSINPOLARCOORDINATES223 Remark. When“ndingpointsofintersectionoftwopolargraphs, r = f ( )and r = g ( ),bysolvingtheequation f ( )= g ( ),one hastokeepinmindthatasinglepointhasmanyrepresentationsas describedin(10.6).Sosomeofthepairs( f ( ) ),where ranges oversolutionsoftheequation f ( )= g ( ),maycorrespondtothe samepoint.Toselectdistinctpoints,allpairs( f ( ) )satisfyingthe intersectionconditioncanbetransformedbymeansof(10.6)sothat r [0 )and [0 2 ).Inthisrangeofpolarcoordinates,thereis aone-to-onecorrespondencebetweenpointsonaplaneandpairs( r, ) withjustoneexceptionwhen r =0;allthepairs(0 )correspondto theoriginofthepolarcoordinatesystem.69.2.ArcLength.Supposethatacurve C istraversedbythepoint ( r, )=( f ( ) )onlyonceas increasesfrom a to b .Choosing asaparameter,thecurveisdescribedbytheparametricequations x = r cos y = r sin ,where r = f ( ).To“ndthearclengthof C onehasto“ndtherelationbetweenthearclengthdierential ds and d .Onehas dx = dr d cos Š r sin d,dy = dr d sin + r cos d Therefore, ds2= dx2+ dy2= dr d cos Š r sin 2+ dr d sin + r cos 2 d2= dr d 2(cos2 +sin2 )+ r2(cos2 +sin2 ) d2= dr d 2+ r2 d2. Thearclengthofthecurve C is L = Cds = b ads d d = b a r2+ dr d 2d. where r = f ( )and b>a Example 10.16 Findthelengthofthecardioid r =1+sin Solution: Onehas r2+ dr d 2=(1+sin )2+(cos )2=2(1+sin ) ,

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22410.PLANARCURVES wherethetrigonometricidentitysin2 +cos2 =1hasbeenused.The cardioidistraversedonceif [ Š ].Therefore,thelengthis L = 2 Š 1+sin d =2 2 / 2 Š / 2 1+sin d, sincethecardioidissymmetricabouttheverticalline(the y axis). Thisintegralcanbeevaluatedbythesubstitution u =1+sin [0 2] sothat du =cos d ,wherecos = 1 Š sin2 = 1 Š ( u Š 1)2= u (2 Š u ).Hence, L =2 2 2 0 u u (2 Š u ) du =2 2 2 0du 2 Š u = Š 4 2 2 Š u 2 0=8 69.3.SurfaceArea.Ifasurfaceisobtainedbyrotatingapolargraph r = f ( )aboutaline,the(10.8)canbeusedto“ndtheareaofthe surfacewherethedistance R ( x,y )andthearclengthdierential ds havetobeexpressedinthepolarcoordinateswith r = f ( ). Example 10.17 Findtheareaofthesurfaceobtainedbyrotating thecardioid r =1+sin aboutitssymmetryaxis. Solution: Thesymmetryaxisofthecardioidisthe y axis.Sothe distancefromthe y axistoapoint( x,y )is R ( x,y )= | x | .Thesurface canbeobtainedbyrotatingthepartofthecardioidthatliesinthe fourthand“rstquadrants,thatis, x 0or [ Š / 2 ,/ 2].Since x = r cos ,thesurfaceareais A =2 C| x | ds =2 / 2 Š / 2r cos ds d d =2 / 2 Š / 2r cos r2+ dr d 2d. Thederivative ds/d hasbeencalculatedinthepreviousexample. Therefore, A =2 2 / 2 Š / 2(1+sin )3 / 2cos d =2 2 1 Š 1(1+ u )3 / 2du =2 2 (1+ u )5 / 2 5 / 2 1 Š 1= 32 5 wherethesubstitution u =sin hasbeenmadetoevaluatethe integral.

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70.CONICSECTIONS225 69.4.Exercises.In(1)…(4),sketchthecurveand“ndtheareathatitencloses. (1) r =4cos(2 )(2) r = a (1+cos ) (3) r =2 Š cos(2 )(4) r2=4cos(2 ) In(5)…(7),sketchthecurveand“ndtheareaofoneloopofthecurve. (5) r =9sin(3 )(6) r =1+2sin (innerloop)(7) r =2cos Š sec In(8)and(9),“ndtheareaoftheregionthatliesinsidethe“rstcurve andoutsidethesecondcurve.Sketchthecurves. (8) r =2sin ,r =1(9) r =3cos ,r =1+cos In(10)…(13),“ndtheareaoftheregionboundedbythecurves.Sketch theregion. (10) r =2 ,r = [0 2 ](11) r2=sin(2 ) ,r2=cos(2 ) (12) r =2 a sin ,r =2 b cos ,a,b> 0 (13) r =3+2cos ,r =3+2sin (14)Findtheareainsidethelargerloopandoutsidethesmaller loopofthelima con r =1 / 2 Š cos In(15)…(17),sketchthecurveand“nditslength. (15) r =2 a sin (16) r = [0 2 ](17) r = a +cos ,a 1 In(18)…(21),“ndtheareaofthesurfaceobtainedbyrotatingthecurve aboutthespeci“edaxis.Sketchthesurface. (18) r = a> 0,aboutalinethroughtheorigin. (19) r =2 a cos a> 0,(i)aboutthe y axisand(ii)aboutthe x axis. (20) r2=cos(2 )aboutthepolaraxis. (21) = a and0 r R ,where0 0,about thepolaraxis. 70.ConicSections Considertwointersectinglinesinspace, L1and L2.Asurfaceswept bytheline L2whenitisrotatedabouttheline L1isacirculardouble cone .Theline L1isthesymmetryaxisofthecone.Thepointofintersectionofthelinesiscalledthe vertex ofacone.Anyplanethatdoes notpassthroughthevertexintersectstheconealongacurve.ItappearsthatallsuchcurvesfallintothreetypesasshowninFigure10.14. Ifthecurveofintersectionisaloop,thenitisanellipse.Iftheplane isparalleltotheline L2,thenthecurveisaparabola.Iftheplane

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22610.PLANARCURVES isparalleltotheaxisofthecone,thenthecurveisahyperbola.The curvesofintersectionofaplaneandaconearecalled conicsections or conics .Theyhaveapuregeometricaldescription,whichwillbe presentedhere. Remark. Atrajectoryofanymassiveobjectinthesolarsystem (e.g.,comet,asteroid,planet)isaconicsection„thatis,aparabola, hyperbola,orellipse.ThisfactfollowsfromNewtonsLawofGravity andwillbeprovedinCalculusIII.70.1.Parabolas.A parabola isthesetofpointsinaplanethatare equidistantfroma“xedpoint F (calledthe focus )anda“xedline (calledthe directrix ).Let P beapointinaplane.Considertheline through P thatisperpendiculartothedirectrixandlet Q bethepoint oftheirintersection.Then P liesonaparabolaif | FP | = | QP | .This conditionisusedtoderivetheequationofaparabola. Aparticularlysimpleequationofaparabolaisobtainedifthecoordinatesystemissetsothatthe y axiscoincideswiththelinethrough thefocusandperpendiculartothedirectrix.Theorigin O ischosen sothat F =(0 ,p )andhencetheparabolacontainstheorigin O ,while thedirectrixistheline y = Š p paralleltothe x axis(theoriginis Figure10.14. Conicsectionsarecurvesthatareintersectionsofaconewithvariousplanes.Theshapeof aconicsectiondependsontheorientationoftheplane relativetotheconesymmetryaxis.

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70.CONICSECTIONS227 Figure10.15. Left:Geometricaldescriptionofa parabolaasasetofpoints P inaplanethatareequidistantfroma“xedpoint F ,calledthe focus ,anda“xed linecalledthe directrix (ahorizontallineinthe“gure). Right:Acircularparaboloidisthesurfaceobtainedby rotatingaparabolaaboutthelinethroughitsfocusand perpendiculartoitsdirectrix. atdistance | p | from F andfromthedirectrix).If P =( x,y ),then | FP | = x2+( y Š p )2,thepoint Q hasthecoordinates( x, Š p ),and | PQ | = ( y + p )2. Anequationoftheparabolawithfocus (0 ,p ) and directrix y = Š p is | FP |2= | PQ |2= x2+( y Š p )2=( y + p )2 x2=4 py. Inthe16thcentury,Galileoshowedthatthepathofaprojectilethatis shotintotheairatanangletothegroundisaparabola.Thesurface obtainedbyrotatingaparabolaaboutitssymmetryaxisiscalleda paraboloid .Ifasourceoflightisplacedatthefocusofaparaboloid mirror,then,afterthere”ection,thelightformsabeamparalleltothe symmetryaxis.Thisfactisusedtodesign”ashlights,headlights,and soon.Conversely,abeamoflightparalleltothesymmetryaxisofa paraboloidmirrorwillbefocusedtothefocuspointafterthere”ection, whichisusedtodesignre”ectingtelescopes.70.2.Ellipses.An ellipse isthesetofpointsinaplane,thesumof whosedistancesfromtwo“xedpoints F1and F2isaconstant.The “xedpointsarecalled foci (pluralof focus ).Let P beapointona plane.Then P belongstoanellipseif | PF1| + | PF2| =2 a ,where a> 0

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22810.PLANARCURVES Figure10.16. Left:Anellipseisthesetofpointsina plane,thesumofwhosedistancesfromtwo“xedpoints F1and F2(thefoci)isaconstant.Right:Acircular ellipsoidisthesurfaceobtainedbyrotatinganellipse aboutthelinethroughitsfoci. isaconstant(thefactor2ischosenforconveniencetobeseenlater). Evidently, | F1F2| < 2 a ;otherwise,noellipseexists. Aparticularlysimpleequationofanellipseisobtainedwhenthe coordinatesystemissetsothatthefocilieonthe x axisandhave thecoordinates F1=( Š c, 0)and F2=( c, 0),where | c |
0.Theellipseintersectsthe x axis at( a, 0)andthe y axisat(0 b )(calledthe vertices ofanellipse). Thelinesegmentjoiningthepoints( a, 0)iscalledthe majoraxis Ifthefociofanellipsearelocatedonthe y axis,then x and y are swappedinthisequation,andthemajoraxisliesonthe y axis.This showsthattherestriction a b canbedroppedintheellipseequation. Inparticular,anellipsebecomesacircleofradius a if a = b .

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70.CONICSECTIONS229 OneofKeplerslawsisthattheorbitsoftheplanetsinthesolar systemareellipseswiththeSunatonefocus.70.3.Hyperbolas.A hyperbola isthesetofallpointsinaplane,the dierenceofwhosedistancesfromtwo“xedpoints F1and F2(thefoci) isaconstant.Foranypoint P onahyperbola, | PF1|Š| PF2| = 2 a (asthedierenceofthedistancescanbenegative).Letthefocibeat ( c, 0).Followingthesameprocedureusedtoderiveanequationofan ellipse, anequationofahyperbolawithfoci ( c, 0) isfoundtobe x2 a2Š y2 b2=1 where c2= a2+ b2.Thedetailsarelefttothereaderasanexercise. Thisequationshowsthat x2/a2 1forany y ,thatis, x a or x Š a .Ahyperbolathereforehastwo branches .Thebranchin x Š a intersectsthe x axisat x = Š a ,whilethebranchin x a doessoat x = a .Thepoints( a, 0)arecalled vertices .Furthermore, intheasymptoticregion | x | ,ahyperbolahasslantasymptotes y = ( b/a ) x .Indeed, y = b x2 a2Š 1= b | x | a 1 Š a2 x2 b | x | a 1 Š a2 2 x2 b | x | a as | x | .Herethelinearization 1+ u 1+ u/ 2hasbeenusedto obtaintheasymptoticbehaviorforsmall u = Š a2/x2 0. Ifthefociofahyperbolaareonthe y axis,then,byreversingthe rolesof x and y ,itfollowsthat thehyperbola y2 a2Š x2 b2=1 hasfoci (0 c ) ,where c2= a2+ b2,vertices (0 a ) ,andslantasymptotes y = ( a/b ) x .70.4.ShiftedConics.Consideracurvede“nedbyaquadraticCartesianequation Ay2+ Bx2+ y + x + =0 Supposethat A =0and B =0.Bycompletingthesquares,this equationcanbetransformedtothestandardform A y Š 2 A 2+ B x Š 2 B 2= 2 4 A + 2 4 B Š = d or ( y Š y0)2 A/d + ( x Š x0)2 B/d =1 ,

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23010.PLANARCURVES Figure10.17. Left:Ahyperbolaisthesetofallpoints inaplane,thedierenceofwhosedistancesfromtwo “xedpoints F1and F2(thefoci)isaconstant.Top right:Acircularhyperboloidofonesheetisthesurface obtainedbyrotatingahyperbolaaboutthelinethrough themidpointofthesegment F1F2andperpendicularto it(theverticallineintheleftpanel).Bottomright:A circularhyperboloidoftwosheetsisthesurfaceobtained byrotatingahyperbolaaboutthelinethroughitsfoci (thehorizontallineintheleftpanel). where x0= / (2 B )and y0= / (2 A ),provided d =0.Depending onthesignsof A/d and B/d ,thisequationdescribeseitheranellipse orahyperbolaasiftheoriginwasmovedtothepoint( x0,y0).If A/d and B/d arebothnegative,thentheequationhasnosolution. Ifeither A or B vanishes,butnotboth,thenthequadraticCartesian equationdescribesaparabola(thedetailsarelefttothereaderasan exercise).If A = B =0,thetheequationdescribesastraightline. If d =0,solutionsoftheequationformasetoftwostraightlines, y Š y0= Š ( B/A )( x Š x0),throughthepoint( x0,y0),provided AB< 0.WhensolutionsoftheCartesianequationformahyperbola ( d =0, AB< 0),theselinesareitsslantasymptotes.

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70.CONICSECTIONS231 70.5.ConicSectionsinPolarCoordinates.Thefollowingtheoremoffersauniformdescriptionofconicsections. Theorem 10.4(ConicSections) Let F bea“xedpoint(calledthe focus )and L bea“xedline(calledthe directrix )inaplane.Let e bea “xedpositivenumber(calledthe eccentricity ).Thesetofpoints P in theplanewhosetheratioofthedistancefrom F tothedistancefrom L istheconstant e isaconicsection.Theconicis (1)anellipseif e< 1 (2)aparabolaif e =1 (3)ahyperbolaif e> 1 e = | PF | | PL | Proof. Setthecoordinatesystemsothat F isattheoriginand thedirectrixisparalleltothe y axisand d unitstotheright.Thus, thedirectrixhastheequation x = d> 0andisperpendiculartothe polaraxis.Ifthepoint P haspolarcoordinates( r, )andrectangular coordinates( x,y ),then | PF | = r = x2+ y2and | PL | = d Š x = d Š r cos .Thecondition | PF | = e | PL | yieldstheequation r = e ( d Š x ). Bysquaringit,oneinfersa quadratic Cartesianequation x2+ y2= e2( d Š x )2 (1 Š e2) x2+ y2+2 e2dx Š e2d =0 whichhasbeeninvestigatedintheprecedingsection.If e =1,then theequationdescribestheshiftedparabola y2= Š 2 d ( x Š 1 / 2).When e =1,bycompletingthesquares,thisequationisbroughttothe standardform x + e2d 1 Š e22+ y2 1 Š e2= e2d2 (1 Š e2)2. If e< 1,thenallthecoecientsarepositive,andtheequationdescribes ashiftedellipse ( x Š x0)2 a2+ y2 b2=1 ,a2= b2 1 Š e2,b2= e2d2 1 Š e2,x0= e2d e2Š 1 = Š c, where c isthedistancefromtheorigintothefocioftheellipse, c2= a2Š b2.Theeccentricityisthen e = c/a .Similarly,if e> 1,thenthe coecientshaveoppositesigns,andtheequationdescribesashifted hyperbola ( x Š x0)2 a2Š y2 b2=1 ,e = c a ,c2= a2+ b2. Inthebeginningoftheproof,thepolarequationforconicsections wasgivenas r = e ( d Š r cos ).Ifthedirectrixischosentobetothe leftofthefocusas x = Š d ,thencos isreplacedby Š cos inthepolar

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23210.PLANARCURVES equation.Ifthedirectrixischosentobeparalleltothepolaraxisas y = d ,thentheconicsectionsare r = e ( d y )= e ( d r sin ).These equationscanbesolvedfor r toobtainconicsectionsaspolargraphs. Corollary 10.5(ConicsinPolarCoordinates) Apolarequation oftheform r = ed 1 e cos or r = ed 1 e sin representsaconicsectionofeccentricity e .Theconicsectionisan ellipseif e< 1 ,aparabolaif e =1 ,andahyperbolaif e> 1 .70.6.Exercises.In(1)…(9),classifytheconicsection.Findthevertices,foci(orfocus), directrix,andasymptotes(ifthecurveisahyperbola).Sketchthe curve. (1) y2=16 x (2) x2= Š 4 y (3) y +12 x Š 2 x2=18 (4) x2+4 y2=16(5)9 x2Š 18 x +4 y2=27 (6) x2+3 y2+2 x Š 12 y +10=0(7)4 x2Š 9 y2=36 (8) y2Š 2 y =4 x2+3(9) y2Š 4 x2+2 y +16 x +3=0 (10)Along-rangeradionavigationsystemusestworadiostations, locatedatpoints A and B alongthecoastline,thattransmitsimultaneoussignalstoashiplocatedatpoint P inthe sea.Theonboardcomputerconvertsthetimedierencein receivingthesesignalsintoadistancedierence | PA |Š| PB | Thislocatestheshipononebranchofahyperbola.Suppose thatstation B islocated D milesfromstation A .Ashipreceivesthesignalfrom B microseconds( s)beforeitreceives thesignalfrom A .Thesignaltravelswiththespeedoflight, c =980ft / s.Howfarothecoastlineistheship?Ifthe coordinatesystemissetsothattheline AB coincideswiththe x axisand A isattheorigin,“ndthecoordinatesoftheship asfunctionsof In(11)…(13),classifytheconicsection.Findtheeccentricity,anequationofthedirectrix,andsketchtheconic. (11) r = 8 4+sin (12) r = 10 2 Š 5cos (13) r = 1 3+3cos (14)Showthattheconicsections r = a/ (1 Š cos )and r = b/ (1+ cos )intersectatrightangles.

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70.CONICSECTIONS233 In(15)…(20),“ndthepolarequationsforthecurve. (15) x2+4 y2=16(16) x2Š 4 y2=9 (17) y2+6 x =0(18) x = Š 4cos t,y =9sin t (19) x = a cosh t,y = b sinh t (20) x = a sec t,y = b tan t (21)TheorbitofHalleysComet,lastseenin1986andduetoreturnin2062,isanellipsewitheccentricity0 97andonefocus attheSun.Thelengthofitsmajoraxisis36.18AU.Anastronomicalunit(AU)isthemeandistancebetweentheEarth andtheSun,about93millionmiles.Findapolarequation fortheorbitofHalleysComet.Whatisthemaximaland minimaldistancefromthecomettotheSun?