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Preface
These learning modules are intended as the primary learning material for the Basic Calculus course in senior high school. It contains the main definitions, theorems, operations, formulas and techniques for the course. The material includes numerous worked-out examples to help you understand the different principles and gain proficiency in the various problem-solving skills and techniques
Calculus is one of the most important inventions in mathematics. Developed in the second half of the 17th century by Isaac Newton and Gottfried Leibniz, it is now a fundamental area of mathematics and is a powerful tool used extensively not only by mathematicians but also by engineers, scientists, economists and others in a wide variety of applications. Calculus provides the language and concepts that allow us to model natural phenome
An introductory course in calculus, such as the Grade 11 Basic Calculus course, is now a standard course in the senior high school curriculum in almost all countries all over the world. It is important that you learn the fundamental concepts and skills now. In particular, you will use the knowledge from this course in learning physics, which will be taught to you in Grade 12 using a calculus-based approach. There are still many things to learn about calculus, and you will encounter them in college
The best way to master the concepts is to study very well, and not just read, these modules. By studying, we mean that you should take your pen and paper, and work carefully through the examples, and solve the exercises given in learning modules until you are comfortable with the ideas and techniques. This is the best way to learn mathematics
The Basic Calculus course will require many concepts and skills that you have already learned in previous math courses, such as equations, functions, polynomials and their graphs. However, there will be some new ideas that you will encounter for the first time. Some of these ideas may appear abstract and complicated, but we expect all students to appreciate and learn how to use them. Because senior high school is a transition to college, mastering this course will prepare you for a higher level of academic rigor and precision. We are confident that you can do it
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Basic Calculus Learner’s Material This learning resource was collaboratively developed and reviewed by educators from public and private schools, colleges, and/or universities. We encourage teachers and other education stakeholders to email their feedback, comments and recommendations to the Department of Education at action@deped.gov.ph. We value your feedback and recommendations. Department of Education Republic of the Philippines
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PrefaceTheselearningmodulesareintendedastheprimarylearningmaterialfortheBasicCalculuscourseinseniorhighschool.Itcontainsthemainde nitions,theorems,operations,formulasandtechniquesforthecourse.Thematerialincludesnumerousworked-outexamplestohelpyouunderstandthedi erentprinciplesandgainpro ciencyinthevariousproblem-solvingskillsandtechniques.Calculusisoneofthemostimportantinventionsinmathematics.Developedinthesecondhalfofthe17thcenturybyIsaacNewtonandGottfriedLeibniz,itisnowafundamentalareaofmathematicsandisapowerfultoolusedextensivelynotonlybymathematiciansbutalsobyengineers,scientists,economistsandothersinawidevarietyofapplications.Calculusprovidesthelanguageandconceptsthatallowustomodelnaturalphenomena.Anintroductorycourseincalculus,suchastheGrade11BasicCalculuscourse,isnowastandardcourseintheseniorhighschoolcurriculuminalmostallcountriesallovertheworld.Itisimportantthatyoulearnthefundamentalconceptsandskillsnow.Inparticular,youwillusetheknowledgefromthiscourseinlearningphysics,whichwillbetaughttoyouinGrade12usingacalculus-basedapproach.Therearestillmanythingstolearnaboutcalculus,andyouwillencounterthemincollege.Thebestwaytomastertheconceptsistostudyverywell,andnotjustread,thesemodules.Bystudying,wemeanthatyoushouldtakeyourpenandpaper,andworkcare-fullythroughtheexamples,andsolvetheexercisesgiveninlearningmodulesuntilyouarecomfortablewiththeideasandtechniques.Thisisthebestwaytolearnmathematics.TheBasicCalculuscoursewillrequiremanyconceptsandskillsthatyouhavealreadylearnedinpreviousmathcourses,suchasequations,functions,polynomialsandtheirgraphs.However,therewillbesomenewideasthatyouwillencounterforthe rsttime.Someoftheseideasmayappearabstractandcomplicated,butweexpectallstudentstoappreciateandlearnhowtousethem.Becauseseniorhighschoolisatransitiontocollege,masteringthiscoursewillprepareyouforahigherlevelofacademicrigorandprecision.Wearecon dentthatyoucandoit.
ContentsPrefacei1LimitsandContinuity1Lesson1:TheLimitofaFunction:TheoremsandExamples…………..2Topic1.1:TheLimitofaFunction……………………..3Topic1.2:TheLimitofaFunctionatcversustheValueoftheFunctionatc..18Topic1.3:IllustrationofLimitTheorems………………….23Topic1.4:LimitsofPolynomial,Rational,andRadicalFunctions………………………….30Lesson2:LimitsofSomeTranscendentalFunctionsandSomeIndeterminateForms.43Topic2.1:LimitsofExponential,Logarithmic,andTrigonometricFunctions…44Topic2.2:SomeSpecialLimits……………………….62Lesson3:ContinuityofFunctions………………………..73Topic3.1:ContinuityataPoint………………………74Topic3.2:ContinuityonanInterval…………………….83Lesson4:MoreonContinuity…………………………..92Topic4.1:Di erentTypesofDiscontinuities…………………93Topic4.2:TheIntermediateValueandtheExtremeValueTheorems…….105Topic4.3:ProblemsInvolvingContinuity………………….115
2Derivatives123Lesson5:TheDerivativeastheSlopeoftheTangentLine……………124Topic5.1:TheTangentLinetotheGraphofaFunctionataPoint……..125Topic5.2:TheEquationoftheTangentLine………………..131Topic5.3:TheDe nitionoftheDerivative…………………139Lesson6:RulesofDi erentiation…………………………151Topic6.1:Di erentiabilityImpliesContinuity………………..152Topic6.2:TheDi erentiationRulesandExamplesInvolvingAlgebraic,Expo-nential,andTrigonometricFunctions…………………158Lesson7:Optimization……………………………..169Topic7.1:OptimizationusingCalculus…………………..170Lesson8:Higher-OrderDerivativesandtheChainRule……………..185Topic8.1:Higher-OrderDerivativesofFunctions………………186Topic8.2:TheChainRule…………………………193Lesson9:ImplicitDi erentiation…………………………200Topic9.1:WhatisImplicitDi erentiation?…………………201Lesson10:RelatedRates…………………………….2123Integration223Lesson11:Integration………………………………224Topic11.1:IllustrationofanAntiderivativeofaFunction………….225Topic11.2:AntiderivativesofAlgebraicFunctions……………..227Topic11.3:AntiderivativesofFunctionsYieldingExponentialFunctionsandLogarithmicFunctions………………………..231Topic11.4:AntiderivativesofTrigonometricFunctions……………234
Lesson12:TechniquesofAntidi erentiation……………………237Topic12.1:Antidi erentiationbySubstitutionandbyTableofIntegrals…..238Lesson13:ApplicationofAntidi erentiationtoDi erentialEquations………258Topic13.1:SeparableDi erentialEquations…………………259Lesson14:ApplicationofDi erentialEquationsinLifeSciences…………267Topic14.1:SituationalProblemsInvolvingGrowthandDecayProblems…..268Lesson15:RiemannSumsandtheDe niteIntegral……………….280Topic15.1:ApproximationofAreausingRiemannSums…………..281Topic15.2:TheFormalDe nitionoftheDe niteIntegral………….299Lesson16:TheFundamentalTheoremofCalculus………………..316Topic16.1:IllustrationoftheFundamentalTheoremofCalculus………317Topic16.2:ComputationofDe niteIntegralsusingtheFundamentalTheoremofCalculus………………….321Lesson17:IntegrationTechnique:TheSubstitutionRuleforDe niteIntegrals….332Topic17.1:IllustrationoftheSubstitutionRuleforDe niteIntegrals……333Lesson18:ApplicationofDe niteIntegralsintheComputationofPlaneAreas….345Topic18.1:AreasofPlaneRegionsUsingDe niteIntegrals…………346Topic18.2:ApplicationofDe niteIntegrals:WordProblems………..362
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LESSON1:TheLimitofaFunction:TheoremsandExamplesLEARNINGOUTCOMES:Attheendofthelesson,thelearnershallbeableto:1.Illustratethelimitofafunctionusingatableofvaluesandthegraphofthefunction;2.Distinguishbetweenlimx!cf(x)andf(c);3.Illustratethelimittheorems;and4.Applythelimittheoremsinevaluatingthelimitofalgebraicfunctions(polynomial,ra-tional,andradical).2
TOPIC1.1:TheLimitofaFunctionLIMITSConsiderafunctionfofasinglevariablex.Consideraconstantcwhichthevariablexwillapproach(cmayormaynotbeinthedomainoff).Thelimit,tobedenotedbyL,istheuniquerealvaluethatf(x)willapproachasxapproachesc.Insymbols,wewritethisprocessaslimx!cf(x)=L:Thisisread,“Thelimitoff(x)asxapproachescisL.”LOOKINGATATABLEOFVALUESToillustrate,letusconsiderlimx!2(1+3x):Here,f(x)=1+3xandtheconstantc,whichxwillapproach,is2.Toevaluatethegivenlimit,wewillmakeuseofatabletohelpuskeeptrackofthee ectthattheapproachofxtoward2willhaveonf(x).Ofcourse,onthenumberline,xmayapproach2intwoways:throughvaluesonitsleftandthroughvaluesonitsright.We rstconsiderapproaching2fromitsleftorthroughvalueslessthan2.Rememberthatthevaluestobechosenshouldbecloseto2.xf(x)141:45:21:76:11:96:71:956:851:9976:9911:99996:99971:99999996:9999997Nowweconsiderapproaching2fromitsrightorthroughvaluesgreaterthanbutcloseto2.3
xf(x)3102:58:52:27:62:17:32:037:092:0097:0272:00057:00152:00000017:0000003Observethatasthevaluesofxgetcloserandcloserto2,thevaluesoff(x)getcloserandcloserto7.Thisbehaviorcanbeshownnomatterwhatsetofvalues,orwhatdirection,istakeninapproaching2.Insymbols,limx!2(1+3x)=7:EXAMPLE1:Investigatelimx! 1(x2+1)byconstructingtablesofvalues.Here,c= 1andf(x)=x2+1:Westartagainbyapproaching 1fromtheleft.xf(x) 1:53:25 1:22:44 1:012:0201 1:00012:00020001Nowapproach 1fromtheright.xf(x) 0:51:25 0:81:64 0:991:9801 0:99991:99980001Thetablesshowthatasxapproaches 1,f(x)approaches2.Insymbols,limx! 1(x2+1)=2:4
EXAMPLE2:Investigatelimx!0jxjthroughatableofvalues.Approaching0fromtheleftandfromtheright,wegetthefollowingtables:xjxj 0:30:3 0:010:01 0:000090:00009 0:000000010:00000001xjxj0:30:30:010:010:000090:000090:000000010:00000001Hence,limx!0jxj=0:EXAMPLE3:Investigatelimx!1×2 5x+4x 1byconstructingtablesofvalues.Here,c=1andf(x)=x2 5x+4x 1:Takenotethat1isnotinthedomainoff,butthisisnotaproblem.Inevaluatingalimit,rememberthatweonlyneedtogoverycloseto1;wewillnotgoto1itself.Wenowapproach1fromtheleft.xf(x)1:5 2:51:17 2:831:003 2:9971:0001 2:9999Approach1fromtheright.xf(x)0:5 3:50:88 3:120:996 3:0040:9999 3:00015
Thetablesshowthatasxapproaches1,f(x)approaches 3.Insymbols,limx!1×2 5x+4x 1= 3:EXAMPLE4:Investigatethroughatableofvalueslimx!4f(x);iff(x)=8<:x+1ifx<4(x 4)2+3ifx4:Thislooksabitdi erent,butthelogicandprocedureareexactlythesame.Westillapproachtheconstant4fromtheleftandfromtheright,butnotethatweshouldevaluatetheappropriatecorrespondingfunctionalexpression.Inthiscase,whenxapproaches4fromtheleft,thevaluestakenshouldbesubstitutedinf(x)=x+1.Indeed,thisisthepartofthefunctionwhichacceptsvalueslessthan4.So,xf(x)3:74:73:854:853:9954:9953:999994:99999Ontheotherhand,whenxapproaches4fromtheright,thevaluestakenshouldbesubstitutedinf(x)=(x 4)2+3.So,xf(x)4:33:094:13:014:0013:0000014:000013:00000000016
Observethatthevaluesthatf(x)approachesarenotequal,namely,f(x)approaches5fromtheleftwhileitapproaches3fromtheright.Insuchacase,wesaythatthelimitofthegivenfunctiondoesnotexist(DNE).Insymbols,limx!4f(x)DNE:Remark1:Weneedtoemphasizeanimportantfact.Wedonotsaythatlimx!4f(x)\equalsDNE",nordowewrite\limx!4f(x)=DNE",because\DNE"isnotavalue.Inthepreviousexample,\DNE"indicatedthatthefunctionmovesindi erentdirectionsasitsvariableap-proachescfromtheleftandfromtheright.Inothercases,thelimitfailstoexistbecauseitisunde ned,suchasforlimx!01xwhichleadstodivisionof1byzero.Remark2:Haveyounoticedapatterninthewaywehavebeeninvestigatingalimit?Wehavebeenspecifyingwhetherxwillapproachavaluecfromtheleft,throughvalueslessthanc,orfromtheright,throughvaluesgreaterthanc.Thisdirectionmaybespeci edinthelimitnotation,limx!cf(x)byaddingcertainsymbols.Ifxapproachescfromtheleft,orthroughvalueslessthanc,thenwewritelimx!c f(x).Ifxapproachescfromtheright,orthroughvaluesgreaterthanc,thenwewritelimx!c+f(x).Furthermore,wesaylimx!cf(x)=Lifandonlyiflimx!c f(x)=Landlimx!c+f(x)=L:Inotherwords,foralimitLtoexist,thelimitsfromtheleftandfromtherightmustbothexistandbeequaltoL.Therefore,limx!cf(x)DNEwheneverlimx!c f(x)6=limx!c+f(x):Theselimits,limx!c f(x)andlimx!c+f(x),arealsoreferredtoasone-sidedlimits,sinceyouonlyconsidervaluesononesideofc.Thus,wemaysay:inourvery rstillustrationthatlimx!2(1+3x)=7becauselimx!2 (1+3x)=7andlimx!2+(1+3x)=7.7
inExample1,limx! 1(x2+1)=2sincelimx! 1 (x2+1)=2andlimx! 1+(x2+1)=2.inExample2,limx!0jxj=0becauselimx!0 jxj=0andlimx!0+jxj=0.inExample3,limx!1x2 5x+4x 1= 3becauselimx!1 x2 5x+4x 1= 3andlimx!1+x2 5x+4x 1= 3.inExample4,limx!4f(x)DNEbecauselimx!4 f(x)6=limx!4+f(x).LOOKINGATTHEGRAPHOFy=f(x)Ifoneknowsthegraphoff(x),itwillbeeasiertodetermineitslimitsasxapproachesgivenvaluesofc.Consideragainf(x)=1+3x.Itsgraphisthestraightlinewithslope3andintercepts(0;1)and( 1=3;0).Lookatthegraphinthevicinityofx=2.Youcaneasilyseethepoints(fromthetableofvaluesinpage4)(1;4);(1:4;5:2);(1:7;6:1),andsoon,approachingthelevelwherey=7.Thesamecanbeseenfromtheright(fromthetableofvaluesinpage4).Hence,thegraphclearlycon rmsthatlimx!2(1+3x)=7:xyy=1+3x(2;7) 10123412345678Letuslookattheexamplesagain,onebyone.RecallExample1wheref(x)=x2+1.Itsgraphisgivenby8
3 2 1012312345678xyy=x2+1( 1;2)Itcanbeseenfromthegraphthatasvaluesofxapproach 1,thevaluesoff(x)approach2.RecallExample2wheref(x)=jxj.xyy=jxj(0;0)Itisclearthatlimx!0jxj=0,thatis,thetwosidesofthegraphbothmovedownwardtotheorigin(0;0)asxapproaches0.RecallExample3wheref(x)=x2 5x+4x 1:9
1234 4 3 2 10xyy=x2 5x+4x 1(1; 3)Takenotethatf(x)=x2 5x+4x 1=(x 4)(x 1)x 1=x 4;providedx6=1.Hence,thegraphoff(x)isalsothegraphofy=x 1,excludingthepointwherex=1.RecallExample4wheref(x)=8<:x+1ifx<4(x 4)2+3ifx4:012345671234567xyy=f(x)(4;5)(4;3)Again,wecanseefromthegraphthatf(x)hasnolimitasxapproaches4.Thetwoseparatepartsofthefunctionmovetowarddi erenty-levels(y=5fromtheleft,y=3fromtheright)inthevicinityofc=4.10
So,ingeneral,ifwehavethegraphofafunction,suchasbelow,determininglimitscanbedonemuchfasterandeasierbyinspection. 3 2 10123456123456xy( 2;1)(0;3)(3;0)(3;2)(3;4)Forinstance,itcanbeseenfromthegraphofy=f(x)that:a.limx! 2f(x)=1.b.limx!0f(x)=3.Here,itdoesnotmatterthatf(0)doesnotexist(thatis,itisunde ned,orx=0isnotinthedomainoff).Alwaysrememberthatwhatmattersisthebehaviorofthefunctionclosetoc=0andnotpreciselyatc=0.Infact,eveniff(0)werede nedandequaltoanyotherconstant(notequalto3),like100or 5000,thiswouldstillhavenobearingonthelimit.Incaseslikethis,limx!0f(x)=3prevailsregardlessofthevalueoff(0),ifany.c.limx!3f(x)DNE.Ascanbeseeninthe gure,thetwopartsofthegraphnearc=3donotmovetowardacommony-levelasxapproachesc=3.SolvedExamplesLOOKINGATTABLESOFVALUESEXAMPLE1:Determinelimx!1 x3 1.Solution.Approachingx=1fromtheleft,11
xf(x)0:9 0:2710:99 0:0297010:999 0:0029970010:9999 0:000299970:99999 0:0000299997Now,takingvaluesfromtherightofx=1,xf(x)1:10:3311:010:0303011:0010:0030030011:00010:000300031:000010:0000300003Thus,limx!1 x3 1=0:.EXAMPLE2:Determinelimx!0jx+2j.Solution.Takingvaluesfromtheleftof0,xf(x) 0:11:9 0:051:95 0:011:99 0:0051:995 0:0011:999Approaching0fromtheright,12
xf(x)0:12:10:052:050:012:010:0052:0050:0012:001Thus,limx!0jx+2j=2:.EXAMPLE3:Givenf(x)=8<:2x 3;x4;x2 x+1;x>4:Evaluatelimx!4f(x).Solution.Approachingx=4fromtheleft,xf(x)3:94:83:954:93:994:983:9954:993:9994:998Takingvaluesfromtherightof4,xf(x)4:113:714:0513:35254:0113:07014:00513:0350254:00113:007001Hence,limx!4f(x)DNE..13
LOOKINGATGRAPHSOFFUNCTIONSEXAMPLE4:RecallExample1,wheref(x)=x3 1andwhosegraphisasfollows:xyy=x3 1(1;0)Solution.Itisclearfromthegraphthatlimx!1 x3 1=0:.EXAMPLE5:RecallExample2,wheref(x)=jx+1j.Itsgraphisasfollows:xyy=jx+2j(0;2)Solution.Itisclearfromthegraphabovethatlimx!0jx+2j=2:.14
EXAMPLE6:RecallExample3,wheref(x)=8<:2x 3;x4;x2 x+1;x>4;andwhosegraphisasfollows:xy(4;5)(4;13)Solution.Fromthegraphabove,limx!4 f(x)=5andlimx!4+f(x)=13.Hence,limx!4f(x)DNE..15
EXAMPLE7:Now,considerthefollowinggraphoffunction:xy( 2;3)(0;5)(3;2)(3;6)(3;3)Solution.Fromthegraph,1.limx! 2f(x)=3,2.limx!0f(x)=5,3.limx!3f(x)DNE..SupplementaryProblems1.Usingtablesofvalues,determinethelimitsofthefollowing.(a)limx!0(x)(b)limx!1)2x)(c)limx!2( x+1)(d)limx! 1(1 x)(e)limx!0(2x 1)(f)limx!3(x 3)(g)limx!2(3×2 2)(h)limx!0(5 x x2)(i)limx!1(x3+1)(j)limx!0(1 x2 x3)(k)limx!1px3 1(l)limx!0px2 x3(m)limx!1px4 x2+1(n)limx!0x1 x(o)limx!11 x2x+116
2.Thegraphoff(x)isgivenbyxy( 4;5)( 1; 2)( 1;4)(1;4)(3; 4)(5; 2)Determinethefollowinglimits.(a)limx! 4f(x)(b)limx! 1f(x)(c)limx!1f(x)(d)limx!3f(x)(e)limx!5f(x)17
TOPIC1.2:TheLimitofaFunctionatcversustheValueoftheFunctionatcWewillmostlyrecallourdiscussionsandexamplesinLesson1.Letusagainconsiderlimx!2(1+3x):Recallthatitstablesofvaluesare:xf(x)141:45:21:76:11:96:71:956:851:9976:9911:99996:99971:99999996:9999997xf(x)3102:58:52:27:62:17:32:037:092:0097:0272:00057:00152:00000017:0000003andwehadconcludedthatlimx!2(1+3x)=7.Incomparison,f(2)=7.So,inthisexample,limx!2f(x)andf(2)areequal.Noticethatthesameholdsforthenextexamplesdiscussed:limx!cf(x)f(c)limx! 1(x2+1)=2f( 1)=2limx!0jxj=0f(0)=0This,however,isnotalwaysthecase.Letusconsiderthefunctionf(x)=8<:jxjifx6=02ifx=0:Incontrasttothesecondexampleabove,theentriesarenowunequal:limx!cf(x)f(c)limx!0jxj=0f(0)=218
Doesthisinanywaya ecttheexistenceofthelimit?Notatall.Thisexampleshowsthatlimx!cf(x)andf(c)maybedistinct.Furthermore,considerthethirdexampleinLesson1wheref(x)=8<:x+1ifx<4(x 4)2+3ifx4:Wehave:limx!cf(x)f(c)limx!4f(x)DNEf(4)=3Onceagainweseethatlimx!cf(x)andf(c)arenotthesame.AreviewofthegraphgiveninLesson1(redrawnbelow)willemphasizethisfact. 3 2 10123456123456xy( 2;1)(0;3)(3;0)(3;2)(3;4)Werestatetheconclusions,addingtherespectivevaluesoff(c):1.limx! 2f(x)=1andf( 2)=1.2.limx!0f(x)=3andf(0)doesnotexist(orisunde ned).19
3.limx!3f(x)DNEandf(3)alsodoesnotexist(orisunde ned).SolvedExamplesEXAMPLE1:Considernowf(x)= x3 1.Comparelimx!1f(x)andf(1).Solution.Inthisexample,f(1)andlimx!1f(x)areequal.limx!cf(x)f(c)limx!1(x3 1)=0f(1)=0.EXAMPLE2:Cinsiderf(x)=jx+2jandcomparelimx!0f(x)andf(0).Solution.Here,f(0)andlimx!0f(x)areequal.limx!cf(x)f(c)limx!0jx+2j=2f(0)=2.EXAMPLE3:Giventhefunctionf(x)=8<:2x 3;x4;x2 x+1;x>4:Comparelimx!4f(x)andf(4).Solution.Here,wetheentriesaredi erent.limx!cf(x)f(c)limx!4(f(x))DNEf(4)=5.20
EXAMPLE4:Giventhefollowinggraphoff(x),determineiflimx!cf(x)=f(c)ati.c= 2,ii.c=0andiii.c=3.xy( 2;3)(0;5)(3;2)(3;6)(3;3)Solution.Itisclearfromthegraphthati.limx! 2f(x)=3=f( 2),ii.limx!0f(x)=5=f(0)andiii.f(3)=6butlimx!3f(x)DNE..SupplementaryProblemsDetermineiflimx!cf(x)=f(c).1.f(x)=x+2;c= 12.f(x)=x 2;c=03.f(x)=x2+2;c=14.f(x)=x2 1;c= 15.f(x)=x3 x;c=06.f(x)=x3 3x;c=17.f(x)=x2 4;c=28.f(x)=x4 1;c=121
9.f(x)=x3 xx;c=010.f(x)=x3 3xx;c=111.f(x)=x2 4x 2;c=212.f(x)=x4 1x 1;c=113.f(x)=px4 1x 1;c=114.f(x)=px x2x2 x;c=015.f(x)=px3 1x 1;c=116.(atc=2)f(x)=8<:x 1ifx<2;(x 2)2+1ifx2;17.(atc= 1)f(x)=8<:x2 1ifx< 1;(x 1)2 4ifx 1;18.(atc=1)f(x)=8<:x3 1ifx<1;x2+4ifx1;19.(atc=2)f(x)=8<:jxjifx<2;x 1ifx2;20.(atc=0)f(x)=8<:x2+xifx<0;x2 1ifx0;22
TOPIC1.3:IllustrationofLimitTheoremsInthefollowingstatements,cisaconstant,andfandgarefunctionswhichmayormaynothavecintheirdomains.1.Thelimitofaconstantisitself.Ifkisanyconstant,thenlimx!ck=k:Forexample,(a)limx!c2=2(b)limx!c 3:14= 3:14(c)limx!c789=7892.Thelimitofxasxapproachescisequaltoc.Thismaybethoughtofasthesubstitutionlawbecausexissimplysubstitutedbyc.limx!cx=c:Forexample,(a)limx!9x=9(b)limx!0:005x=0:005(c)limx! 10x= 10Fortheremainingtheorems,wewillassumethatthelimitsoffandgbothexistasxapproachescandthattheyareLandM,respectively.Inotherwords,limx!cf(x)=L;andlimx!cg(x)=M:3.TheConstantMultipleTheorem:Thissaysthatthelimitofamultipleofafunctionissimplythatmultipleofthelimitofthefunction.limx!ckf(x)=klimx!cf(x)=kL:Forexample,iflimx!cf(x)=4,then(a)limx!c8f(x)=8limx!cf(x)=84=32:23
(b)limx!c 11f(x)= 11limx!cf(x)= 114= 44:(c)limx!c32f(x)=32limx!cf(x)=324=6:4.TheAdditionTheorem:Thissaysthatthelimitofasumoffunctionsisthesumofthelimitsoftheindividualfunctions.Subtractionisalsoincludedinthislaw,thatis,thelimitofadi erenceoffunctionsisthedi erenceoftheirlimits.limx!c(f(x)+g(x))=limx!cf(x)+limx!cg(x)=L+M:limx!c(f(x) g(x))=limx!cf(x) limx!cg(x)=L M:Forexample,iflimx!cf(x)=4andlimx!cg(x)= 5,then(a)limx!c(f(x)+g(x))=limx!cf(x)+limx!cg(x)=4+( 5)= 1:(b)limx!c(f(x) g(x))=limx!cf(x) limx!cg(x)=4 ( 5)=9:5.TheMultiplicationTheorem:ThisissimilartotheAdditionTheorem,withmultiplicationreplacingadditionastheoperationinvolved.Thus,thelimitofaproductoffunctionsisequaltotheproductoftheirlimits.limx!c(f(x)g(x))=limx!cf(x)limx!cg(x)=LM:Again,letlimx!cf(x)=4andlimx!cg(x)= 5.Thenlimx!cf(x)g(x)=limx!cf(x)limx!cg(x)=4( 5)= 20:Remark1:TheAdditionandMultiplicationTheoremsmaybeappliedtosums,di erences,andproductsofmorethantwofunctions.Remark2:TheConstantMultipleTheoremisaspecialcaseoftheMultiplicationTheorem.Indeed,intheMultiplicationTheorem,ifthe rstfunctionf(x)isreplacedbyaconstantk,theresultistheConstantMultipleTheorem.24
6.TheDivisionTheorem:Thissaysthatthelimitofaquotientoffunctionsisequaltothequotientofthelimitsoftheindividualfunctions,providedthedenominatorlimitisnotequalto0.limx!cf(x)g(x)=limx!cf(x)limx!cg(x)=LM;providedM6=0:Forexample,(a)Iflimx!cf(x)=4andlimx!cg(x)= 5,limx!cf(x)g(x)=limx!cf(x)limx!cg(x)=4 5= 45:(b)Iflimx!cf(x)=0andlimx!cg(x)= 5,limx!cf(x)g(x)=0 5=0:(c)Iflimx!cf(x)=4andlimx!cg(x)=0,itisnotpossibletoevaluatelimx!cf(x)g(x);orwemaysaythatthelimitDNE.7.ThePowerTheorem:Thistheoremstatesthatthelimitofanintegerpowerpofafunctionisjustthatpowerofthelimitofthefunction.Iflimx!cf(x)=L,thenlimx!c(f(x))p=(limx!cf(x))p=Lp:Forexample,(a)Iflimx!cf(x)=4,thenlimx!c(f(x))3=(limx!cf(x))3=43=64:(b)Iflimx!cf(x)=4,thenlimx!c(f(x)) 2=(limx!cf(x)) 2=4 2=142=116:25
8.TheRadical/RootTheorem:Thistheoremstatesthatifnisapositiveinteger,thelimitofthenthrootofafunctionisjustthenthrootofthelimitofthefunction,providedthenthrootofthelimitisarealnumber.Thus,itisimportanttokeepinmindthatifniseven,thelimitofthefunctionmustbepositive.Iflimx!cf(x)=L,thenlimx!cnpf(x)=nqlimx!cf(x)=npL:Forexample,(a)Iflimx!cf(x)=4,thenlimx!cpf(x)=qlimx!cf(x)=p4=2:(b)Iflimx!cf(x)= 4,thenitisnotpossibletoevaluatelimx!cpf(x)becausethen,qlimx!cf(x)=p 4;andthisisnotarealnumber.SolvedExamplesEXAMPLE1:Evaluatethefollowinglimits.1.limx! 10(Ans.0)2.limx!32910 1(Ans.-1)3.limx!0:000011(Ans.1)EXAMPLE2:Evaluatethegivenlimits.1.limx!1x(Ans.1)2.limx! 0:01x(Ans.-0.01)3.limx!300x(Ans.300)EXAMPLE3:Solvethefollowingcompletely.1.Givenlimx!3f(x)= 1,evaluatelimx!3(5f(x)).26
2.Givenlimx!0g(x)=2,determinelimx!0( 2g(x)).Solution.Usinglimittheorems,wehave1.limx!3(5f(x))=5limx!3f(x)=5 1= 5;2.limx!0( 2g(x))= 2limx!0g(x)= 22= 4:.EXAMPLE4:Uselimittheoremstoevaluatethefollowinglimits.1.Determinelimx!1(f(x)+g(x))iflimx!1f(x)=2andlimx!1g(x)= 1.2.Evaluatelimx! 1(f(x) g(x))giventhatlimx! 1f(x)=0andlimx! 1g(x)= 1.Solution.Weuselimittheoremstoget1.limx!1(f(x)+g(x))=limx!1f(x)+limx!1g(x)=2+( 1)=1;2.limx! 1(f(x) g(x))=limx! 1f(x) limx! 1g(x)=0 ( 1)=1:.EXAMPLE5:Evaluatethefollowinglimits.1.Givenlimx!2f(x)=3andlimx!2g(x)= 1,determinelimx!2f(x)g(x).2.Iflimx! 3f(x)=9andlimx! 3g(x)= 3,evaluatelimx! 3f(x)g(x).Solution.Usingthetheoremsabove,wehave1.limx!2f(x)g(x)=limx!2f(x)limx!2g(x)=3( 1)= 3;2.limx! 3f(x)g(x)=limx! 3f(x)limx! 3g(x)=9 3= 3:.27
EXAMPLE6:Solvethefollowingcompletely.1.Evaluatelimx!1(f(x))3iflimx!1f(x)= 1.2.Evaluatelimx!0(g(x)) 2iflimx!0g(x)=2.Solution.Usingthelimittheorems,weobtain1.limx!1(f(x))3=(limx!1f(x))3=( 1)3= 1;2.limx!0(g(x)) 2=(limx!0g(x)) 2=2 2=14:.EXAMPLE7:Evaluatethefollowinglimits.1.Determinelimx! 1pf(x)giventhatlimx! 1f(x)=1.2.Evaluatelimx! 2pg(x)iflimx! 2g(x)=9.Solution.Weuselimittheoremstoget1.limx! 1pf(x)=qlimx! 1f(x)=p1=1,2.limx! 2pg(x)=qlimx! 2g(x)=p9=3..SupplementaryProblems1.Givenlimx!1f(x)=3andlimx!1g(x)= 1,evaluatethefollowinglimits.(a)limx!12f(x)(b)limx!1 3g(x)(c)limx!1(f(x)+g(x))(d)limx!1(f(x) g(x))(e)limx!1(f(x)g(x))(f)limx!1f(x)g(x)28
2.Givenlimx! 1f(x)=2andlimx! 1g(x)= 2,evaluatethefollowinglimits.(a)limx! 112f(x)(b)limx! 1(f(x)+g(x))(c)limx! 1(f(x) g(x))(d)limx! 1(f(x)g(x))(e)limx! 1f(x)g(x)(f)limx! 1(f(x))2(g)limx! 1(g(x)) 33.Givenlimx!0f(x)=0andlimx!0g(x)=1,evaluatethefollowinglimits.(a)limx!0139401f(x)(b)limx!0(f(x)+g(x))(c)limx!0(f(x) g(x))(d)limx!0(f(x)g(x))(e)limx!0f(x)g(x)(f)limx!0pf(x)(g)limx!0pg(x)29
TOPIC1.4:LimitsofPolynomial,Rational,andRadicalFunctionsInthepreviouslesson,wepresentedandillustratedthelimittheorems.Westartbyrecallingtheselimittheorems.Theorem1.Letc,k,LandMberealnumbers,andletf(x)andg(x)befunctionsde nedonsomeopenintervalcontainingc,exceptpossiblyatc.1.Iflimx!cf(x)exists,thenitisunique.Thatis,iflimx!cf(x)=Landlimx!cf(x)=M,thenL=M.2.limx!cc=c.3.limx!cx=c4.Supposelimx!cf(x)=Landlimx!cg(x)=M.(a)(ConstantMultiple)limx!c[kg(x)]=kM.(b)(Addition)limx!c[f(x)g(x)]=LM.(c)(Multiplication)limx!c[f(x)g(x)]=LM.(d)(Division)limx!cf(x)g(x)=LM,providedM6=0.(e)(Power)limx!c[f(x)]p=Lpforp,apositiveinteger.(f)(Root/Radical)limx!cnpf(x)=npLforpositiveintegersn,andprovidedthatL>0whenniseven.Inthislesson,wewillshowhowtheselimittheoremsareusedinevaluatingalgebraicfunctions.Particularly,wewillillustratehowtousethemtoevaluatethelimitsofpolynomial,rationalandradicalfunctions.LIMITSOFALGEBRAICFUNCTIONSWestartwithevaluatingthelimitsofpolynomialfunctions.30
EXAMPLE1:Determinelimx!1(2x+1).Solution.Fromthetheoremsabove,limx!1(2x+1)=limx!12x+limx!11(Addition)=2limx!1x+1(ConstantMultiple)=2(1)+1limx!cx=c=2+1=3:.EXAMPLE2:Determinelimx! 1(2×3 4×2+1).Solution.Fromthetheoremsabove,limx! 1(2×3 4×2+1)=limx! 12×3 limx! 14×2+limx! 11(Addition)=2limx! 1×3 4limx! 1×2+1(ConstantMultiple)=2( 1)3 4( 1)2+1(Power)= 2 4+1= 5:.EXAMPLE3:Evaluatelimx!0(3×4 2x 1).Solution.Fromthetheoremsabove,limx!0(3×4 2x 1)=limx!03×4 limx!02x limx!01(Addition)=3limx!0x4 2limx!0x2 1(ConstantMultiple)=3(0)4 2(0) 1(Power)=0 0 1= 1:.Wewillnowapplythelimittheoremsinevaluatingrationalfunctions.Inevaluatingthelimitsofsuchfunctions,recallfromTheorem1theDivisionRule,andalltherulesstatedinTheorem1whichhavebeenusefulinevaluatinglimitsofpolynomialfunctions,suchastheAdditionandProductRules.31
EXAMPLE4:Evaluatelimx!11x.Solution.First,notethatlimx!1x=1:Sincethelimitofthedenominatorisnonzero,wecanapplytheDivisionRule.Thus,limx!11x=limx!11limx!1x(Division)=11=1:.EXAMPLE5:Evaluatelimx!2xx+1.Solution.Westartbycheckingthelimitofthepolynomialfunctioninthedenominator.limx!2(x+1)=limx!2x+limx!21=2+1=3:Sincethelimitofthedenominatorisnotzero,itfollowsthatlimx!2xx+1=limx!2xlimx!2(x+1)=23(Division).EXAMPLE6:Evaluatelimx!1(x 3)(x2 2)x2+1.Solution.First,notethatlimx!1(x2+1)=limx!1×2+limx!11=1+1=26=0:Thus,usingthetheorem,limx!1(x 3)(x2 2)x2+1=limx!1(x 3)(x2 2)limx!1(x2+1)(Division)=limx!1(x 3)limx!1(x2 2)2(Multipication)=limx!1x limx!13limx!1×2 limx!122(Addition)=(1 3)(12 2)2=1:32
.Theorem2.Letfbeapolynomialoftheformf(x)=anxn+an 1xn 1+an 2xn 2+:::+a1x+a0:Ifcisarealnumber,thenlimx!cf(x)=f(c):EXAMPLE7:Evaluatelimx! 1(2×3 4×2+1).Solution.Note rstthatourfunctionf(x)=2×3 4×2+1;isapolynomial.Computingforthevalueoffatx= 1,wegetf( 1)=2( 1)3 4( 1)2+1=2( 1) 4(1)+1= 5:Therefore,fromTheorem2,limx! 1(2×3 4×2+1)=f( 1)= 5:.Notethatwegetthesameanswerwhenweuselimittheorems.Theorem3.Lethbearationalfunctionoftheformh(x)=f(x)g(x)wherefandgarepolynomialfunctions.Ifcisarealnumberandg(c)6=0,thenlimx!ch(x)=limx!cf(x)g(x)=f(c)g(c):EXAMPLE8:Evaluatelimx!11 5×1+3×2+4×4.Solution.Sincethedenominatorisnotzerowhenevaluatedatx=1,wemayapplyTheorem3:limx!11 5×1+3×2+4×4=1 5(1)1+3(1)2+4(1)4= 48= 12:.33
Wewillnowevaluatelimitsofradicalfunctionsusinglimittheorems.EXAMPLE9:Evaluatelimx!1px.Solution.Notethatlimx!1x=1>0.Therefore,bytheRadical/RootRule,limx!1px=qlimx!1x=p1=1:.EXAMPLE10:Evaluatelimx!0px+4:Solution.Notethatlimx!0(x+4)=4>0:Hence,bytheRadical/RootRule,limx!0px+4=qlimx!0(x+4)=p4=2:.EXAMPLE11:Evaluatelimx! 23px2+3x 6:Solution.Sincetheindexoftheradicalsignisodd,wedonothavetoworrythatthelimitoftheradicandisnegative.Therefore,theRadical/RootRuleimpliesthatlimx! 23px2+3x 6=3rlimx! 2(x2+3x 6)=3p4 6 6=3p 8= 2:.EXAMPLE12:Evaluatelimx!2p2x+51 3x.Solution.First,notethatlimx!2(1 3x)= 56=0:Moreover,limx!2(2x+5)=9>0:Thus,usingtheDivisionandRadicalRulesofTheorem1,weobtainlimx!2p2x+51 3x=limx!2p2x+5limx!21 3x=qlimx!2(2x+5) 5=p9 5= 35:.34
INTUITIVENOTIONSOFINFINITELIMITSWeinvestigatethelimitatapointcofarationalfunctionoftheformf(x)g(x)wherefandgarepolynomialfunctionswithf(c)6=0andg(c)=0.NotethatTheorem3doesnotcoverthisbecauseitassumesthatthedenominatorisnonzeroatc.Now,considerthefunctionf(x)=1×2.Notethatthefunctionisnotde nedatx=0butwecancheckthebehaviorofthefunctionasxapproaches0intuitively.We rstconsiderapproaching0fromtheleft. 4 3 2 10123412345678xyf(x)=1x2xf(x) 0:91:2345679 0:54 0:1100 0:0110;000 0:0011;000;000 0:0001100;000;000Observethatasxapproaches0fromtheleft,thevalueofthefunctionincreaseswithoutbound.Whenthishappens,wesaythatthelimitoff(x)asxapproaches0fromtheleftispositivein nity,thatis,limx!0 f(x)=+1:xf(x)0:91:23456790:540:11000:0110;0000:0011;000;0000:0001100;000;00035
Again,asxapproaches0fromtheright,thevalueofthefunctionincreaseswithoutbound,so,limx!0+f(x)=+1.Sincelimx!0 f(x)=+1andlimx!0+f(x)=+1,wemayconcludethatlimx!0f(x)=+1.Now,considerthefunctionf(x)= 1×2.Notethatthefunctionisnotde nedatx=0butwecanstillcheckthebehaviorofthefunctionasxapproaches0intuitively.We rstconsiderapproaching0fromtheleft. 4 3 2 101234 1 2 3 4 5 6 7 8xyf(x)= 1x2xf(x) 0:9 1:2345679 0:5 4 0:1 100 0:01 10;000 0:001 1;000;000 0:0001 100;000;000Thistime,asxapproaches0fromtheleft,thevalueofthefunctiondecreaseswithoutbound.So,wesaythatthelimitoff(x)asxapproaches0fromtheleftisnegativein nity,thatis,limx!0 f(x)= 1:xf(x)0:9 1:23456790:5 40:1 1000:01 10;0000:001 1;000;0000:0001 100;000;000Asxapproaches0fromtheright,thevalueofthefunctionalsodecreaseswithoutbound,thatis,limx!0+f(x)= 1.36
Sincelimx!0 f(x)= 1andlimx!0+f(x)= 1,weareabletoconcludethatlimx!0f(x)= 1.Wenowstatetheintuitivede nitionofin nitelimitsoffunctions:Thelimitoff(x)asxapproachescispositivein nity,denotedby,limx!cf(x)=+1ifthevalueoff(x)increaseswithoutboundwheneverthevaluesofxgetcloserandclosertoc.Thelimitoff(x)asxapproachescisnegativein nity,denotedby,limx!cf(x)= 1ifthevalueoff(x)decreaseswithoutboundwheneverthevaluesofxgetcloserandclosertoc.Letusconsiderf(x)=1x.Thegraphontherightsuggeststhatlimx!0 f(x)= 1whilelimx!0+f(x)=+1:Becausetheone-sidedlimitsarenotthesame,wesaythatlimx!0f(x)DNE: 4 3 2 101234 5 4 3 2 112345xyf(x)=1xRemark1:Rememberthat1isNOTanumber.Itholdsnospeci cvalue.So,limx!cf(x)=+1orlimx!cf(x)= 1describesthebehaviorofthefunctionnearx=c,butitdoesnotexistasarealnumber.Remark2:Wheneverlimx!c+f(x)=1orlimx!c f(x)=1,wenormallyseethedashedverticallinex=c.Thisistoindicatethatthegraphofy=f(x)isasymptotictox=c,meaning,thegraphsofy=f(x)andx=careveryclosetoeachotherasx-valuesapproachc.Inthiscase,wecallx=caverticalasymptoteofthegraphofy=f(x).37
SolvedExamplesEvaluatethefollowinglimits.EXAMPLE1:limx!2(3x 5)Solution.limx!2(3x 5)=limx!2(3x) limx!25=limx!2(3x) 5=3(limx!2x) 5=3(2) 5=1:.EXAMPLE2:limx! 1(2×4 4×3+x 2)Solution.limx! 1(2×4 4×3+x 2)=2( 1)4 4( 1)3+( 1) 2=3:.EXAMPLE3:limx!22xSolution.Notethatlimx!2x=26=0:Therefore,limx!22x=limx!22limx!2x=22=1:.EXAMPLE4:limx!12x 12xSolution.Notethatlimx!122x=2(12)=16=0:Thus,wehavelimx!12x 12x=limx!12(x 1)limx!122x=12 11= 12:.EXAMPLE5:limx! 3(2x+1)(x2+3)x+438
Solution.Wenotethatlimx! 3(x+4)= 3+4=16=0:Hence,weobtainlimx! 3(2x+1)(x2+3)x+4=limx! 3(2x+1)(x2+3)limx! 3x+4=limx! 3(2x+1)limx! 3(x2+3)limx! 3(x+4)=(2( 3)+1)(( 3)2+3)1=( 7)(12)= 84:.EXAMPLE6:limx!0(5×3 3×2+1)Solution.Weremarkthatx=0isinthedomainofthepolynomialfunctionf(x)=5×3 3×2+1.Therefore,limx!0(5×3 3×2+1)=f(0)=5(0)3 3(0)2+1=1:.EXAMPLE7:limx! 2×2+x 2x+1Solution.Notethatx= 2isinthedomainofg(x)=x2+x 2x+1.Thus,limx! 2×2+x 2x+1=g( 2)=( 2)2+( 2) 2 2+1=0 1=0:.EXAMPLE8:limx!12 3x2x3 3x+13Solution.Notethatx=1isinthedomainofthepolynomialfunctionh(x)=2 3x2x3 3x+1andh(1)=2 3(1)2(1)3 3(1)+1= 1 1=1:39
Thus,wehavelimx!12 3x2x3 3x+12=limx!12 3x2x3 3x+12=(h(1))2=1:.EXAMPLE9:limx! 1px+1Solution.Notethatlimx! 1(x+1)=0:Hence,wegetlimx! 1px+1=qlimx! 1(x+1)=p0=0:.EXAMPLE10:limx!3px2+2x+1Solution.Notethatlimx!3(x2+2x+1)=(3)2+2(3)+1=16>0.Thus,limx!3px2+2x+1=rlimx!3×2+2x+1=p16=4:.EXAMPLE11:limx!01x4Solution.Approaching0fromtheleft,xf(x) 0:91:524157903 0:516 0:110;000 0:01100;000;000 0:0011;000;000;000;000 0:000110;000;000;000;000;000Approaching0fromtheright,40
xf(x)0:91:5241579030:5160:110;0000:01100;000;0000:0011;000;000;000;0000:000110;000;000;000;000;000Hence,limx!01×4=+1..EXAMPLE12:limx!01x3Solution.Approaching0fromtheleft,xf(x) 0:9 1:371742112 0:5 8 0:1 1000 0:01 1;000;000 0:001 1;000;000;000 0:0001 1;000;000;000;000Approaching0fromtheright,xf(x)0:91:3717421120:580:11;0000:011;000;0000:0011;000;000;0000:00011;000;000;000;000Hence,limx!01x3DNE..41
SupplementaryProblems1.Evaluatethefollowinglimitsusingthelimittheorems.(a)limx!1(x+1)(b)limx!2(x 2)(c)limx! 1(2x+3)(d)limx!0(1 3x)(e)limx! 2(x2+3x 5)(f)limx! 1(2x 1 5×2)(g)limx!2(x3+x2 1)(h)limx!1(x4 x2 x+1)(i)limx!0xx+1(j)limx!3x 12x 4(k)limx! 3 x2 1x+1(l)limx!21+x2x+3(m)limx!5px 1(n)limx!1(px2 1)2.Determinethefollowinglimitsusingtablesofvalues.(a)limx! 21x+2(b)limx!11×2 1(c)limx!11x 1(d)limx! 11x+1(e)limx!01×2 x(f)limx!01×3 x42
LESSON2:LimitsofSomeTranscendentalFunctionsandSomeIndeterminateFormsLEARNINGOUTCOMES:Attheendofthelesson,thelearnershallbeableto:1.Computethelimitsofexponential,logarithmic,andtrigonometricfunctionsusingtablesofvaluesandgraphsofthefunctions;2.Evaluatethelimitsofexpressionsinvolvingsintt,1 costt,andet 1tusingtablesofvalues;and3.Evaluatethelimitsofexpressionsresultingintheindeterminateform\00″.43
TOPIC2.1:LimitsofExponential,Logarithmic,andTrigono-metricFunctionsReal-worldsituationscanbeexpressedintermsoffunctionalrelationships.Thesefunctionalrelationshipsarecalledmathematicalmodels.Inapplicationsofcalculus,itisquiteimportantthatonecangeneratethesemathematicalmodels.Theysometimesusefunctionsthatyouen-counteredinprecalculus,liketheexponential,logarithmic,andtrigonometricfunctions.Hence,westartthislessonbyrecallingthesefunctionsandtheircorrespondinggraphs.1.Ifb>0,b6=1,theexponentialfunctionwithbasebisde nedbyf(x)=bx;x2R:2.Letb>0;b6=1.Ifby=xthenyiscalledthelogarithmofxtothebaseb,denotedy=logbx.EVALUATINGLIMITSOFEXPONENTIALFUNCTIONSFirst,weconsiderthenaturalexponentialfunctionf(x)=ex,whereeiscalledtheEulernumber,andhasvalue2.718281….EXAMPLE1:Evaluatethelimx!0ex.Solution.Wewillconstructthetableofvaluesforf(x)=ex.Westartbyapproachingthenumber0fromtheleftorthroughthevalueslessthanbutcloseto0.xf(x) 10:36787944117 0:50:60653065971 0:10:90483741803 0:010:99004983374 0:0010:99900049983 0:00010:999900049983 0:000010:99999000005Intuitively,fromthetableabove,limx!0 ex=1.Nowweconsiderapproaching0fromitsrightorthroughvaluesgreaterthanbutcloseto0.44
xf(x)12:718281828460:51:64872127070:11:105170918080:011:010050167080:0011:001000500170:00011:0001000050:000011:00001000005Fromthetable,asthevaluesofxgetcloserandcloserto0,thevaluesoff(x)getcloserandcloserto1.So,limx!0+ex=1.Combiningthetwoone-sidedlimitsallowsustoconcludethatlimx!0ex=1:.Wecanusethegraphoff(x)=extodetermineitslimitasxapproaches0.The gurebelowisthegraphoff(x)=ex.LookingatFigure1.1,asthevaluesofxapproach0,eitherfromtherightortheleft,thevaluesoff(x)willgetcloserandcloserto1.Wealsohavethefollowing:1.limx!1ex=e=2:718:::2.limx!2ex=e2=7:389:::3.limx! 1ex=e 1=0:367::: 3 2 10123xy1y=exEVALUATINGLIMITSOFLOGARITHMICFUNCTIONSNow,considerthenaturallogarithmicfunctionf(x)=lnx.Recallthatlnx=logex.Moreover,itistheinverseofthenaturalexponentialfunctiony=ex.45
EXAMPLE2:Evaluatelimx!1lnx.Solution.Wewillconstructthetableofvaluesforf(x)=lnx.We rstapproachthenumber1fromtheleftorthroughvalueslessthanbutcloseto1.xf(x)0:1 2:302585092990:5 0:693147180560:9 0:105360515650:99 0:010050335850:999 0:001000500330:9999 0:0001000050:99999 0:00001000005Intuitively,limx!1 lnx=0.Nowweconsiderapproaching1fromitsrightorthroughvaluesgreaterthanbutcloseto1.xf(x)20:693147180561:50:40546510811:10:09531017981:010:009950330851:0010:000999500331:00010:0000999951:000010:00000999995Intuitively,limx!1+lnx=0.Asthevaluesofxgetcloserandcloserto1,thevaluesoff(x)getcloserandcloserto0.Insymbols,limx!1lnx=0:.Wenowconsiderthecommonlogarithmicfunctionf(x)=log10x.Recallthatf(x)=log10x=logx.46
EXAMPLE3:Evaluatelimx!1logx.Solution.Wewillconstructthetableofvaluesforf(x)=logx.We rstapproachthenumber1fromtheleftorthroughthevalueslessthanbutcloseto1.xf(x)0:1 10:5 0:301029995660:9 0:045757490560:99 0:00436480540:999 0:000434511770:9999 0:000043431610:99999 0:00000434296Nowweconsiderapproaching1fromitsrightorthroughvaluesgreaterthanbutcloseto1.xf(x)20:301029995661:50:176091259051:10:041392685151:010:004321373781:0010:000434077471:00010:000043427271:000010:00000434292Asthevaluesofxgetcloserandcloserto1,thevaluesoff(x)getcloserandcloserto0.Insymbols,limx!1logx=0:.Considernowthegraphsofboththenaturalandcommonlogarithmicfunctions.Wecanusethefollowinggraphstodeterminetheirlimitsasxapproaches1.47
x1234567f(x)=lnx0xf(x)=logx0The gurehelpsverifyourobservationsthatlimx!1lnx=0andlimx!1logx=0.Also,basedonthe gure,wehave1.limx!elnx=12.limx!10logx=13.limx!3lnx=ln3=1:09:::4.limx!3logx=log3=0:47:::5.limx!0+lnx= 16.limx!0+logx= 1TRIGONOMETRICFUNCTIONSEXAMPLE4:Evaluatelimx!0sinx:Solution.Wewillconstructthetableofvaluesforf(x)=sinx.We rstapproach0fromtheleftorthroughthevalueslessthanbutcloseto0.xf(x) 1 0:8414709848 0:5 0:4794255386 0:1 0:09983341664 0:01 0:00999983333 0:001 0:00099999983 0:0001 0:00009999999 0:00001 0:0000099999948
Nowweconsiderapproaching0fromitsrightorthroughvaluesgreaterthanbutcloseto0.xf(x)10:84147098480:50:47942553860:10:099833416640:010:009999833330:0010:000999999830:00010:000099999990:000010:00000999999Asthevaluesofxgetcloserandcloserto1,thevaluesoff(x)getcloserandcloserto0.Insymbols,limx!0sinx=0:.Wecanalso ndlimx!0sinxbyusingthegraphofthesinefunction.Considerthegraphoff(x)=sinx.1 12 2 322523ThegraphvalidatesourobservationinExample4thatlimx!0sinx=0.Also,usingthegraph,wehavethefollowing:1.limx!2sinx=1.2.limx!sinx=0.3.limx! 2sinx= 1.4.limx! sinx=0.49
SolvedExamplesEXAMPLE1:Evaluatelimx!0ex+1.Solution.Thefollowingtableshowsthevaluesoff(x)=ex+1atvaluesfromtheleftofx=0.xf(x) 11 0:51:648721271 0:12:459603111 0:012:691234472 0:0012:715564905 0:00012:718010014 0:000012:718254646Approachingx=0fromtheright,xf(x)17:3890560990:54:481689070:13:0041660240:012:7456010150:0012:721001470:00012:718553670:000012:718309011Thus,limx!0ex+1=2:71:::.50
xyey=ex+1The guretellsusthat:1.limx!1ex+1=7:38905:::2.limx!2ex+1=20:08553:::3.limx! 1ex+1=14.limx! 2ex+1=0:36787:::.EXAMPLE2:Evaluatelimx!1ex 1.Solution.Approachingx=1fromtheleft,xf(x)00:367879441170:50:606530659710:10:904837418030:010:990049833740:0010:999000499830:00010:999000499830:000010:9999900000551
Takingvaluesclosetox=1fromtheright,xf(x)22:718281828461:51:64872127071:11:105170918081:011:010050167081:0011:001000500171:00011:0001000051:000011:00001000005xy1y=ex 1Fromthe gureabove,wehave1.limx!0ex 1=0:36787:::2.limx!2ex 1=2:71828:::3.limx! 1ex 1=0:13533:::4.limx! 2ex 1=0:04978:::.EXAMPLE3:Evaluatelimx!12x.Solution.Approachingx=1fromtheleft,52
xf(x)010:51:4142135620:91:8660659830:991:9861849910:9991:9986141860:99991:9998613750:999991:999986137Takingvaluesclosetox=1fromtheright,xf(x)241:52:8284271251:12:1435469251:012:01391111:0012:0013867751:00012:0001386341:000012:000013863Thus,limx!12x=2.xy1y=2xUsingthe gureabove,1.limx!02x=12.limx!22x=43.limx! 12x=0:54.limx! 22x=0:2553
.EXAMPLE4:Evaluatelimx!1ln(x+1).Solution.Approachingvaluesfromtheleftofx=1,xf(x)000:50:4054651080:90:6418538860:990:6881346380:9990:6926470550:99990:6930971790:999990:69314218Approachingx=1fromtheright,xf(x)21:0986122891:50:9162907311:10:7419373441:010:6981347221:0010:6936470551:00010:6931971791:000010:69315218Therefore,limx!1ln(x+1)=0:69:::.xf(x)=ln(x+1)0x54
Basedonthe gure,1.limx!0ln(x+1)=02.limx!2ln(x+1)=1:09861:::3.limx!3ln(x+1)=1:38629:::4.limx!4ln(x+1)=1:60943:::.EXAMPLE5:Evaluatelimx!0ln(x2+1).Solution.Approachingvaluesfromtheleftofx=0,xf(x) 10:69314718 0:50:223143551 0:10:00995033 0:010:000099995 0:0010:000000999 0:00010:000000009 0:000010:0000000009Approachingx=0fromtheright,xf(x)10:693147180:50:2231435510:10:009950330:010:0000999950:0010:0000009990:00010:0000000090:000010:0000000009Therefore,limx!0ln(x2+1)=0.55
xf(x)=ln(x2+1)0xThe gureabovegivesus:1.limx!1ln(x2+1)=0:69314:::2.limx!2ln(x2+1)=1:60943:::3.limx! 1ln(x2+1)=0:69314:::4.limx! 2ln(x2+1)=1:60943:::.EXAMPLE6:Evaluatelimx!1log(x+1).Solution.Approachingvaluesfromtheleftofx=1,xf(x)000:50:1760912590:90:2787536010:990:2988530760:9990:3008127940:99990:301008280:999990:301027824Approachingx=1fromtheright,56
xf(x)20:4771212541:50:3979400081:10:3222192941:010:3031960571:0010:3012470881:00010:3010517091:000010:301032167Hence,limx!1log(x+1)=0:301:::.xf(x)=log(x+1)0xFromthe gureabove,1.limx!2log(x+1)=0:47712:::2.limx!3log(x+1)=0:60205:::3.limx!4log(x+1)=0:69897:::4.limx!5log(x+1)=0:77815:::.EXAMPLE7:Evaluatelimx! 1log(x2).Solution.Approachingvaluesfromtheleftofx= 1,57
xf(x) 20:603059991 1:50:352182518 1:10:08278537 1:010:008642747 1:0010:00868154 1:00010:000086854 1:000010:000008685Approachingx= 1fromtheright,xf(x) 0:1 2 0:5 0:602059991 0:9 0:091514981 0:99 0:00872961 0:999 0:000869023 0:9999 0:000086863 0:99999 0:000008685Hence,limx! 1log(x2)=0.xf(x)=log(x2)0xUsingthe gure,wehave1.limx!2log(x2)=0:60205:::2.limx!3log(x2)=0:95424:::3.limx!4log(x2)=1:20411:::4.limx!5log(x2)=1:39794:::.58
EXAMPLE8:Evaluatelimx!0cosx.Solution.Approachingvaluesfromtheleftofx=0,xf(x) 10:999847695 0:50:999961923 0:10:999998476 0:010:999999984 0:0010:999999999 0:00010:9999999999 0:000010:99999999999Approachingx=0fromtheright,xf(x)10:9998476950:50:9999619230:10:9999984760:010:9999999840:0010:9999999990:00010:99999999990:000010:99999999999Hence,limx!0cosx=1.1 12 2 32252359
Basedonthe gure,1.limx!2cosx=02.limx!4cosx=0:70710:::3.limx!3cosx=0:54.limx!cosx= 1.EXAMPLE9:Evaluatelimx!0cosx+2.Solution.Approachingvaluesfromtheleftofx=0,xf(x) 10:841470984 0:50:479425538 0:10:099833416 0:010:009999833 0:0010:000999999 0:00010:000099999 0:000010:000009999Approachingx=0fromtheright,xf(x)1 0:8414709840:5 0:4794255380:1 0:0998334160:01 0:0099998330:001 0:0009999990:0001 0:0000999990:00001 0:000009999Hence,limx!0cosx+2=0.60
1 12 2 322523Usingthegraphabove,1.limx!2cosx+2= 12.limx!4cosx+2= 0:70710:::3.limx!3cosx+2= 0:86602:::4.limx!cosx+2=0:.SupplementaryProblemsEvaluatethefollowinglimits.1.limx! 1ex2.limx!2ex3.limx!1ex 24.limx!3ex 15.limx!02×6.limx!03×7.limx! 13x+18.limx!02x 19.limx!1ln(x2+x 1)10.limx!2ln(x2 2)11.limx!0ln(x2 x+1)12.limx! 1ln(x+2)13.limx!32sinx14.limx!4sinx15.limx!4sinx+216.limx!2sin4 x17.limx!4cosx18.limx!2cosx19.limx!32cosx+220.limx!34cos2 x61
TOPIC2.2:SomeSpecialLimitsWewilldeterminethelimitsofthreespecialfunctions;namely,f(t)=sintt,g(t)=1 costt,andh(t)=et 1t.Thesefunctionswillbevitaltothecomputationofthederivativesofthesine,cosine,andnaturalexponentialfunctionsinChapter2.THREESPECIALFUNCTIONSWestartbyevaluatingthefunctionf(t)=sintt.EXAMPLE1:Evaluatelimt!0sintt:Solution.Wewillconstructthetableofvaluesforf(t)=sintt.We rstapproachthenumber0fromtheleftorthroughvalueslessthanbutcloseto0.tf(t) 10:84147099848 0:50:9588510772 0:10:9983341665 0:010:9999833334 0:0010:9999998333 0:00010:99999999983Nowweconsiderapproaching0fromtherightorthroughvaluesgreaterthanbutcloseto0.tf(t)10:84147098480:50:95885107720:10:99833416650:010:99998333340:0010:99999983330:00010:999999998362
Sincelimt!0 sinttandlimt!0+sinttarebothequalto1,weconcludethatlimt!0sintt=1:.Thegraphoff(t)=sinttbelowcon rmsthatthey-valuesapproach1astapproaches0. 8 6 4 224680:510y=sinttNow,considerthefunctiong(t)=1 costt:EXAMPLE2:Evaluatelimt!01 costt.Solution.Wewillconstructthetableofvaluesforg(t)=1 costt.We rstapproachthenumber1fromtheleftorthroughthevalueslessthanbutcloseto0.tg(t) 1 0:4596976941 0:5 0:2448348762 0:1 0:04995834722 0:01 0:0049999583 0:001 0:0004999999 0:0001 0:000005Nowweconsiderapproaching0fromtherightorthroughvaluesgreaterthanbutcloseto0.63
tg(t)10:45969769410:50:24483487620:10:049958347220:010:00499995830:0010:00049999990:00010:000005Sincelimt!0 1 costt=0andlimt!0+1 costt=0,weconcludethatlimt!01 costt=0:.Belowisthegraphofg(t)=1 costt.Weseethatthey-valuesapproach0asttendsto0. 8 6 4 22468 0:50:510y=1 costtWenowconsiderthespecialfunctionh(t)=et 1t.EXAMPLE3:Evaluatelimt!0et 1t.Solution.Wewillconstructthetableofvaluesforh(t)=et 1t.We rstapproachthenumber0fromtheleftorthroughthevalueslessthanbutcloseto0.64
th(t) 10:6321205588 0:50:7869386806 0:10:9516258196 0:010:9950166251 0:0010:9995001666 0:00010:9999500016Nowweconsiderapproaching0fromtherightorthroughvaluesgreaterthanbutcloseto0.th(t)11:7182818280:51:2974425410:11:0517091810:011:0050167080:0011:0005001670:00011:000050002Sincelimx!0 et 1t=1andlimx!0+et 1t=1,weconcludethatlimx!0et 1t=1:.Thegraphofh(t)=et 1tbelowcon rmsthatlimt!0h(t)=1. 8 6 4 220:511:50y=et 1t65
INDETERMINATEFORM\00″TherearefunctionswhoselimitscannotbedeterminedimmediatelyusingtheLimitTheoremswehavesofar.Inthesecases,thefunctionsmustbemanipulatedsothatthelimit,ifitexists,canbecalculated.Wecallsuchlimitexpressionsindeterminateforms.Inthislesson,wewillde neaparticularindeterminateform,\00″,anddiscusshowtoevaluatealimitwhichwillinitiallyresultinthisform.INDETERMINATEFORMOFTYPE\00″Iflimx!cf(x)=0andlimx!cg(x)=0,thenlimx!cf(x)g(x)iscalledanindeterminateformoftype\00″.Remark1:Alimitthatisindeterminateoftype\00″mayexist.To ndtheactualvalue,oneshould ndanexpressionequivalenttotheoriginal.Thisiscommonlydonebyfactoringorbyrationalizing.Hopefully,theexpressionthatwillemergeafterfactoringorrationalizingwillhaveacomputablelimit.EXAMPLE4:Evaluatelimx! 1×2+2x+1x+1.Solution.Thelimitofboththenumeratorandthedenominatorasxapproaches 1is0.Thus,thislimitascurrentlywrittenisanindeterminateformoftype00.However,observethat(x+1)isafactorcommontothenumeratorandthedenominator,andx2+2x+1x+1=(x+1)2x+1=x+1;whenx6= 1:Therefore,limx! 1×2+2x+1x+1=limx! 1(x+1)=0:.EXAMPLE5:Evaluatelimx!1×2 1px 1.66
Solution.Sincelimx!1×2 1=0andlimx!1px 1=0,thenlimx!1×2 1px 1isanindeterminateformoftype\00″.To ndthelimit,observethatifx6=1,thenx2 1px 1px+1px+1=(x 1)(x+1)(px+1)x 1=(x+1)(px+1):So,wehavelimx!1×2 1px+1=limx!1(x+1)(px+1)=4:.SolvedExamplesEXAMPLE1:Evaluatelimt!1sin(1 t)1 t.Solution.Approachingvaluesfromtheleftof1,tf(t)00:8414709840:50:9588510770:90:9983341660:990:9999833330:9990:9999998330:99990:9999999998Approachingx=1fromtheright,tf(t)20:8414709841:50:9588510771:10:9983341661:010:9999833331:0010:9999998331:00010:999999998Thus,limt!1sin(1 t)1 t=1:.67
EXAMPLE2:Evaluatelimt!1sin(1 t2)1 t2.Solution.Approachingvaluesfromtheleftof1,tf(t)00:8414709840:50:908851680:90:9939941840:990:9999339990:9990:0000003340:99990:999999993Approachingx=1fromtheright,tf(t)20:0470400021:50:7591876951:10:9926661891:010:9999326661:0010:9999993321:00010:999999993Thus,limt!1sin(1 t2)1 t2=1:.EXAMPLE3:Evaluatelimt!11 cos(t+1)t+1.Solution.Approachingvaluesfromtheleftof1,tf(t)0 0:4596976940:5 0:2448348760:9 0:0499583470:99 0:0049999580:999 0:0004999990:9999 0:00004999968
Approachingx=1fromtheright,tf(t)20:4596976941:50:2448348761:10:0499583471:010:0049999581:0010:0004999991:00010:000049999Thus,limt!11 cos(t+1)t+1=0.EXAMPLE4:Evaluatelimt! 11 cos(t2 1)t2 1.Solution.Approachingvaluesfromtheleftofx= 1,tf(t) 20:663330832 1:50:54774211 1:10:104614691 1:010:010049661 1:0010:001000499 1:00010:000100004Approachingx= 1fromtheright,tf(t)0 0:45697694 0:5 0:357748174 0:9 0:094714552 0:99 0:009949671 0:999 0:000999499 0:9999 0:000099994Thus,limt! 11 cos(t2 1)t2 1=0:69
.EXAMPLE5:Evaluatelimt!1et 1 1t 1.Solution.Approachingvaluesfromtheleftof1,tf(t)00:6321205580:50:786938680:90:9516258190:990:9950166250:9990:9995001660:99990:999950002Approachingx=1fromtheright,tf(t)21:7182818281:51:2974425411:11:0517091811:011:0050167081:0011:0005001671:00011:000050001Thus,limt!1et 1 1t 1=1:.EXAMPLE6:Evaluatelimt!1et2 1 1t2 1.Solution.Approachingvaluesfromtheleftof1,tf(t)00:6321205580:50:7035112630:90:91074140:990:9901156740:9990:9990011650:99990:999000170
Approachingx=1fromtheright,tf(t)26:3618456411:51:9922743661:11:1127526661:011:0101176751:0011:1:0010011681:00011:000100011Thus,limt!1et2 1 1t2 1=1:.EXAMPLE7:Evaluatelimx!0xx2 x.Solution.limx!0xx2 x00=limx!0xx(x 1)=limx!01x 1=1 1= 1:.EXAMPLE8:Evaluatelimx!1px2+x 1 1x 1.Solution.limx!1px2+x 1 1x 100=limx!1px2+x 1 1x 1px2+x 1+1px2+x 1+1=limx!1×2+x 1 1(x 1)(px2+x 1+1)=limx!1×2+x 2(x 1)(px2+x 1+1)=limx!1(x 1)(x+2)(x 1)(px2+x 1+1)=limx!1x+2px2+x 1+1=32:.71
SupplementaryProblemsEvaluatethefollowinglimits.1.limt!3sin(t 3)t 32.limt! 2sin(t+2)t+23.limt!0sin(t2+t)t2+t4.limt!1sin(t3+2t 3)t3+2t 35.limt! 1sin(t3 t)t3 t6.limt!21 cos(t 2)t 27.limt! 21 cos(t+2)t+28.limt!01 cos(t2+t)t2+t9.limt!31 cos(t2 9)t2 910.limt!21 cos(t2 4)t2 411.limt!3et 3 1t 312.limt!1et 1 1t 113.limt!0et2 1t214.limt! 1et2 1 1t2 115.limx!0xx16.limx! 1×2 1x+117.limx!1×2 1x 118.limx!0x2 xx19.limx!3px 3×2 920.limx!1px 1×2 172
LESSON3:ContinuityofFunctionsLEARNINGOUTCOMES:Attheendofthelesson,thelearnershallbeableto:1.Illustratecontinuityofafunctionatapoint;2.Determinewhetherafunctioniscontinuousatapointornot;3.Illustratecontinuityofafunctiononaninterval;and4.Determinewhetherafunctioniscontinuousonanintervalornot.LESSONOUTLINE:1.Continuityatapoint2.Determiningwhetherafunctioniscontinuousornotatapoint3.Continuityonaninterval4.Determiningwhetherafunctioniscontinuousornotonaninterval73
TOPIC3.1:ContinuityataPointAswehaveobservedinourdiscussionoflimitsinTopics1.1and1.2,therearefunctionswhoselimitsarenotequaltothefunctionvalueatx=c,meaning,limx!cf(x)6=f(c).Thisleadsustothestudyofcontinuityoffunctions.Inthissection,wewillbefocusingonthecontinuityofafunctionataspeci cpoint.LIMITSANDCONTINUITYATAPOINTWhatdoes\continuityatapoint”mean?Intuitively,thismeansthatindrawingthegraphofafunction,thepointinquestionwillbetraversed.Westartbygraphicallyillustratingwhatitmeanstobecontinuityatapoint.EXAMPLE1:Considerthegraphbelow.123 112340f(x)=3x 1Isthefunctioncontinuousatx=1?Solution.Tocheckifthefunctioniscontinuousatx=1,usethegivengraph.Notethatoneisabletotracethegraphfromtheleftsideofthenumberx=1goingtotherightsideofx=1,withoutliftingone’spen.Thisisthecasehere.Hence,wecansaythatthefunctioniscontinuousatx=1..74
EXAMPLE2:Considerthegraphofthefunctiong(x)below.123 112340g(x)=3×2 4x+1x 1Isthefunctioncontinuousatx=1?Solution.Wefollowtheprocessinthepreviousexample.Tracingthegraphfromtheleftofx=1goingtorightofx=1,one ndsthats/hemustlifther/hispenbrie
yuponreachingx=1,creatingaholeinthegraph.Thus,thefunctionisdiscontinuousatx=1..EXAMPLE3:Considerthegraphofthefunctionh(x)=1x. 3 2 1123 3 2 11230h(x)=1xIsthefunctioncontinuousatx=0?75
Solution.Ifwetracethegraphfromtheleftofx=0goingtorightofx=0,wehavetoliftourpensinceattheleftofx=0,thefunctionvalueswillgodownwardinde nitely,whileattherightofx=0,thefunctionvalueswillgotoupwardinde nitely.Inotherwords,limx!0 1x= 1andlimx!0+1x=1Thus,thefunctionisdiscontinuousatx=0..EXAMPLE4:Consideragainthegraphofthefunctionh(x)=1x.Isthefunctioncontinuousatx=2?Solution.Ifwetracethegraphofthefunctionh(x)=1xfromtheleftofx=2totherightofx=2,youwillnotliftyourpen.Therefore,thefunctionhiscontinuousatx=2..Supposewearenotgiventhegraphofafunctionbutjustthefunctionitself.Howdowedetermineifthefunctioniscontinuousatagivennumber?Inthiscase,wehavetocheckthreeconditions.THREECONDITIONSOFCONTINUITYAfunctionf(x)issaidtobecontinuousatx=cifthefollowingthreeconditionsaresatis ed:(i)f(c)exists;(ii)limx!cf(x)exists;and(iii)f(c)=limx!cf(x).Ifatleastoneoftheseconditionsisnotmet,fissaidtobediscontinuousatx=c.EXAMPLE5:Determineiff(x)=x3+x2 2iscontinuousornotatx=1.Solution.Wehavetocheckthethreeconditionsforcontinuityofafunction.1.Ifx=1,thenf(1)=0.2.limx!1f(x)=limx!1(x3+x2 2)=13+12 2=0.3.f(1)=0=limx!1f(x).Therefore,fiscontinuousatx=1..76
EXAMPLE6:Determineiff(x)=x2 x 2x 2iscontinuousornotatx=0.Solution.Wehavetocheckthethreeconditionsforcontinuityofafunction.1.Ifx=0,thenf(0)=1.2.limx!0f(x)=limx!0x2 x 2x 2=limx!0(x 2)(x+1)x 2=limx!0(x+1)=1.3.f(0)=1=limx!0f(x).Therefore,fiscontinuousatx=0..EXAMPLE7:Determineiff(x)=x2 x 2x 2iscontinuousornotatx=2.Solution.Notethatfisnotde nedatx=2since2isnotinthedomainoff.Hence,the rstconditioninthede nitionofacontinuousfunctionisnotsatis ed.Therefore,fisdiscontinuousatx=2..EXAMPLE8:Determineiff(x)=8<:x+1ifx<4;(x 4)2+3ifx4iscontinuousornotatx=4.(ThisexamplewasgiveninTopic1.1.)Solution.Notethatfisde nedatx=4sincef(4)=3.However,limx!4 f(x)=5whilelimx!4+f(x)=3.Thereforelimx!4 f(x)DNE,andfisdiscontinuousatx=4..77
SolvedExamplesEXAMPLE1:Considerthegraphbelow.12 1120f(x)=1 2xSolution.Thefunctionf(x)iscontinuousatx=0becausewecantracethegraphfromtheleftof1andfromtherightof1withoutliftingthepen..EXAMPLE2:Considerthegraphofafunctionbelow.123 2 1123450g(x)=3x2 12x+12x 2Solution.Ifwetracethegraphofg(x),weneedtoliftthepenattheholeat(2;4).Thus,g(x)isnotcontinuousatx=2..EXAMPLE3:Considerthegraphofthefunctionh(x)= 1x+2.78
6 5 4 3 2 1123 3 2 11230h(x)= 1x+2Solution.Thefunctionvaluesfromtheleftof 2goto1whilefromtheleft,theyapproach 1.Therefore,h(x)isdiscontinuousatx= 2..EXAMPLE4:Consideragainthegraphofh(x)= 1x+2.Solution.Thefunctionh(x)iscontinuousatx=0sinceweneednotliftthepenaswetracethegraphfromtheleftandrightof0..EXAMPLE5:Determineiff(x)=x4 x2+1iscontinuousatx= 1.Solution.Notethat1.f( 1)=12.limx! 1f(x)=limx! 1(x4 x2+1)=( 1)4 ( 1)2+1=13.f( 1)=1=limx! 1f(x)Thus,f(x)iscontinuousatx= 1..EXAMPLE6:Determineifg(x)=x2 9x+3iscontinuousatx= 3.Solution.Notethatg( 3)isunde nedandhence,g(x)isdiscontinuousatx= 3..EXAMPLE7:Determineifh(x)=px 1x 1iscontinuousatx=1.79
Solution.Notethath(1)isunde nedandtherefore,hisdiscontinuousatx=1..EXAMPLE8:Determineifj(x)=px 1x+1iscontinuousatx=1.Solution.Notethat1.j(1)=0;2.limx!1j(x)=limx!1px 1x+1=p1 11+1=01=0;3.j(1)=0=limx!1j(x).Hence,jiscontinuousatx=1..EXAMPLE9:Determineiff(x)=8<:2x 3;x4;x2 x+1;x>4;iscontinuousatx=4.Solution.Notethatf(4)=5butlimx!4 f(x)=56=13=limx!4+f(x),thatis,limx!4f(x)DNE.Thus,fisdiscontinuousatx=4..EXAMPLE10:Thegraphofafunctionkisgivenbelow.xy( 2;3)(0;5)(3;2)(3;6)(3;3)80
Fromthegraph,wehave1.k( 2)=3=limx! 2k(x);2.k(0)=5=limx!0k(x);3.k(3)=6butlimx!3k(x)DNE.Thus,k(x)iscontinuousatx= 2andatx=0,butisdiscontinuousatx=3.SupplementaryProblems1.Determineifthefollowingfunctionsarecontinuousatx=c.(a)f(x)=x+2;c= 1(b)f(x)=x 2;c=0(c)f(x)=x2+2;c=1(d)f(x)=x2 1;c= 1(e)f(x)=x3 x;c=0(f)f(x)=x3 3x;c=1(g)f(x)=x2 4;c=2(h)f(x)=x4 1;c=1(i)f(x)=x3 xx;c=0(j)f(x)=x3 3xx;c=1(k)f(x)=x2 4x 2;c=2(l)f(x)=x4 1x 1;c=1(m)f(x)=px4 1x 1;c=1(n)f(x)=px x2x2 x;c=0(o)f(x)=px3 1x 1;c=181
2.Thegraphoff(x)isgivenbyxy( 4;5)( 1; 2)( 1;4)(1;4)(3; 4)(5; 2)Determineisfiscontinuousatx=c.(a)x= 4(b)x= 1(c)x=1(d)x=3(e)x=582
TOPIC3.2:ContinuityonanIntervalAfunctioncanbecontinuousonaninterval.Thissimplymeansthatitiscontinuousateverypointontheinterval.Equivalently,ifweareabletodrawtheentiregraphofthefunctiononanintervalwithoutliftingourtracingpen,orwithoutbeinginterruptedbyaholeinthemiddleofthegraph,thenwecanconcludethatthethefunctioniscontinuousonthatinterval.Webeginourdiscussionwithtwoconceptswhichareimportantindeterminingwhetherafunctioniscontinuousattheendpointsofclosedintervals.ONE-SIDEDCONTINUITY1.Afunctionfissaidtobecontinuousfromtheleftatx=ciff(c)=limx!c f(x):2.Afunctionfissaidtobecontinuousfromtherightatx=ciff(c)=limx!c+f(x):Hereareknownfactsoncontinuitiesoffunctionsonintervals:CONTINUITYOFPOLYNOMIAL,ABSOLUTEVALUE,RATIONALANDSQUAREROOTFUNCTIONS1.Polynomialfunctionsarecontinuouseverywhere.2.Theabsolutevaluefunctionf(x)=jxjiscontinuouseverywhere.3.Rationalfunctionsarecontinuousontheirrespectivedomains.4.Thesquarerootfunctionf(x)=pxiscontinuouson[0;1).LIMITSANDCONTINUITYONANINTERVALWe rstlookatgraphsoffunctionstoillustratecontinuityonaninterval.83
EXAMPLE1:Considerthegraphofthefunctionfgivenbelow. 2 112 2 1120Usingthegivengraph,determineifthefunctionfiscontinuousonthefollowingintervals:1.( 1;1)2.( 1;0)3.(0;+1)Solution.Rememberthatwhenwesay\tracefromtherightsideofx=c”,wearetracingnotfromx=conthex-axis,butfromthepoint(c;f(c))alongthegraph.1.Wecantracethegraphfromtherightsideofx= 1totheleftsideofx=1withoutliftingthepenweareusing.Hence,wecansaythatthefunctionfiscontinuousontheinterval( 1;1).2.Ifwetracethegraphfromanynegativelylargenumberuptotheleftsideof0,wewillnotliftourpenandso,fiscontinuouson( 1;0).3.Fortheinterval(0;+1),wetracethegraphfromtherightsideof0toanylargenumber,and ndthatwewillnotliftourpen.Thus,thefunctionfiscontinuouson(0;+1)..84
EXAMPLE2:Considerthegraphofthefunctionhbelow. 2 112 2 1120Determineusingthegivengraphifthefunctionfiscontinuousonthefollowingintervals:a.( 1;1)b.[0:5;2]Solution.Becausewearealreadygiventhegraphofh,wecharacterizethecontinuityofhbythepossibilityoftracingthegraphwithoutliftingthepen.1.Ifwetracethegraphofthefunctionhfromtherightsideofx= 1totheleftsideofx=1,wewillbeinterruptedbyaholewhenwereachx=0.Weareforcedtoliftourpenjustbeforewereachx=0toindicatethathisnotde nedatx=0andcontinuetracingagainstartingfromtherightofx=0.Therefore,wearenotabletotracethegraphofhon( 1;1)withoutliftingourpen.Thus,thefunctionhisnotcontinuouson( 1;1).2.Fortheinterval[0:5;2],ifwetracethegraphfromx=0:5tox=2,wedonothavetoliftthepenatall.Thus,thefunctionhiscontinuouson[0:5;2]..85
Now,ifafunctionisgivenwithoutitscorrespondinggraph,wemust ndothermeanstodetermineifthefunctioniscontinuousornotonaninterval.Herearede nitionsthatwillhelpus:Afunctionfissaidtobecontinuous1.everywhereiffiscontinuousateveryrealnumber.Inthiscase,wealsosayfiscontinuousonR.2.on(a;b)iffiscontinuousateverypointxin(a;b).3.on[a;b)iffiscontinuouson(a;b)andfromtherightata.4.on(a;b]iffiscontinuouson(a;b)andfromtheleftatb.5.on[a;b]iffiscontinuouson(a;b]andon[a;b).6.on(a;1)iffiscontinuousatallx>a.7.on[a;1)iffiscontinuouson(a;1)andfromtherightata.8.on( 1;b)iffiscontinuousatallx
1.( 2;2).2.( 1;0).3.(0;+1).Solution.Notethattracingthegraphofffromtheleftofx= 2totherightofx=2,weneednotliftourpen.Thus,fiscontinuouson( 2;2).Moreover,fiscontinuouson( 1;0)andon(0;+1)..EXAMPLE2:Thegraphofafunctionhisgivenbelow.xy( 2;3)(0;5)(3;2)(3;6)(3;3)Solution.Fromthegivengraph,1.hiscontinuouson( 1;2].2.hiscontinuouson( 1;0).3.hiscontinuouson( 3;0).4.hiscontinuouson[ 2;2].5.hisdiscontinuouson(2;+1).6.hisdiscontinuouson[2;+1).7.hisdiscontinuouson(0;4).8.hisdiscontinuouson[ 4;4]..EXAMPLE3:Determinethelargestintervalinwhichf(x)=x4 x3+1iscontinuous.Solution.Sincefisapolynomialfunction,thenitiscontinuouseverywhere.Hence,fiscontinuousonR..88
EXAMPLE4:Determinethelargestintervalwhereg(x)=x2 4x 2iscontinuous.Solution.Notethatg(x)isunde nedatx=2.Thus,ifx6=2,g(x)=x+2,whichisapolynomialfunction.Therefore,g(x)=x2 4x 2iscontinuousforallx2Rnf2g..EXAMPLE5:Determinethelargestintervalwhereh(x)=px2+1iscontinuous.Solution.Sincex2+10,foranyx2R,hisde nedonR.Moreover,foranyc2R,h(c)=pc2+1=limx!ch(x):Hence,hiscontinuousonR.EXAMPLE6:Determinethelargestintervalwherej(x)=px2 1iscontinuous.Solution.Notethatjisde nedonlywheneverx2 10,thatis,x 1orx1.Also,forallc2( 1; 1)[(1;+1),j(c)=pc2 1=limx!cj(x):Therefore,jiscontinuouson( 1; 1)[(1;+1)..EXAMPLE7:Considerthefunctiongivenbyf(x)=8<:x 1ifx<4;px+1ifx4:Solution.Wehavethefollowing:1.Forx<4,f(x)=x 1isapolynomialfunctionandtherefore,continuouseverywhere.Thus,fiscontinuouson( 1;4).2.Forx>4,f(x)=px+1isalwaysde ned.Moreover,f(a)=pa+1=limx!af(x),foranya>4.Thus,f(x)iscontinuousforallx>4..89
EXAMPLE8:Considerthefunctiongivenbyg(x)=8>>><>>>:1x+1ifx<0;x2 x+1if0x2;5ifx>2:Solution.Wehavethefollowing:1.Notethatforx<0,g(x)=1x 1isunde nedatx= 1.Thus,gisdiscontinuouson( 1;0).2.Forx2(0;2),g(x)isapolynomialfunction,whichiscontinuousonR.Therefore,giscontinuouson(0;2).3.Sinceg(x)=5isaconstantfunctionforanyx>2,g(x)iscontinuousforallx2(2;+1)..SupplementaryProblems1.DetermineifthefollowingfunctionsarecontinuousontheintervalI.(a)f(x)=x 1;I=[ 2;2](b)f(x)=x2 5;I=[ 5;5](c)f(x)=x3 x+1;I=(2;+1)(d)f(x)=x3 xx;I=( 1;1)(e)f(x)=x3 3xx;I=( 2;2)(f)f(x)=x2 4x 2;I=(2;3)(g)f(x)=x4 1x 1;I=(1;5)(h)f(x)=px4 1x 1;I=( 2;2)(i)f(x)=px x2x2 x;I=(0;1)(j)f(x)=px3 1x 1;I=(2;+1)(k)f(x)=px2 1x 3;I=[ 4;4]2.Considerf(x)=8>>>>>><>>>>>>:x2 2x+1ifx< 1;xx+1if 1x0;px 1if0
Solution.Fromthegraphabove,limx! 1 g(x)=1andlimx! 1+g(x)=0.Hence,limx! 1g(x)DNE,i.e.,g(x)hasajumpessentialdiscontinuityatx= 1..SupplementaryProblemsDetermineifthefollowingfunctionsarecontinuousatthepointx=c.Ifnot,classifythediscontinuityastoremovable,jumpessentialorini niteessential.1.f(x)=1x 3;x=32.f(x)=1x+5;x= 53.f(x)=x2 1x 1;x=14.f(x)=x2 1x+1;x= 15.f(x)=x 2x 1;x=16.f(x)=x 1x+1;x= 17.f(x)=x3 1x 1;x=18.f(x)=x3+1x+1;x= 19.f(x)=x2+2x 1;x=110.f(x)=x2 3x+1;x= 111.f(x)=x3 8x 2;x=212.f(x)=x3+8x+2;x= 213.f(x)=2×3 16x 2;x=214.f(x)=x3+27x+3;x= 315.f(x)=8<:3x2 1;x0;x 2;x>0;16.f(x)=8<:x 1;x1;x2+1;x>1;17.f(x)=8><>:jx 1j;x12; (x 1);x>12;18.f(x)=8<:x3 8;x< 2;x;x 2;19.f(x)=8<:3x2 1;x<0;x 2;x0;20.f(x)=8<: x2 3 ;x<0;x 3;x0;104
TOPIC4.2:TheIntermediateValueandtheExtremeValueTheoremsTheIntermediateValueTheoremThe rsttheoremwewillillustratesaysthatafunctionf(x)whichisfoundtobecontinuousoveraclosedinterval[a;b]willtakeanyvaluebetweenf(a)andf(b).Theorem4(IntermediateValueTheorem(IVT)).Ifafunctionf(x)iscontinuousoveraclosedinterval[a;b],thenforeveryvaluembetweenf(a)andf(b),thereisavaluec2[a;b]suchthatf(c)=m.xyPy=f(x)abcf(a)f(b)mLookatthegraphasweconsidervaluesofmbetweenf(a)andf(b).Imaginemovingthedottedlineformupanddownbetweenthedottedlinesforf(a)andf(b).Correspondingly,thedotPwillmovealongthethickenedcurvebetweenthetwopoints,(a;f(a))and(b;f(b)).Wemakethefollowingobservations:Asthedarkdotmoves,sowilltheverticaldottedlineoverx=cmove.Inparticular,thesaidlinemovesbetweentheverticaldottedlinesoverx=aandx=b.Moreinparticular,foranyvaluethatweassignminbetweenf(a)andf(b),theconse-quentpositionofthedarkdotassignsacorrespondingvalueofcbetweenaandb.ThisillustrateswhattheIVTsays.105
EXAMPLE1:Considerthefunctionf(x)=2x 5.Sinceitisalinearfunction,weknowitiscontinuouseverywhere.Therefore,wecanbesurethatitwillbecontinuousoveranyclosedintervalofourchoice.Taketheinterval[1;5].TheIVTsaysthatforanymintermediateto,orinbetween,f(1)andf(5),wecan ndavalueintermediateto,orinbetween,1and5.Startwiththefactthatf(1)= 3andf(5)=5.Then,chooseanm2[ 3;5],toexhibitacorrespondingc2[1;5]suchthatf(c)=m.xy154114 350312y=2x 5Choosem=12.ByIVT,thereisac2[1;5]suchthatf(c)=12.Therefore,12=f(c)=2c 5=)2c=112=)c=114:Indeed,1142(1;5).Wecantryanotherm-valuein( 3;5).Choosem=3.ByIVT,thereisac2[1;5]suchthatf(c)=3.Therefore,3=f(c)=2c 5=)2c=8=)c=4:Again,theanswer,4,isin[1;5].TheclaimofIVTisclearlyseeninthegraphofy=2x 5.106
EXAMPLE2:Considerthesimplestquadraticfunctionf(x)=x2.xyy=x2m=9 4 3216Beingapolynomialfunction,itiscontinuouseverywhere.Thus,itisalsocontinuousoveranyclosedintervalwemayspecify.Wechoosetheinterval[ 4;2].Foranyminbetweenf( 4)=16andf(2)=4,thereisavaluecinsidetheinterval[ 4;2]suchthatf(c)=m.Supposewechoosem=92[4;16].ByIVT,thereexistsanumberc2[ 4;2]suchthatf(c)=9.Hence,9=f(c)=c2=)c=3:However,weonlychoosec= 3becausetheothersolutionc=3isnotinthespeci edinterval[ 4;2].Note:Inthepreviousexample,iftheintervalthatwasspeci edwas[0,4],thenthe nalanswerwouldinsteadbec=+3.Remark1:Thevalueofc2[a;b]intheconclusionoftheIntermediateValueTheoremisnotnecessarilyunique.107
EXAMPLE3:Considerthepolyno-mialfunctionf(x)=x3 4x2+x+7overtheinterval[ 1:5;4]Notethatf( 1:5)= 6:875andf(4)=11:Wechoosem=1.ByIVT,thereex-istsc2[ 1:5;4]suchthatf(c)=1:Thus,f(c)=c3 4c2+c+7=1=)c3 4c2+c+6=0=)(c+1)(c 2)(c 3)=0=)c= 1orc=2orc=3:Weseethattherearethreevaluesofc2[ 1:5;4]whichsatisfytheconclu-sionoftheIntermediateValueTheo-rem. 3 2 112345 6:87517110y=1y=x3 4x2+x+7TheExtremeValueTheoremThesecondtheoremwewillillustratesaysthatafunctionf(x)whichisfoundtobecontinuousoveraclosedinterval[a;b]isguaranteedtohaveextremevaluesinthatinterval.Anextremevalueoff,orextremum,iseitheraminimumoramaximumvalueofthefunction.Aminimumvalueoffoccursatsomex=ciff(c)f(x)forallx6=cintheinterval.Amaximumvalueoffoccursatsomex=ciff(c)f(x)forallx6=cintheinterval.Theorem5(ExtremeValueTheorem(EVT)).Ifafunctionf(x)iscontinuousoveraclosedinterval[a;b],thenf(x)isguaranteedtoreachamaximumandaminimumon[a;b].Note:Inthissection,welimitourillustrationofextrematographicalexamples.Moredetailedandcomputationalexampleswillfollowoncederivativeshavebeendiscussed.108
EXAMPLE4:Considerthefunctionf(x)= 2x4+4x2over[ 1;1].Fromthegraph,itisclearthatontheinterval,fhasThemaximumvalueof2,occurringatx=1;andTheminimumvalueof0,occurringatx=0. 112 10123xyy= 2x4+4x2Remark2:SimilartotheIVT,thevaluec2[a;b]atwhichaminimumoramaximumoccursisnotnecessarilyunique.Herearemoreexamplesexhibitingtheguaranteedexistenceofextremaoffunctionscontinuousoveraclosedinterval.EXAMPLE5:ConsiderExample1.Observethatf(x)=2x 5on[1;5]exhibitstheextremaattheendpoints:Theminimumoccursatx=1,givingtheminimumvaluef(1)= 3;andThemaximumoccursatx=5,givingthemaximumvaluef(5)=5.EXAMPLE6:ConsiderExample2.f(x)=x2on[ 4;2]exhibitsanextremumatoneendpointandanotheratapointinsidetheinterval(or,aninteriorpoint):Theminimumoccursatx=0,givingtheminimumvaluef(0)=0;andThemaximumoccursatx= 4,givingthemaximumvaluef( 4)=16.109
EXAMPLE7:Considerf(x)=2x4 8x2.Ontheinterval[ 2; p2],theextremaoccurattheendpoints.{Endpointx= 2yieldsthemaxi-mumvaluef( 2)=0:{Endpointx= p2yieldsthemini-mumvaluef( p2)= 8.Ontheinterval[ 2; 1],oneextremumoccursatanendpoint,anotherataninte-riorpoint.{Endpointx= 2yieldsthemaxi-mumvaluef( 2)=0:{Interiorpointx= p2yieldstheminimumvaluef( p2)= 8.xyy=2x4 8x2 22( p2; 8)(p2;8)Ontheinterval[ 1:5;1],theextremaoccuratinteriorpoints.{Interiorpointx= p2yieldstheminimumvaluef( p2)= 8.{Interiorpointx=0yieldsthemaximumvaluef(0)=0.Ontheinterval[ 2;2],theextremaoccuratboththeendpointsandseveralinteriorpoints.{Endpointsx=2andinteriorpointx=0yieldthemaximumvalue0.{Interiorpointsx=p2yieldtheminimumvalue 8.Remark3:KeepinmindthattheIVTandtheEVTareexistencetheorems(\thereisavaluec..."),andtheirstatementsdonotgiveamethodfor ndingthevaluesstatedintheirrespectiveconclusions.Itmaybedicultorimpossibleto ndthesevaluesalgebraicallyespeciallyifthefunctioniscomplicated.110
SolvedExamplesEXAMPLE1:Considerf(x)=3x 1on[ 1;1].Solution.Sincefisapolynomialfunction,itiscontinuouseverywhere.Notethatf( 1)= 4andf(1)=2.IVTon[ 1;1]=)foranym2( 4;2),thereisac2[ 1;1]suchthatf(c)=m.Takem= 12( 4;2).Notec=02( 1;1)andf(0)= 1=m.Choosem=12( 4;2).Notec=232( 1;1)andf23=1=m.Takem=02( 4;2).Notec=132( 1;1)andf13=0=m..EXAMPLE2:Considerg(x)=2x 3on[0;3].Solution.g,polynomial=)g,continuouseverywhere.Noteg(0)= 3andg(3)=3Thus,IVTon[0;3]=)forallm2( 3;3),thereisac2(0;3),withg(c)=m.Takem=02( 3;3).Thenc=322(0;3)andg32=0=m.Takem= 12( 3;3).Thenc=12(0;3)andg(1)= 1=m.Takem=12( 3;3).Thenc=22(0;3)andg(2)=1=m..EXAMPLE3:Considerh(x)=x2 1on[ 1;1].Solution.h,polynomial=)h,continuouseverywhere.Noteh( 1)=0=h(1).Thus,IVTcannotbeappliedtohon[ 1;1]..EXAMPLE4:Consideragainh(x)=x2 1,buton[ 1;2].Solution.h,polynomial=)h,continuouseverywhere.Noteh( 1)=0andh(2)=3Thus,IVTon[ 1;2]=)forallm2(0;3),thereisac2( 1;2),withh(c)=m.111
Takem=12(0;3).Thenc=02( 1;2)andh(0)= 1=m.Takem=02(0;3).Thenc=12( 1;2)andh(1)=0=m.Takem=22(0;3).Thenc=p32( 1;2)andh(p3)=2=m..EXAMPLE5:Considerj(x)=x4 2on[ 2;1].Solution.j,polynomial=)j,continuouseverywhere.Notej( 2)=14andj(1)= 1Thus,IVTon[ 2;1]=)forallm2( 1;14),thereisac2( 2;1),withj(c)=m.Takem=12( 1;14).Thenc=4p32( 2;1)andj(4p3)=1=m.Takem=02( 1;14).Thenc=4p22( 2;1)andj(c)=0=m..EXAMPLE6:Considerf(x)=3x4 2x2.Itsgraphisthefollowing.xyy=3x4 2x2 22 p33; 13! p33; 13!112
Solution..Ontheinterval[ 2; p3]{fattainsitsmaximumatx= 2(f( 2)=40){fattainsitsminimumatx= p33 f p33!= 13!Ontheinterval[ p3;0]{fattainsitsminimumatx=p33 f p33!= 13!{fattainsitsmaximumatx=0(f(0)=0)Ontheinterval[0;p3]{fattainsitsminimumatx=p33 f p33!= 13!{fattainsitsmaximumatx=0(f(0)=0)Ontheinterval[p3;2]{fattainsitsminimumatx=p33 f p33!= 13!{fattainsitsmaximumatx=2(f(40)=0).SupplementaryProblems1.Findatleastonevalueofc2I=[a;b]suchthattheIVTholdsforthefunctionfform2(f(a);f(b)).(a)f(x)=x;I=[1;2](b)f(x)=x 2;I=[ 1;1](c)f(x)= x2;I=[0;1](d)f(x)= x2 1;I=[ 1;2](e)f(x)=x2 x;I=[0;2](f)f(x)= x2+x;I=[ 1;0](g)f(x)=x3 2;I=[ 1;0](h)f(x)=x4+x;I=[ 1;1]113
2.DetermineiftheIVTisapplicabletothefollowingfunctionsontheintervalI.(a)f(x)=x;I=[1;2](b)f(x)=x2+1;I=[ 1;1](c)f(x)=3 x3;I=[0;1](d)f(x)=x+1x 1;I=[0;2](e)f(x)=px2 1;I=[ 1;0](f)f(x)=x2 1;I=[ 1;1](g)f(x)=px2+x+1;I=[0;2](h)f(x)=3x+1x+2;I=[ 1;1]3.GivethemaximumandminimumpointsofthefunctionsontheintervalI.(a)f(x)=x 2;I=[ 1;1](b)f(x)=x2;I=[ 2;2](c)f(x)=1 x3;I=[0;1](d)f(x)=sinx;I=[ ;]114
TOPIC4.3:ProblemsInvolvingContinuityForeveryproblemthatwillbepresented,wewillprovideasolutionthatmakesuseofcontinuityandtakesadvantageofitsconsequences,suchastheIntermediateValueTheorem(IVT).APPROXIMATINGROOTS(MethodofBisection)Findingtherootsofpolynomialsiseasyiftheyarespecialproductsandthuseasytofactor.Sometimes,withalittleaddede ort,rootscanbefoundthroughsyntheticdivision.However,formostpolynomials,roots,canatbest,justbeapproximated.Sincepolynomialsarecontinuouseverywhere,theIVTisapplicableandveryusefulinapprox-imatingrootswhichareotherwisedicultto nd.Inwhatfollows,wewillalwayschooseaclosedinterval[a;b]suchthatf(a)andf(b)di erinsign,meaning,f(a)>0andf(b)<0,orf(a)<0andf(b)>0.IninvokingtheIVT,wetakem=0.Thisisclearlyanintermediatevalueoff(a)andf(b)sincef(a)andf(b)di erinsign.TheconclusionoftheIVTnowguaranteestheexistenceofc2[a;b]suchthatf(c)=0.Thisistantamounttolookingfortherootsofpolynomialf(x).EXAMPLE1:Considerf(x)=x3 x+1.Itsrootscannotbefoundusingfactoringandsyntheticdivision.WeapplytheIVT.Chooseanyinitialpairofnumbers,say 3and3.Evaluatefatthesevalues.f( 3)= 23<0andf(3)=25>0:Sincef( 3)andf(3)di erinsign,arootmustliebetween 3and3.Toapproachtheroot,wetrimtheinterval.{Try[0;3].However,f(0)=1>0likef(3)sonoconclusioncanbemadeaboutarootexistingin[0;3].{Try[ 3;0].Inthiscase,f(0)andf( 3)di erinsignsoweimprovethesearchspacefortherootfrom[ 3;3]to[ 3;0].115
Wetrimfurther.{f( 1)=1>0sotherootisin[ 3; 1].{f( 2)= 5<0sotherootisin[ 2; 1].{f 32= 78<0sotherootisin 32; 1.{f 54=1964>0sotherootisin 32; 54.FurthertrimmingandapplicationoftheIVTwillyieldtheapproximaterootwhichisx= 5340= 1:325.Thisgivesf(x) 0:0012.Thejust-concludedproceduregaveoneroot,anegativeone.Therearetwomorepossiblerealroots.FINDINGINTERVALSFORROOTSWhen ndinganexactrootofapolynomial,orevenanapproximateroot,provestootedious,someproblem-solversarecontentwith ndingasmallintervalcontainingthatrootEXAMPLE2:Consideragainf(x)=x3 x+1.Ifwejustneedanintervaloflength1,wecanalreadystopat[ 2; 1].Ifweneedanintervaloflength1/2,wecanalreadystopat 32; 1.Ifwewantanintervaloflength1/4,westopat 32; 54.EXAMPLE3:Considerf(x)=x3 x2+4.Findthreedistinctintervalsoflength1,orless,containingarootoff(x).Whenapproximating,wemaychooseassharpanestimateaswewant.Thesamegoesforaninterval.Whilesomeproblem-solverswillmakedowithanintervaloflength1,somemaywanta nerinterval,say,oflength1/4.Weshouldnotforgetthatthistypeofsearchispossiblebecausewearedealingwithpolynomials,andthecontinuityofpolynomialseverywhereallowsusrepeateduseoftheIVT.116
SOMECONSEQUENCESOFTHEIVTSomeinterestingapplicationsariseoutofthelogicusedintheIVT.EXAMPLE4:Wealreadyknowfromour rstlessonsonpolynomialsthatthedegreeofapolynomialisanindicatorofthenumberofrootsithas.Furthermore,didyouknowthatapolynomialofodddegreehasatleastonerealroot?Recallthatapolynomialtakestheform,f(x)=a0xn+a1xn 1+:::+an 1x+anwherea0;a1;:::;anarerealnumbersandnisanoddinteger.Takeforexamplea0=1.So,f(x)=xn+a1xn 1+:::+an 1x+an:Imaginextakingbiggerandbiggervalues,liketenthousandoramillion.Forsuchvalues,the rsttermwillfaroutweighthetotalofalltheotherterms.See,ifxispositive,forbignthevalueoff(x)willbepositive.Ifxisnegative,forbignthevalueoff(x)willbenegative.WenowinvoketheIVT.Remember,nisodd.Letabealarge-enoughnegativenumber.Then,f(a)<0.Letbbealarge-enoughpositivenumber.Then,f(b)>0.BytheIVT,thereisanumberc2(a;b)suchthatf(c)=0.Inotherwords,f(x)doeshavearealroot!117
SolvedExamplesEXAMPLE1:Findanapproximationofasolutionoff(x)=x3+x 1.Solution.Notethatfisapolynomialfunction,continuouseverywhere,sowecanapplyIVTforanyx2R.Choose[0;1].f(0)= 1<0andf(1)=1>0.ByIVT,fhasarootbetween0and1.Trimtheinterval:{f12= 38<0=)rootisin12;1{f34=1164>0=)rootisin12;34{f516 0:13<0=)rootisin516;34{f11160:01>0=)rootisin516;1116Furthertrimminggivesusanapproximaterootx0:683whichgivesf(x)0:002.EXAMPLE2:Findanapproximationofarootofg(x)=x3 2x+1.Solution.Notethatgisapolynomialfunction,continuouseverywhere,sowecanapplyIVTforanyx2R.Choose[ 2; 1].g( 2)= 3<0andg( 1)=2>0.ByIVT,fhasarootbetween-2and-1.Trimtheinterval:{g 74<0=)rootisin 74; 1{g 54>0=)rootisin 74; 54{g 32<0=)rootisin 32; 54Furthertrimminggivesusanapproximaterootx 1:615whichgivesf(x)0:01:::.118
EXAMPLE3:Findthreedistinctintervalswhichcontaintherootsofh(x)=x3 3x+1.Solution.Sincehisapolynomialfunction,thenhiscontinuousonR.Notethath( 2)= 1<0andh(0)=1>0=)hhasarootin[-1,0];h(0)=1>0andh(1)= 1<0=)hhasarootin[0,1];andh(1)= 1<0andh(2)=3>0=)hhasarootin[1,2]..EXAMPLE4:Showthatj(x)=cos(x+1) sin(x 1)hasarootbetween0and34.Solution.Sincejiscontinuouseverywhere,wecanapplyIVTon0;34.Notethatj(0)=1:382>0andj34= 1:954<0.Thus,byIVT,jhasasolutionin0;34:.EXAMPLE5:Showthatj(x)=cos(x+1) sin(x 1)hasanegativerootgreaterthan 54.Solution.ConsiderI= 54;0.SincejiscontinuousonR,wecanapplyIVTonI.Now,j 54= 1:954<0andj(0)=1:382>0givesourdesiredconclusion..EXAMPLE6:UseIVTtoshowthatk(x)=x 1xhasapositiverootlessthan2.Solution.Notethatk(0)isunde ned.Thus,kisdiscontinuousatx=0.Consider[0:5;2]so,k(0:5)= 1<0andk(2)=0:5>0.Hence,byIVT,khasasolutionon[0.5,2]..119
SupplementaryProblems1.Approximatetherootsofthefollowingfunctions.(a)f(x)=x3+x 1(b)f(x)=x3+x 3(c)f(x)=x3 2x 4(d)f(x)=x3 x+2(e)f(x)=x3+3x+1(f)f(x)=x3 x 22.Findanintervalcontainingasolutionofthefollowing.(a)f(x)=x3 3x+5(b)f(x)=2×3 x+1(c)f(x)=x3 2x 2(d)f(x)=x3+x+1(e)f(x)=x4 x3 1(f)f(x)=1 x x4(g)f(x)=sin(x+1)x+1(h)f(x)=cosxx+13.UseIVTtoshowthefollowing.(a)f(x)=x2 4hasazeroon[-3,-1](b)f(x)=x2 81hasazeroon[8,10](c)f(x)=cos(x 1) xhasarootbetween0and2(d)f(x)=sinxx 1hasarootbetween-4and-2(e)f(x)=sinx 2cosxhasapositiverootlessthan3(f)f(x)=x+cosxhasanegativerootgreaterthan-2120
CHAPTER1REVIEWI.Evaluatethefollowinglimits.1.limx!0x4 3×2+12.limx!1×3 1x 13.limx! 1px2 1x+1II.Givenf(x)=8>><>>: jx+6jx;x2( 1; 5];x2+8x+16×2+4x;x2( 5;+1):Discusscontinuityoffatx= 5andx= 4.Classifyeachdiscontinuity,ifany.III.Doasindicated.1.UsetheIntermediateValueTheoremtoshowthatthefunctionf(x)=x2 x 5hasazerobetween2and3.2.Giventhatlimx!1sin(x)=0andlimx!1ex 1=1Determinelimx!1sin(x)ex 1.3.Identifythepointswherethefunctiong(x)=x4 1willattainitsmaximumandminimumvalueson[ 1;1].IV.Fromthegraphofthefunctionhgivenbelow,evaluatethefollowinglimits.1.limx! 4h(x)2.limx! 1h(x)3.limx!1h(x)4.limx!3h(x)5.limx!5h(x)xy( 4;5)( 1; 2)( 1;4)(1;4)(3; 4)(5; 2)121
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Chapter2Derivatives
LESSON5:TheDerivativeastheSlopeoftheTangentLineLEARNINGOUTCOMES:Attheendofthelesson,thelearnershallbeableto:1.Illustratethetangentlinetothegraphofafunctionatagivenpoint;2.Applythede nitionofthederivativeofafunctionatagivennumber;and3.Relatethederivativeofafunctiontotheslopeofthetangentline.124
TOPIC5.1:TheTangentLinetotheGraphofaFunctionataPointTANGENTLINESTOCIRCLESRecallfromgeometrythatatangentlinetoacirclecenteredatOisalineintersectingthecircleatexactlyonepoint.Itisfoundbyconstructingtheline,throughapointAonthecircle,thatisperpendiculartothesegment(radius)OA.Asecantlinetoacircleisalineintersectingthecircleattwopoints.HOWTODRAWTANGENTLINESTOCURVESATAPOINTThede nitionofatangentlineisnotveryeasytoexplainwithoutinvolvinglimits.Onecanimaginethatlocally,thecurvelookslikeanarcofacircle.Hence,wecandrawthetangentlinetothecurveaswewouldtoacircle.xyy=f(x)cf(c)Onemorewaytoseethisistochoosethelinethroughapointthatlocallylooksmostlikethecurve.Amongallthelinesthroughapoint(c;f(c)),theonewhichbestapproximatesthecurvey=f(x)nearthepoint(c;f(c))isthetangentlinetothecurveatthatpoint.125
xyy=f(x)tangentlinecf(c)Amongalllinespassingthrough(c;f(c)),thetangentlineistheclosesttothecurvelocally.Anotherwayofqualitativelyunderstandingthetangentlineistovisualizethecurveasarollercoaster.Thetangentlinetothecurveatapointisparalleltothelineofsightofthepassengerslookingstraightaheadandsittingerectinoneofthewagonsoftherollercoaster.P(c;f(c))xytangentlinecxylineofsighttangentlinecEXAMPLE1:Whatdoyouthinkarethetangentlinesatthe\peaks”and\troughs”ofasmoothcurve?AB126
Wheneverthegraphissmooth(meaning,therearenosharpcorners),thetangentlinesatthe\peaks”and\troughs”arealwayshorizontal.EXAMPLE2:Thefollowingisthegraphofy=2 12(x 3)3.DrawnarethetangentlinesateachofthegivenpointsA,B,andC.y=2 12(x 3)3ABCTHETANGENTLINEDEFINEDMOREFORMALLYTheprecisede nitionofatangentlinereliesonthenotionofasecantline.LetCbethegraphofacontinuousfunctiony=f(x)andletPbeapointonC.Asecantlinetoy=f(x)throughPisanylineconnectingPandanotherpointQonC.Inthe gureontheright,theline !PQisasecantlineofy=f(x)throughP.y=f(x)PQWenowconstructthetangentlinetoy=f(x)atP.ChooseapointQontherightsideofP,andconnectthetwopointstoconstructthesecantline !PQ.ChooseanotherpointQ1inbetweenPandQ.ConnectthetwopointsPandQ1toconstructthesecantline !PQ1.ChooseanotherpointQ2inbetweenPandQ1.Constructthesecantline !PQ2.y=f(x)PQQ1Q2127
ConsideralsothecasewhenQistotheleftofPandperformthesameprocess.Intuitively,wecande nethetangentlinethroughPtobethelimitingpositionofthesecantlines !PQasthepointQ(whethertotheleftorrightofP)approachesP.Ifthesequenceofsecantlinestothegraphofy=f(x)throughPapproachesonelimitingposition(inconsiderationofpointsQtotheleftandfromtherightofP),thenwede nethislinetobethetangentlinetoy=f(x)atP.y=f(x)PQ`Wesummarizebelowthede nitionsofthesecantlinethroughapoint,andthetangentlineatapointofthegraphofy=f(x).De nitionLetCbethegraphofacontinuousfunctiony=f(x)andletPbeapointonC.1.Asecantlinetoy=f(x)throughPisanylineconnectingPandanotherpointQonC.2.Thetangentlinetoy=f(x)atPisthelimitingpositionofallsecantlines !PQasQ!P.EXAMPLE3:Thetangentlinetoanotherlineatanypointisthelineitself.(Thisdebunkstheideathatatangentlinetouchesthegraphatonlyonepoint!)Indeed,let`bealineandletPbeon`.ObservethatnomatterwhatpointQon`wetake,thesecantline !PQis`itself.Hence,thelimitingpositionofaline`is`itself.P`Thetangentlineto`atanypointPisitself!128
EXAMPLE4:Ourde nitionofthetangentlineallowsforaverticaltangentline.Wehaveseenthisontheunitcircleatpoints(1;0)and( 1;0).Averticaltangentlinemayalsoexistevenforcontinuousfunctions.Drawthecurvey=3pxandmarkthepointP(0;0).AllowtheclasstodeterminethetangentlinetothegraphatPusingtheformalde nition.Considerthetwocases:whenQistotherightofPandwhenQistotheleftofP.P`y=3pxCURVESTHATDONOTHAVETANGENTLINESItispossiblethatthetangentlinetoagraphofafunctionatapointP(x0;f(x0))doesnotexist.Thereareonlytwocaseswhenthishappens:1.Thecasewhenthefunctionisnotcontinuousatx0:Itisclearfromthede nitionofthetangentlinethatthefunctionmustbecontinuous.2.Thecasewhenthefunctionhasasharpcorner/cuspatP:Thiscaseproducesdi erentlimitingpositionsofthesecantlinesPQdependingonwhetherQistotheleftortotherightofP.Remark1:Theword\sharpcorner”ismorecommonlyusedforjointswhereonlylinesareinvolved.Forexample,theabsolutevaluefunctiony=jxjhasasharpcornerattheorigin.Incontrast,theterm\cusp”isoftenusedwhenatleastonegraphinvolvedrepresentsanonlinearfunction.Seethegraphsbelow.PcorneratPPcuspatPIntheaboveexamples,eachhasasharpcorner/cuspatP.ChoosingQtobepointstotheleftofPproducesadi erentlimitingpositionthanfromchoosingQtotherightofP.Sincethetwolimitingpositionsdonotcoincide,thenthetangentlineatPdoesnotexist.(Thisisthesamethingthathappenswhenthelimitfromtheleftofcdi ersfromthelimitfromtherightofc,wherewethenconcludethatthelimitdoesnotexist.)129
SupplementaryProblemsConstructtangentlinesatthelabeledpoints.1.. 112345 112340ABCDE2.. 2: 1:1:2: 1:1:2:3:0FED130
TOPIC5.2:TheEquationoftheTangentLineRecall:SlopeofaLineAline`passingthroughdistinctpoints(x0;y0)and(x;y)hasslopem`=y y0x x0:EXAMPLE1:GivenA(1; 3);B(3; 2),andC( 1;0),whataretheslopesofthelines !AB, !ACand !BC?Solution.Theslopeof !ABism !AB= 2 ( 3)3 1=12:Theslopeof !ACism !AC=0 ( 3) 1 1= 32:Theslopeof !BCism !BC=0 ( 2) 1 3= 24= 12:.Recall:Point-SlopeFormThelinepassingthrough(x0;y0)withslopemhastheequationy y0=m(x x0):EXAMPLE2:FromExample1above,sincem !AB=12,thenusingA(1; 3)asourpoint,thenthepoint-slopeformoftheequationof !ABisy ( 3)=12(x 1)ory+3=12(x 1):THEEQUATIONOFTHETANGENTLINEGivenafunctiony=f(x),howdowe ndtheequationofthetangentlineatapointP(x0;y0)?Considerthegraphofafunctiony=f(x)whosegraphisgivenbelow.LetP(x0;y0)beapointonthegraphofy=f(x).Ourobjectiveisto ndtheequationofthetangentline(TL)tothegraphatthepointP(x0;y0).131
y=f(x)Py0QxyTLx0FindanypointQ(x;y)onthecurve.Gettheslopeofthissecantline !PQ.m !PQ=y y0x x0:ObservethatlettingQapproachPisequivalenttolettingxapproachx0.Weusetheformalde nitionofthetangentline.SincethetangentlineisthelimitingpositionofthesecantlinesasQapproachesP,itfollowsthattheslopeofthetangentline(TL)atthepointPisthelimitoftheslopesofthesecantlines !PQasxapproachesx0.Insymbols,mTL=limx!x0y y0x x0=limx!x0f(x) f(x0)x x0:Finally,sincethetangentlinepassesthroughP(x0;y0),thenitsequationisgivenbyy y0=mTL(x x0):EquationoftheTangentLineTo ndtheequationofthetangentlinetothegraphofy=f(x)atthepointP(x0;y0),followthis2-stepprocess:Gettheslopeofthetangentlinebycomputingm=limx!x0y y0x x0orm=limx!x0f(x) f(x0)x x0:SubstitutethisvalueofmandthecoordinatesoftheknownpointP(x0;y0)intothepoint-slopeformtogety y0=m(x x0):132
EXAMPLE3:Findtheequationofthetangentlinetoy=x2atx=2.Solution.Togettheequationoftheline,weneedthepointP(x0;y0)andtheslopem.Weareonlygivenx0=2.However,they-coordinateofx0iseasyto ndbysubstitutingx0=2intoy=x2.Thisgivesusy0=4.Hence,Phasthecoordinates(2;4).Now,welookfortheslope:limx!x0y y0x x0=limx!2×2 4x 2=4:Finally,theequationofthetangentlinewithslopem=4andpassingthroughP(2;4)isy 4=4(x 2)ory=4x 4:.2Py=x2EXAMPLE4:Findtheslope-interceptformofthetangentlinetof(x)=pxatx=4.4Py=pxSolution.Again,we ndthey-valuecorrespondingtox0=4:y0=f(x0)=px0=p4=2:Hence,Phascoordinates(4;2).Now,welookfortheslopeofthetangentline.Noticethatwehavetorationalizethenumeratortoevaluatethelimit.m=limx!x0f(x) f(x0)x x0=limx!4px 2x 4px+2px+2=limx!4x 4(x 4)(px+2)=limx!41px+2=14:Finally,withpointP(4;2)andslopem=14;theequationofthetangentlineisy 2=14(x 4)ory=x4+1:.133
Thenextexampleshowsthatourprocessof ndingthetangentlineworksevenforhorizontallines.EXAMPLE5:Showthatthetangentlinetoy=3×2 12x+1atthepoint(2; 11)ishorizontal.Solution.Computingfortheslope,weget:m=limx!x0y y0x x0=limx!2(3×2 12x+1) ( 11)x 2=limx!23(x2 4x+4)x 2=limx!2(3(x 2))=0:Recallthatahorizontallinehaszeroslope.Sincetheslopeofthetangentlineis0,itmustbehorizontal.Itsequationisy ( 11)=0(x 2)ory= 11:.EXAMPLE6:Verifythatthetangentlinetotheliney=2x+3at(1;5)isthelineitself.Solution.We rstcomputefortheslopeofthetangentline.Notethatx0=1andy0=5.m=limx!x0y y0x x0=limx!1(2x+3) 5x 1=limx!12x 2x 1=2:Therefore,substitutingthisintothepoint-slopeformwithP(1;5)andm=2,wegety 5=2(x 1)or,y=2x+3:Thisisthesameequationasthatofthegivenline..SolvedExamplesEXAMPLE1:GivenA(2; 1),B(0;5),andC( 3;1),whataretheslopesofthesegmentsAB,BC,andAC?Solution.TheslopeofABismAB=5 ( 1)0 2=6 2= 3:134
TheslopeofBCismBC=1 5 3 0= 4 3=43:TheslopeofACismAC=1 ( 1) 3 2=2 5= 25:.EXAMPLE2:FindtheequationofthelinecontainingsegmentBC.Solution.Theslopeofthislinewasalreadysolvedinthepreviousexample,i.e.,mBC=43.SincethelinepassesthroughpointB(0;5),applyingthepoint-slopeform,wegety 5=43(x 0)ory=43x+5:OnealsogetsthesameequationifthepointC( 3;1)isusedinstead..EXAMPLE3:Showthattheequationofthetangentlinetotheliney=1 2xat(1; 1)isitself.Solution.We rstcomputefortheslopeofthetangentlineatthegivenpoint.Here,wehavex0=1andy0= 1.m=limx!x0y y0x x0=limx!1(1 2x) ( 1)x 1=limx!12 2xx 1=limx!1 2(x 1)x 1= 2:Thus,substitutingthisinthepoint-slopeformwithP(1; 1)andm=2,wegety ( 1)= 2(x 1)ory= 2x+1=1 2x:.EXAMPLE4:Findtheequationofthetangentlinetothecurvey=2×2 1atx=2.Solution.We rstneedtogetthecoordinatesofthepointoftangencyP(x0;y0),andtheslopeatP(x0;y0).Inthisproblem,x0=2.Therefore,y0=2(2)2 1=7.Thus,Phascoordinates135
(2;7).Wenowcomputefortheslope:m=limx!x0y y0x x0=limx!2 2×2 1 7x 2=limx!22×2 8x 2=limx!22 x2 4x 2=limx!22(x+2)(x 2)x 2=limx!22(x+2)=8:Theequationofthetangentlinewithslopem=8passingthroughP(2;7)isy 7=8(x 2)ory=8x 9:.EXAMPLE5:Findtheequationofthetangentlinetothecurvef(x)=p3x 1atx=53.Solution.They-coordinateofthepointonthegraphwhenx0=53isy0=f(x0)=s353 1=p5 1=p4=2:Wenow ndtheslopeofthetangentlineat53;2:m=limx!x0f(x) f(x0)x x0=limx!53p3x 1 2x 53p3x 1+2p3x 1+2=limx!533x 1 4 x 53 p3x 1+2=limx!533 x 53 x 53 p3x 1+2=3q3 53 1+2=34:Theequationofthetangentlinewithslopem=34passingthrough53;2isy 2=34x 53:.136
EXAMPLE6:Showthatthetangentlinetothecurvey=2×2 12x+19atthepoint(3;1)ishorizontal.Solution.Weneedtoshowthattheslopeofthetangentlineis0.m=limx!x0y y0x x0=limx!3 2×2 12x+19 1x 3=limx!32×2 12x+18x 3=limx!32(x 3)2x 3=limx!32(x 3)=0:Sincetheslopeis0,thetangentlineat(3;1)ishorizontal..EXAMPLE7:Showthatthetangentlinetothecurvef(x)=px 1atx=1isvertical.Whatistheequationofthisline?Solution.Observethatm=limx!x0f(x) f(x0)x x0=limx!1px 1 p1 1x 1=limx!1px 1x 1px 1px 1=limx!1x 1(x 1)px 1=limx!11px 1Thislimitassumestheform10andisthusunde ned.Inthiscase,thetangentlineisverticalwithequationx=1..137
SupplementaryProblemsFindthestandard(slope-interceptform)equationofthetangentlinetothefollowing:1.f(x)=x2 3x+1atx=12.f(x)=52x 23atx= 13.f(x)=x3+8atx= 24.f(x)=2×2 3x+4atx=55.f(x)=1 2x x2atx= 16.f(x)=x3+3×2+3x+1atx= 17.f(x)=x3 x2+1atx=18.f(x)=x4 3×3+5×2 2x+1atx= 19.f(x)=x5 x4+x3 x2+x+1atx=110.f(x)=6×2 5x+1atx=1211.f(x)=px+2atx=412.f(x)=p2 3xatx= 213.f(x)=p1+2xatx=314.f(x)=p5 9x2atx= 1315.f(x)=px2+2xatx=016.f(x)=p3x2+5atx=117.f(x)=3px3+1atx=118.f(x)=3p1 x3atx= 119.f(x)=3px3+3×2+3x+1atx=320.f(x)=px+1 (x+1)atx=0138
TOPIC5.3:TheDe nitionoftheDerivativeDe nitionoftheDerivativeLetfbeafunctionde nedonanopenintervalIR,andletx02I.Thederivativeoffatx0isde nedtobef0(x0)=limx!x0f(x) f(x0)x x0;ifthislimitexists.Thatis,thederivativeoffatx0istheslopeofthetangentlineat(x0;f(x0)),ifitexists.Notations:Ify=f(x),thederivativeoffiscommonlydenotedbyf0(x);Dx[f(x)];ddx[f(x)];ddx[y];dydx:Remark1:Notethatthelimitde nitionofthederivativeisinherentlyindeterminate!Hence,theusualtechniquesforevaluatinglimitswhichareindeterminateoftype00areapplied,e.g.,factoring,rationalization,orusingoneofthefollowingestablishedlimits:(i)limx!0sinxx=1(ii)limx!01 cosxx=0(iii)limx!0ex 1x=1.EXAMPLE1:Computef0(1)foreachofthefollowingfunctions:1.f(x)=3x 12.f(x)=2×2+43.f(x)=2xx+14.f(x)=px+8Solution.Here,x0is xedtobeequalto1.Usingthelimitde nitionofthederivative,f0(1)=limx!1f(x) f(1)x 1:Rememberthatwhatwearecomputing,f0(1),isjusttheslopeofthetangentlinetoy=f(x)atx=1.139
1.Notethatf(1)=2,sobyfactoringf0(1)=limx!1(3x 1) 2x 1=limx!13(x 1)x 1=limx!13=3:2.Here,f(1)=6soagain,byfactoring,f0(1)=limx!1(2×2+4) 6x 1=limx!12(x+1)(x 1)x 1=limx!12(x+1)=4:3.Weseethatf(1)=1.So,fromthede nition,f0(1)=limx!12xx+1 1x 1:Wemultiplyboththenumeratorandthedenominatorbyx+1tosimplifythecomplexfraction:f0(1)=limx!12xx+1 1x 1x+1x+1=limx!12x (x+1)(x 1)(x+1)=limx!1x 1(x 1)(x+1)=limx!11x+1=12:4.Notethatf(1)=3.Therefore,byrationalizingthenumerator(meaning,multiplyingbypx+8+3),f0(1)=limx!1px+8 3x 1px+8+3px+8+3=limx!1(x+8) 9(x 1)(px+8+3)=limx!11px+8+3=16:140
.AlternativeDe nitionoftheDerivativeLetfbeafunctionde nedonanopenintervalIR,andletx2I.Thederivativeoffatxisde nedtobef0(x)=limx!0f(x+x) f(x)x;(2.1)ifthislimitexists.EXAMPLE2:Letf(x)=sinx,g(x)=cosx,ands(x)=ex.Findf0(2);g0();ands0(3).Solution.Weusethealternativede nitionofthederivative.1.Here,wesubstitutex0=2.f0(2)=limh!0f(2+h) f(2)h=limh!0sin(2+h) 0h:Usingthesumidentityofthesinefunction:sin( + )=sin cos +cos sin ;andnotingthatsin(2)=0andcos(2)=1,wegetf0(2)=limh!0sin(2)cosh+cos(2)sinhh=limh!0sinhh=1:(Why?)2.g0()=limh!0g(+h) g()h=limh!0cos(+h) ( 1)h:Usingthesumidentityofthecosinefunction:cos( + )=cos cos sin sin ;andnotingthatcos= 1andsin=0,wegetg0()=limh!0coscosh sinsinh+1h=limh!0 cosh+1h=0:(Why?)141
3.s0(3)=limh!0s(3+h) s(3)h=limh!0e3+h e3h:Usingtheexponentlaws,e3+h=e3eh.Moreover,sincee3isjustaconstant,wecanfactoritoutofthelimitoperator.So,s0(3)=limh!0e3eh e3h=e3limh!0eh 1h=e3:(Why?).INSTANTANEOUSVELOCITYOFAPARTICLEINRECTILINEARMOTIONThederivativeofafunctionisalsointerpretedastheinstantaneousrateofchange.Wediscusshereaparticularquantitywhichisimportantinphysics{theinstantaneousvelocityofamov-ingparticle.Supposethatanobjectoraparticlestartsfroma xedpointAandmovesalongastraightlinetowardsapointB.Supposealsothatitspositionalongline !ABattimetiss.Thenthemotionoftheparticleiscompletelydescribedbythepositionfunctions=s(t);t0andsincetheparticlemovesalongaline,itissaidtobeinrectilinearmotion.EXAMPLE3:Supposethataparticlemovesalongalinewithpositionfunctions(t)=2t2+3t+1wheresisinmetersandtisinseconds.a.Whatisitsinitialposition?b.Whereisitlocatedaftert=2seconds?c.Atwhattimeistheparticleatpositions=6?142
Solution..a.Theinitialpositioncorrespondstotheparticle’slocationwhent=0.Thus,s(0)=2(0)2+3(0)+1=1:Thismeansthattheparticlecanbefound1metertotherightoftheorigin.b.After2seconds,itcannowbefoundatpositions(2)=2(2)2+3(2)+1=15meters.c.Weequates(t)=2t2+3t+1=6.So,2t2+3t 5=0()(2t+5)(t 1)=0()t= 52ort=1:Sincetimecannotbenegative,wechooset=1second..1234567891011121314150t=0t=1t=2Now,weask:Whatistheparticle’svelocityattheinstantwhentimet=1?Recallthattheformulafortheaveragevelocityofaparticleisaveragevelocity=displacementtimeelapsed=s(t nal) s(tinitial)t nal tinitial:Thisposesaproblembecauseattheinstantwhent=1,thereisnoelapsedtimesincenot nalisgiven.Weremedythisbycomputingthevelocityatshorttimeintervalswithanend-pointatt=1.Forexample,onthetimeinterval[1;2],thevelocityoftheparticleisv=s(2) s(1)2 1=15 62 1=9m/s:143
Wecomputetheparticle’svelocityonshorterintervals:TimeIntervalAverageVelocity[1;1:5]8[1;1:1]7:2[1;1:01]7:02[1;1:001]7:002TimeIntervalAverageVelocity[0:5;1]6[0:9;1]6:8[0:99;1]6:98[0:999;1]6:998Weseefromthetablesabovethatthevelocitiesoftheparticleonshortintervalsendingorstartingatt=1approach7m/sasthelengthsofthetimeintervalsapproach0.Thislimitlimt!1s(t) s(1)t 1iswhatwerefertoastheinstantaneousvelocityoftheparticleatt=1.However,thelimitexpressionaboveispreciselythede nitionofthederivativeofsatt=1,andtheinstantaneousvelocityisactuallytheslopeofthetangentlineatthepointt=t0ifthefunctioninconsiderationisthepositionfunction.Wemaketheconnectionbelow.InstantaneousvelocityLets(t)denotethepositionofaparticlethatmovesalongastraightlineateachtimet0.Theinstantaneousvelocityoftheparticleattimet=t0iss0(t0)=limt!t0s(t) s(t0)t t0;ifthislimitexists.EXAMPLE4:Aballisshotstraightupfromabuilding.Itsheight(inmeters)fromthegroundatanytimet(inseconds)isgivenbys(t)=40+35t 5t2.Finda.theheightofthebuilding.b.thetimewhentheballhitstheground.c.theaveragevelocityontheinterval[1;2].d.theinstantaneousvelocityatt=1and2.e.theinstantaneousvelocityatanytimet0.BuildingBall’sPathGroundlevel144
Solution..a.Theheightofthebuildingistheinitialpositionoftheball.So,thebuildingiss(0)=40meterstall.b.Theballisonthegroundwhentheheightsoftheballfromthegroundiszero.Thuswesolvethetimetwhens(t)=0:30+40t 5t2=0()5(8 t)(1+t)=0()t=8ort= 1:Sincetimeispositive,wechooset=8seconds.c.Theaveragevelocityoftheballon[1;2]iss(2) s(1)2 1=90 702 1=20m/s:d.Theninstantaneousvelocityattimet=1islimt!1s(t) s(1)t 1=limt!1(40+35t 5t2) 70t 1=limt!1 5(t 6)(t 1)t 1=25m/s:Attimet=2,limt!2s(t) s(2)t 2=limt!2(40+35t 5t2) 90t 2=limt!2 5(t 5)(t 2)t 2=15m/s:e.Theinstantaneousvelocityatanytimet0islimt!t0s(t) s(t0)t t0=limt!1(40+35t 5t2) (40+35t0 5t20)t t0=limt!15(t t0)(7 (t+t0))t t0=(35 10t0)m/s:.SolvedExamplesEXAMPLE1:Computef0(2)forthefollowingfunctions:1.f(x)=x2+12.f(x)=x3 13.f(x)=x2x+14.f(x)=px+8145
Solution.Inthisexample,x0is xedandisequalto2.Usingthede nitionofthederivative,f0(2)=limx!2f(x) f(2)x 2:1.Notethatf(2)=22+1=5.Weget,f0(2)=limx!2 x2+1 5x 2=limx!2×2 4x 2=limx!2(x+2)=4:2.Here,f(2)=23 1=7.Thus,f0(2)=limx!2 x3 1 7x 2=limx!2×3 8x 2=limx!2 x2+2x+4=12:3.Weseethatf(2)=22(2)+1=25.So,f0(2)=limx!2x2x+1 25x 2=limx!25x 4x 25(2x+1)x 2=limx!2(x 2)5(2x+1)(x 2)=15(2(2)+1)=125:146
4.Notethatf(2)=p2+8=p10.Therefore,f0(2)=limx!2px+8 p10x 2px+8+p10px+8+p10=limx!2x+8 10(x 2) px+8+p10=limx!2(x 2)(x 2) px+8+p10=1p10+p10=12p10:.EXAMPLE2:Letf(x)=sin2x,g(x)=cos(x+),andh(x)=ex 1.Findf0(),g0(2),andh0(1).Solution.Usingthealternativede nitionofthederivative,weobtainf0()=limh!0f(+h) f()h=limh!0sin(2(+h)) 0h=limh!0sin(2+2h)h:Weapplythesumidentityforthesinefunction:sin( + )=sin cos +cos sin andnotingthatsin(2)=0andcos(2)=1,weobtainf0()=limh!0sin(2)cos(2h)+cos(2)sin(2h)h=limh!0sin(2h)h22=limh!0sin(2h)2h2=2limh!0sin(2h)2h=21=2:Now,notethatg(2)=cos(2+)= 1.Thus,g0(2)=limx!2cos(x+) ( 1)x 2:147
Weusetheidentity,cos( + )=cos cos sin sin .Andso,g0(2)=limx!2cosxcos sinxsin+1x 2=limx!21 cosxx 2:Atthispoint,wemakeasubstitution,t=x 2.Thus,x=t+2,andasx!2,t!0.Thus,g0(2)=limt!01 cos(t+2)t=limt!01 (costcos2 sintsin2)t=limt!01 costt=0:Lastly,h(1)=e1 1=1.Calculatingthederivative,h0(1)=limx!1ex 1 1x 1:Applyingthesubstitutiont=x 1,andnotingthatasx!1,t!0,weobtain,h0(1)=limt!0et 1t=1:.EXAMPLE3:Supposethecosty(inpesos)ofmanufacturingxkilosofacertainproductisgivenbyy=C(x)=x4 2×3+x2+1:Computefortheaveragerateofchangeofthecostyofproducingxproductsover[1;2],[1;5],and[0;5].Solution.Tosolvethis,weneedthevaluesofC(1),C(2),C(5),andC(0).Computing,wegetC(1)=1,C(2)=5,C(5)=401,andC(0)=1.Thus,C(2) C(1)2 1=5 11=Php4=kgC(5) C(1)5 1=401 15 1=Php100=kgC(5) C(0)5 0=401 15 0=Php80=kg.148
EXAMPLE4:SolvefortheinstantaneousrateofchangeofthecostfunctioninEx.3atx=1.Solution.Theinstantaneousrateofchangeatx=1ispreciselythederivativeofthecostfunctionatx=1.SinceC(1)=1,wegetC0(1)=limx!1C(x) C(1)x 1=limx!1(x4 2×3+x2+1) 1x 1=limx!1×4 2×3+x2x 1=limx!1×2 x2 2x+1x 1=limx!1 x2(x 1)=0:.SupplementaryProblems1.Foreachofthefollowingfunctions, ndtheindicatedderivative.(a)f(x)=2×2+x 1;f0(1)(b)f(x)=2 x3;f0( 1)(c)f(x)=x5 x4+x3 x2+x+1;f0(0)(d)f(x)=p3x 5;f0(2)(e)f(x)=1 px2 3x+4;f0(3)(f)f(x)=x2x 1;f0( 3)(g)f(x)=x2+1x 4;f0(0)(h)f(x)=p3 2x x2+1x 2;f0( 5)(i)f(x)=2sinx4;f0(2)(j)f(x)=xe x;f0(1)(k)f(x)=3px3+1;f0(2)(l)f(x)=cot2x+2;f02149
2.Supposethatthetotalrevenue,R(x)(inpesos),fromthesaleofxproductsisgivenbyR(x)=2×2 3x+5:Findthefollowing:(a)Theaveragerateofchangeoftherevenueintheinterval:i.[10;50]ii.[10;100]iii.[0;100](b)Findtheinstantaneousrateofchangeoftherevenueat:i.x=10ii.x=50iii.x=100(c)Findtheinstantaneousrateofchangeoftheslopeofthetangentlineateachpointofthegraphoff(x)=x3 2×2+x+1,wheretheslopeofthetangentlineiszero.150
LESSON6:RulesofDi erentiationLEARNINGOUTCOMES:Attheendofthelesson,thelearnershallbeableto:1.Determinetherelationshipbetweendi erentiabilityandcontinuity;2.Derivethedi erentiationrules;and3.Applythedi erentiationrulesincomputingthederivativesofalgebraic,exponential,andtrigonometricfunctions.151
TOPIC6.1:Di erentiabilityImpliesContinuityThedi erencebetweencontinuityanddi erentiabilityisacriticalissue.Most,butnotall,ofthefunctionsweencounterincalculuswillbedi erentiableovertheirentiredomain.Beforewecancon dentlyapplytherulesregardingderivatives,weneedtobeabletorecognizetheexceptionstotherule.Recallthefollowingde nitions:De nition1(ContinuityataNumber).Afunctionfiscontinuousatanumbercifallofthefollowingconditionsaresatis ed:(i)f(c)isde ned;(ii)limx!cf(x)exists;and(iii)limx!cf(x)=f(c).Ifatleastoneofthetheseconditionsisnotsatis ed,thefunctionissaidtobediscontinuousatc.De nition2(ContinuityonR).Afunctionfissaidtobecontinuouseverywhereiffiscontinuousateveryrealnumber.De nition3.Afunctionfisdi erentiableatthenumberciff0(c)=limh!0f(c+h) f(c)hexists.Wenowpresentseveralexamplesofdeterminingwhetherafunctioniscontinuousordi eren-tiableatanumber.152
EXAMPLE1:1.Thepiecewisefunctionde nedbyf(x)=8><>:x2+2x 3x 1ifx6=1;4ifx=1;iscontinuousatc=1.Thisisbecausef(1)=4,limx!1f(x)=limx!1(x 1)(x+3)x 1=4;andf(1)=limx!1f(x).2.Thefunctionde nedbyf(x)=8<: x2ifx<2;3 xifx2:isnotcontinuousatc=2sincelimx!2 f(x)= 46=1=limx!2+f(x),thatis,thelimx!2f(x)doesnotexist.3.Considerthefunctionf(x)=3px:Byde nition,itsderivativeisf0(x)=limh!0f(x+h) f(x)h=limh!03px+h 3pxh3p(x+h)2+3p(x+h)(x)+3px23p(x+h)2+3p(x+h)(x)+3px2=limh!0(x+h) xh(3p(x+h)2+3p(x+h)(x)+3px2)=limh!013p(x+h)2+3p(x+h)(x)+3px2=133px2:Sincef0(1)=133p12=13,thenfisdi erentiableatx=1.Ontheotherhand,f0(0)doesnotexist.Hencefisnotdi erentiableatx=0.4.Thefunctionde nedbyf(x)=(5xifx<12x+3ifx1iscontinuousbutnotdi erentiableatx=1.Indeed,f(1)=2(1)+3=5.Now,Ifx<1,thenf(x)=5xandsolimx!1 5x=5.Ifx>1,thenf(x)=2x+3andsolimx!1+(2x+3)=5:153
Sincetheone-sidedlimitsexistandareequaltoeachother,thelimitexistsandequals5.So,limx!1f(x)=5=f(1):Thisshowsthatfiscontinuousatx=1.Ontheotherhand,computingforthederivative,Forx<1,f(x)=5xandlimh!0 5(x+h) (5x)h=5.Forx>1,f(x)=2x+3andlimh!0+(2(x+h)+3) (2x+3)h=2:Sincetheone-sidedlimitsatx=1donotcoincide,thelimitatx=1doesnotexist.Sincethislimitisthede nitionofthederivativeatx=1,weconcludethatthederivativedoesnotexist.Therefore,fisnotdi erentiableatx=1.5.Anotherclassicexampleofafunctionthatiscontinuousatapointbutnotdi erentiableatthatpointistheabsolutevaluefunctionf(x)=jxjatx=0.Clearly,f(0)=0=limx!0jxj.However,ifwelookatthelimitde nitionofthederivative,limh!0f(0+h) f(0)h=limh!0j0+hj j0jh=limh!0jhjh:Notethattheabsolutevaluefunctionisde neddi erentlytotheleftandrightof0soweneedtocomputeone-sidedlimits.Notethatifhapproaches0fromtheleft,thenitapproaches0throughnegativevalues.Sinceh<0=)jhj= h,itfollowsthatlimh!0 jhjh=limh!0 hh=limh!0 1= 1:Similarly,ifhapproaches0fromtheright,thenhapproaches0throughpositivevalues.Sinceh>0=)jhj=h,weobtainlimh!0+jhjh=limh!0+hh=limh!0+1=1:Hence,thederivativedoesnotexistatx=0sincetheone-sidedlimitsdonotcoincide.Theprevioustwoexamplesprovethatcontinuitydoesnotnecessarilyimplydi erentiability.Thatis,therearefunctionswhicharecontinuousatapoint,butnotdi erentiableatthatpoint.ThenexttheoremhoweversaysthattheconverseisalwaysTRUE.Theorem6.Ifafunctionfisdi erentiableata,thenfiscontinuousata.154
Remark1:1.Iffiscontinuousatx=a,itdoesnotmeanthatfisdi erentiableatx=a.2.Iffisnotcontinuousatx=a,thenfisnotdi erentiableatx=a.3.Iffisnotdi erentiableatx=a,itdoesnotmeanthatfisnotcontinuousatx=a.4.Afunctionfisnotdi erentiableatx=aifoneofthefollowingistrue:(a)fisnotcontinuousatx=a:(b)thegraphoffhasaverticaltangentlineatx=a:(c)thegraphoffhasacornerorcuspatx=a:SolvedExamplesEXAMPLE1:Thefunctionf(x)=8><>:x2 9x 3ifx6=36ifx=3iscontinuousatx=3sincef(3)=6andlimx!3f(x)=6.EXAMPLE2:Thefunctionf(x)=8<:x3ifx2x+1ifx<2isnotcontinuousatx=2sincelimx!2+f(x)=8whilelimx!2 f(x)=3.Sincefisnotcontinuousatx=2,itcannotbedi erentiableatx=2.EXAMPLE3:Thederivativeofthefunctionf(x)=1xisf0(x)= 1x2.Observethatfisdi erentiableatx=1sincef0(1)= 1.However,fisnotdi erentiableatx=0sincef0(0)isunde ned.EXAMPLE4:Considerthefunctionf(x)=8<:2xifx0;5xifx<0:Thisfunctioniscontinuousatx=0sincef(0)=limx!0+f(x)=limx!0 f(x)=0.However,thisfunctionisnotdi erentiableatx=0.Toseethis,notethatf0(x)=8<:2ifx>0;5ifx<0;155
bytakingthederivativesoff(x)=2xandf(x)=5x,respectively.Notealsothechangeintheinequalitiesinvolvedintheintervalspresent.Sincelimx!0+f0(x)=26=5=limx!0 f0(x),thenthelimitdoesnotexist.Thederivative,beingde nedasthislimit,wouldalsonotexistatx=0.Graphically,acornerisfoundatx=0.Notealsothatfailuretobedi erentiableatapointdoesnotimplydiscontinuityatthatpoint.Thisisshowninthisexample,whereasfisnotdi erentiableatx=0howeveritiscontinuousatx=0.SupplementaryProblems1.Supposefisafunctionsuchthatf0(1)isunde ned.WhichofthefollowingstatementsisALWAYStrue?(a)fmustbecontinuousatx=1.(b)fisNOTcontinuousatx=1.(c)Thereisnotenoughinformationtodeterminewhetherornotfiscontinuousatx=1.2.Whichofthefollowingstatementsis/areALWAYStrue?(a)Afunctionthatiscontinuousatx=amustbedi erentiableatx=a.(b)AfunctionthatisNOTcontinuousatx=amustNOTbedi erentiableatx=a.(c)AfunctionthatisNOTdi erentiableatx=amustNOTbecontinuousatx=a.(d)Afunctionthatisdi erentiableatx=amustNOTbecontinuousatx=a.3.Supposethatfisafunctionwhosedomainisthesetofallrealnumbers,R;whichisNOTcontinuousatx= 1.Whichofthefollowingstatementsis/areFALSE?(a)fisNOTdi erentiableatx= 1.(b)Eitherf( 1)6=limx! 1+f(x)orf( 1)6=limx! 1 f(x).(c)f( 1)isde ned.4.Considerthefunctionde nedbyf(x)=8<:x2+1ifx<56x 4ifx5:(a)Isfcontinuousatx=5?(b)Isfdi erentiableatx=5?156
5.Determinethevaluesforwhichfiscontinuous.(a)f(x)=x+5x 5(b)f(x)=px 1x 2(c)f(x)=2x p3 xx2 5x+6(d)f(x)=r3 xx+2(e)f(x)=px2 7x+126.Letf(x)=8>>><>>>:2x 13x+2;x>1;px+5;0x1;x;x<0:AtwhichpointsisfNOTcontinuous?7.Supposeaandbarerealnumbers.Consider,f(x)=8<:ax+b;ifx2;x2 1;ifx<2:(a)Isitpossibleto ndaandbsuchthatfisdi erentiableatx=2?Ifso,whatarethesenumbers?(b)Ifweonlyrequirecontinuityatx=2,whatarethepossiblevaluesforaandbsuchthatthisissatis ed?8.Isthefunctionde nedbyg(x)=x3 tanx4+1continuousatx=2?9.Considerthefunctionf(x)= x2+x 12 :(a)Forwhichvaluesisthisfunctioncontinuous?(b)Forwhichvaluesisthisfunctiondi erentiable?157
TOPIC6.2:TheDi erentiationRulesandExamplesInvolvingAlgebraic,Exponential,andTrigonometricFunctionsWe rstrecallthede nitionofthederivativeofafunction.Thederivativeofthefunctionfthefunctionf0whosevalueatanumberxinthedomainoffisgivenbyf0(x)=limh!0f(x+h) f(x)h(2.2)ifthelimitexists.Forexample,letuscomputethederivativeofthe rstfunctionoftheseatworkabove:f(x)=3x2+4.Letus rstcomputethenumeratorofthequotientin(2.2):f(x+h) f(x)=(3(x+h)2+4) (3x2+4)=(3x2+6xh+3h2+4) (3x2+4)=6xh+3h2=h(6x+3h):Therefore,f0(x)=limh!0f(x+h) f(x)h=limh!03(x+h)2+4 (3x2+4)h=limh!0h(6x+3h)h=limh!0(6x+3h)=6x:Weseethatcomputingthederivativeusingthede nitionofevenasimplepolynomialisalengthyprocess.Whatfollowsnextarerulesthatwillenableusto ndderivativeseasily.WecallthemDIFFERENTIATIONRULES.DIFFERENTIATINGCONSTANTFUNCTIONSThegraphofaconstantfunctionisahorizontallineandahorizontallinehaszeroslope.Thederivativemeasurestheslopeofthetangent,andsothederivativeiszero.158
RULE1:TheConstantRuleIff(x)=cwherecisaconstant,thenf0(x)=0.Thederivativeofaconstantisequaltozero.EXAMPLE1:1.Iff(x)=10,thenf0(x)=0.2.Ifh(x)= p3,thenh0(x)=0.3.Ifg(x)=5,theng0(x)=0.DIFFERENTIATINGPOWERFUNCTIONSRULE2:ThePowerRuleIff(x)=xnwheren2N,thenf0(x)=nxn 1.EXAMPLE2:1.Iff(x)=x3,thenf0(x)=3x3 1=3x2.2.Findg0(x)whereg(x)=1x2.Solution.Insomecases,thelawsofexponentsmustbeusedtorewriteanexpressionbeforeapplyingthepowerrule.Thus,we rstwriteg(x)=1x2=x 2beforeweapplythePowerRule.Wehave:g0(x)=( 2)x 2 1= 2x 3or 2x3:.3.Ifh(x)=px,thenwecanwriteh(x)=x12.Sowehave,h0(x)=12x12 1=12x 12or12px159
DIFFERENTIATINGACONSTANTTIMESAFUNCTIONRULE3:TheConstantMultipleRuleIff(x)=kh(x)wherekisaconstant,thenf0(x)=kh0(x).EXAMPLE3:Findthederivativesofthefollowingfunctions.1.f(x)=5x342.g(x)=133px3.h(x)= p3xSolution.WeuseRule3inconjunctionwithRule2.1.f0(x)=534x34 1=154x 14:2.g(x)=13x13=)g0(x)=1313x13 1=19x 23:3.h0(x)= p3x1 1= p3:.DIFFERENTIATINGSUMSANDDIFFERENCESOFFUNCTIONSRULE4:TheSumRuleIff(x)=g(x)+h(x)wheregandharedi erentiablefunctions,thenf0(x)=g0(x)+h0(x).EXAMPLE4:1.Di erentiatethefollowing:(i)f(x)+g(x)(ii)g(x)+h(x)(iii)f(x)+h(x)160
2.UseRules3and4todi erentiatethefollowing:(Hint:f(x) g(x)=f(x)+( 1)g(x).)(i)f(x) g(x)(ii)g(x) h(x)(iii)f(x) h(x)Solution..1.CopyingthederivativesinthesolutionofExample(3),andsubstitutingthemintotheformulaoftheSumRule,weobtain(i)154x 14+19x 23:(ii)19x 23+( p3):(iii)154x 14+( p3):2.UsingRules3and4,wededucethatthederivativeoff(x) g(x)isequaltothedi erenceoftheirderivatives:f0(x) g0(x).Thereforeweobtain(i)154x 14 19x 23:(ii)19x 23 ( p3):(iii)154x 14 ( p3):.RULE5:TheProductRuleIffandgaredi erentiablefunctions,thenDx[f(x)g(x)]=f(x)g0(x)+g(x)f0(x):Rule5statesthatthederivativeoftheproductoftwodi erentiablefunctionsisthe rstfunctiontimesthederivativeofthesecondfunctionplusthesecondfunctiontimesthederivativeofthe rstfunction.EXAMPLE5:1.Findf0(x)iff(x)=(3x2 4)(x2 3x)Solution.f0(x)=(3x2 4)Dx(x2 3x)+(x2 3x)Dx(3x2 4)=(3x2 4)(2x 3)+(x2 3x)(6x)=6x3 9x2 8x+12+6x3 18x2=12x3 27x2 8x+12:.161
Remark1:Intheaboveexample,wecouldhavealsomultipliedthetwofactorsandgetf(x)=3x4 9x3 4x2+12x:Then,bytheRules2,3and4,thederivativeoffisf0(x)=12x3 27x2 8x+12whichisconsistentwiththeonederivedfromusingtheproductrule.2.Findf0(x)iff(x)=px(6x3+2x 4).Solution.Usingproductrule,f0(x)=x1=2Dx(6x3+2x 4)+Dx(x1=2)(6x3+2x 4)=x1=2(18x2+2)+12x 1=2(6x3+2x 4)=18x5=2+2x1=2+3x5=2+x1=2 2x 1=2=21x5=2+3x1=2 2x 1=2:.DIFFERENTIATINGQUOTIENTSOFTWOFUNCTIONSRULE6:TheQuotientRuleLetf(x)andg(x)betwodi erentiablefunctionswithg(x)6=0.ThenDxf(x)g(x)=g(x)f0(x) f(x)g0(x)[g(x)]2:Theruleabovestatesthatthederivativeofthequotientoftwofunctionsisthefractionhavingasitsdenominatorthesquareoftheoriginaldenominator,andasitsnumeratorthede-nominatortimesthederivativeofthenumeratorminusthenumeratortimesthederivativeofthedenominator.EXAMPLE6:1.Leth(x)=3x+5x2+4.Computeh0(x).162
Solution.Ifh(x)=3x+5x2+4,thenf(x)=3x+5andg(x)=x2+4andthereforef0(x)=3andg0(x)=2x.Thus,h(x)=g(x)f0(x) f(x)g0(x)[g(x)]2=(x2+4)(3) (3x+5)(2x)(x2+4)2=3x2+12 6x2 10x(x2+4)2=12 10x 3x2(x2+4)2:.2.Findg0(x)ifg(x)=2x4+7x2 43x5+x4 x+1:Solution.g0(x)=(3x5+x4 x+1)Dx(2x4+7x2 4) (2x4+7x2 4)Dx(3x5+x4 x+1)(3x5+x4 x+1)2=(3x5+x4 x+1)(8x3+14x) (2x4+7x2 4)(15x4+4x3 1)(3x5+x4 x+1)2:.DIFFERENTIATINGTRIGONOMETRICFUNCTIONSRULE7:Derivativesoftrigonometricfunctions1.Dx(sinx)=cosx2.Dx(cosx)= sinx3.Dx(tanx)=sec2x4.Dx(cotx)= csc2x5.Dx(secx)=secxtanx6.Dx(cscx)= cscxcotxEXAMPLE7:Di erentiatethefollowingfunctions:1.f(x)=secx+3cscx2.g(x)=x2sinx 3xcosx+5sinxSolution.Applyingtheformulasabove,weget1.Iff(x)=secx+3cscx,thenf0(x)=secxtanx+3( cscxcotx)=secxtanx 3cscxcotx:163
2.Ifg(x)=x2sinx 3xcosx+5sinx,theng0(x)=[(x2)(cosx)+(sinx)(2x)] 3[(x)( sinx)+(cosx)(1)]+5cosx=x2cosx+2xsinx+3xsinx 3cosx+5cosx=x2cosx+5xsinx+2cosx:.DIFFERENTIATINGANEXPONENTIALFUNCTIONRULE8:DerivativeofanexponentialfunctionIff(x)=ex,thenf0(x)=ex.EXAMPLE8:1.Findf0(x)iff(x)=3ex.Solution.ApplyingRules3and7,wehavef0(x)=3Dx[ex]=3ex:.2.Findg0(x)ifg(x)= 4x2ex+5xex 10ex.Solution.ApplyingRule5tothe rsttwotermsandRule3tothethirdterm,wehaveg0(x)=[( 4x2)(ex)+(ex)( 8x)]+[(5x)(ex)+(ex)(5) 10ex]= 4x2ex 3xex 5ex:.3.Findh0(x)ifh(x)=exsinx 3excosx.Solution.WeapplytheProductRuletoeachterm.h0(x)=[(ex)(cosx)+(sinx)(ex)] 3[(ex)( sinx)+(cosx)(ex)]=excosx+exsinx+3exsinx 3excosx=ex(4sinx 2cosx):.164
4.Finddydxwherey=17exxe+2x 3px.Solution.UsingQuotientRule(alsoProductRulewhendi erentiatingxeex),weobtaindydx=(exxe+2x 3px)(0) (17)([(ex)(exe 1)+(xe)(ex)]+2 32px)(exxe+2x 3px)2= 17ex+1xe 1 17exxe 34+512px(exxe+2x 3px)2:.SolvedExamplesEXAMPLE1:1.Iff(x)=25,thenf0(x)=0.2.Ifg(x)= p73p2,theng0(x)=0.3.Ifh(x)=e1:2345,thenh0(x)=0.4.Ifr(x)=c,wherec2R,thenr0(x)=0.EXAMPLE2:1.Iff(x)=x3,thenf0(x)=3x3 1=3x2.2.Iff(x)=5px3,thenwecanwritefasf(x)=x35.Thus,f0(x)=35x35 1=35x 25.3.Iff(x)= 1x4,thenfcanbewrittenasf(x)= x 4.Thus,f0(x)= 4x 4 1= 4x 5.EXAMPLE3:1.Iff(x)= 4x43,thenf0(x)= 4 43x43 1= 163x13.2.Ifg(x)=24px3,thengcanbewrittenasg(x)=2x34.Thus,g0(x)=2 34x34 1=32x 14.3.Ifh(x)= p7x2,thenh0(x)= p7 2x2 1= 2p7x.165
EXAMPLE4:1.Iff(x)=3x2 6px+1,thenf0(x)=3 2x1 1 612px+0=6x 3px.2.Iff(x)=x 2 x,thenf0(x)= 2x 3 1.EXAMPLE5:1.Iff(x)=3xsinx.Thenf0(x)=Dx(3x)sinx+(3x)Dx(sinx)=3sinx+3xcosx.2.Iff(x)= 4x3 2x2+3x+1 x 2 px,thenf0(x)= x 2 pxDx 4x3 2x2+3x+1+ 4x3 2x2+3x+1Dx x 2 px= x 2 px 12x2 4x+3+ 4x3 2x2+3x+1 2x 3 12px:EXAMPLE6:Findf0(x)iff(x)=x3 12x2+3x.f0(x)= 2x2+3xDx x3 1 x3 1Dx 2x2+3x(2x2+3x)2= 2x2+3x 3x2 x3 1(4x+3)(2x2+3x)2=6x4+9x3 4x4+3x3 4x 3(2x2+3x)2=2x4+6x3+4x+3(2x2+3x)2:EXAMPLE7:Findg0(x)ifg(x)=2x3ex 4xex+ex.Solution.Notethatgcanbewrittenasg(x)=ex 2x3 4x+1.Thus,byapplyingtheProductRule,wegetg0(x)=ex 2x3 4x+1+ex 6x2 4=ex 2x3 4x+1+6x2 4=ex 2x3+6x2 4x 3=2x3ex+6x2ex 4xex 3ex:.166
EXAMPLE8:Findf0(x)iff(x)=xexx2+2ex.Solution.WeapplytheQuotientRuletoobtainf0(x)= x2+2ex(ex+xex) xex(2x+2ex)(x2+2ex)2=x2ex+x3ex+2e2x+2xe2x 2x2ex 2xe2x(x2+2ex)2=x3ex x2ex+2e2x(x2+2ex)2:.EXAMPLE9:Di erentiatethefollowingtrigonometricfunctions:1.f(x)=cotx+2tanxsecxf0(x)= csc2x+2 sec2xsecx+tanx(secxtanx)= csc2x+2sec3x+2secxtan2x:2.g(x)=3x3cosx 2xsinx+5cscxg0(x)= 9x2cosx+3x3( sinx) 2(sinx+xcosx) 5cscxcotx=9x2cosx 3x3sinx 2sinx 2xcosx 5cscxcotx:SupplementaryProblems1.Findthederivativeofthefollowingfunctions:(a)f(x)= 2x2 3x 4x 2+5(b)f(x)=4x4 5x 32+6x2 3(c)g(x)=5x 3 px 2+x2(d)h(x)=3x2 4x12+3x(e)f(x)=x5 x4+x3 x2+x 1(f)f(x)=x6+x5+x4 x3 x2 x 1(g)h(t)=t+2t2 3t(h)g(u)=pu3 cosu(i)f(x)=sinxsecx+x2(j)g(x)=xxtanx 2xpx167
2.Finddydxandsimplifyifpossible.(a)y=px3 1px3(b)y=x4+ ex(c)y=exx(d)y=sinxcosxtanx(e)y=x2 exsinx3.Findtheequationofthetangentlinestothefollowingcurvesatthegivenpoints.(a)f(x)=3x2 2x+1;x=2(b)g(x)=sinx+cosx;x=4(c)h(x)=x2ex+secx;at(0;1)(d)Findallpointsonthegraphofy=(2x+1)2atwhichthetangentlineisparalleltothelinewithequationy+3x 5=0.(e)Findallpointsonthegraphofy=x33 x2 2xsuchthatthetangentlineisperpendiculartothelinewithequationy+x=2.168
LESSON7:OptimizationLEARNINGOUTCOME:Attheendofthelesson,thelearnershallbeabletosolveopti-mizationproblems.169
TOPIC7.1:OptimizationusingCalculusREVIEWOFMATHEMATICALMODELINGBeforewestartwithproblemsolving,werecallkeyconceptsinmathematicalmodeling.Func-tionsareusedtodescribephysicalphenomena.Forexample:Thenumberofpeopleyinacertainareathatisinfectedbyanepidemicaftersometimet;Theconcentrationcofadruginaperson'sbloodstreamthoursafteritwastaken;Themicepopulationyasthesnakepopulationxchanges,etc.Wemodelphysicalphenomenatohelpuspredictwhatwillhappeninthefuture.Wedothisby ndingorconstructingafunctionthatexhibitsthebehaviorthathasalreadybeenobserved.Inthe rstexampleabove,wewantto ndthefunctiony(t).Forexample,ify(t)=10002 t,thenweknowthatinitially,therearey(0)=1000a ectedpatients.Afteronehour,therearey(1)=10002 1=500a ectedpatients.Observethattheindependentvariablehereistimetandthatthequantityydependsont.Sinceyisdependentont,itnowbecomespossibletooptimizethevalueofybycontrollingatwhichtimetyouwillmeasurey.Wenowlookatsomeexamples.EXAMPLE1:Foreachofthefollowing,determinethedependentquantityQ(x)andtheindependentquantityxonwhichitdepends.Then, ndthefunctionQ(x)thataccuratelymodelsthegivenscenario.1.TheproductPofagivennumberxandthenumberwhichisoneunitbigger..Answer:P(x)=x(x+1)=x2+x2.ThevolumeVofasphereofagivenradiusr.Answer:V(r)=4=3r33.ThevolumeVofarightcircularconewithradius3cmandagivenheighth.Answer:V(h)=3h170
4.Thelength`ofarectanglewithwidth2cmandagivenareaA.Answer:`(A)=A=25.ThetotalmanufacturingcostCofproducingxpencilsifthereisanoverheadcostofP100andproducingonepencilcostsP2.Answer:C(x)=100+2x6.Thevolumeoftheresultingopen-topboxwhenidenticalsquareswithsidexarecuto fromthefourcornersofa3meterby5meteraluminumsheetandthesidesarethenturnedup.Answer:V(x)=x(3 2x)(5 2x):CRITICALPOINTSANDPOINTSWHEREEXTREMAOCCURDe nitionLetfbeafunctionthatiscontinuousonanopenintervalIcontainingx0.Wesaythatx0isacriticalpointoffiff0(x0)=0orf0(x0)doesnotexist(thatis,fhasacorneroracuspat(x0;f(x0))).Wesaythatthemaximumoccursatx0ifthevaluef(x0)isthelargestamongallotherfunctionalvaluesonI,thatis,f(x0)f(x)forallx2I:Wesaythattheminimumoffoccursatx0ifthevaluef(x0)isthesmallestamongalltheotherfunctionalvaluesonI,thatis,f(x0)f(x)forallx2I:Wesaythatanextremumoffoccursatx0ifeitherthemaximumortheminimumoccursatx0.EXAMPLE2:Findallcriticalpointsofthegivenfunctionf.1.f(x)=3x2 3x+42.f(x)=x3 9x2+15x 203.f(x)=x3 x2 x 104.f(x)=x 3x1=35.f(x)=x5=4+10x1=4171
Solution.Wedi erentiatefand ndallvaluesofxsuchthatf0(x)becomeszeroorunde ned.1.Notethatfisdi erentiableeverywhere,socriticalpointswillonlyoccurwhenf0iszero.Di erentiating,wegetf0(x)=6x 3.Therefore,f0(x)=0,6x 3=0,x=1=2.Sox=1=2isacriticalpoint.2.f0(x)=3x2 18x+15=3(x2 6x+5)=3(x 5)(x 1).Hencethecriticalpointsare1and5.3.f0(x)=3x2 2x 1=(3x+1)(x 1):So,thecriticalpointsare 1=3and1.4.f0(x)=1 x 2=3=x2=3 1x2=3:Observethatf0iszerowhenthenumeratoriszero,orwhenx=1.Moreover,f0isunde nedwhenthedenominatoriszero,i.e.,whenx=0.So,thecriticalpointsare0and1.5.f0(x)=54x1=4+104x 3=4=5(x+2)4x3=4:Notethatthedomainoffis[0;1);therefore 2cannotbeacriticalpoint.Theonlycriticalpointis0..FERMAT'STHEOREMANDTHEEXTREMEVALUETHEOREMTheorem7(Fermat'sTheorem).LetfbecontinuousonanopenintervalIcontainingx0.Iffhasanextremumatx0,thenx0mustbeacriticalpointoff.Theorem8(ExtremeValueTheorem).Letfbeafunctionwhichiscontinuousonaclosedandboundedinterval[a;b].Thentheextremevalues(maximumandminimum)offalwaysexist,andtheyoccureitherattheendpointsoratthecriticalpointsoff.EXAMPLE3:Findtheextremaofthegivenfunctionsontheinterval[ 1;1].(Thesefunctionsarethesameasinthepreviousexample.)1.f(x)=3x2 3x+42.f(x)=x3 9x2+15x 203.f(x)=x3 x2 x 104.f(x)=x 3x1=3172
Solution.Wehadalreadysolvedallcriticalpointsoffinthepreviousexercise.Wewillonlyconsiderthosecriticalpointsontheinterval[ 1;1].BytheExtremeValueTheorem,wealsohavetoconsidertheendpoints.So,whatremainstobedoneisthefollowing:Getthefunctionalvaluesofallthecriticalpointsinside[ 1;1];Getthefunctionalvaluesattheendpoints;andComparethevalues.Thehighestoneisthemaximumvaluewhilethelowestoneistheminimumvalue.1.Thereisonlyonecriticalpoint,x=1=2,andtheendpointsarex=1.Wepresentthefunctionalvaluesinatable.x 11=21f(x)1013=44Clearly,themaximumoffoccursatx= 1andhasvalue10.Theminimumoffoccursatx=1=2andhasvalue13=4.2.Thecriticalpointsoffare1and5,butsincewelimitedourdomainto[ 1;1],weareonlyinterestedinx=1.Belowisthetableoffunctionalvaluesforthiscriticalpoint,aswellasthoseattheendpoints.x 11f(x) 45 13Therefore,themaximumvalue 13occursatx=1whiletheminimumvalue 45occursatx= 1.3.Weconsiderthefunctionalvaluesat 1=3; 1and1:x 1 1=31f(x) 11 25=27 11Thus,themaximumpointis( 1=3; 25=27)whiletheminimumpointsare(1; 11)and( 1; 11).4.f(0)=0,f(1)= 2andf( 1)=2.So,themaximumpointis( 1;2)whiletheminimumpointis(1; 2)..173
OPTIMIZATION:APPLICATIONOFEXTREMATOWORDPROBLEMSManyreal-lifesituationsrequireusto ndavaluethatbestsuitsourneeds.Ifwearegivenseveraloptionsforthevalueofavariablex,howdowechoosethe\bestvalue?"Suchaproblemisclassi edasanoptimizationproblem.Wenowapplyourpreviousdiscussionto ndingextremumvaluesofafunctiontosolvesomeoptimizationproblems.SuggestedStepsinSolvingOptimizationProblems1.Ifpossible,drawadiagramoftheproblem.2.Assignvariablestoallunknownquantitiesinvolved.3.Specifytheobjectivefunction.Thisfunctionmustbecontinuous.(a)Identifythequantity,sayq,tobemaximizedorminimized.(b)Formulateanequationinvolvingqandotherquantities.Expressqintermsofasinglevariable,sayx.Ifnecessary,usetheinformationgivenandrelationshipsbetweenquantitiestoeliminatesomevariables.(c)Theobjectivefunctionismaximizeq=f(x)orminimizeq=f(x):4.Determinethedomainorconstraintsofqfromthephysicalrestrictionsoftheproblem.Thedomainmustbeaclosedandboundedinterval.5.Useappropriatetheoremsinvolvingextrematosolvetheproblem.Makesuretogivetheexactanswer(withappropriateunits)tothequestion.EXAMPLE4:Findthenumberintheinterval[ 2;2]sothatthedi erenceofthenumberfromitssquareismaximized.Solution.Letxbethedesirednumber.Wewanttomaximizef(x)=x2 xwherex2[ 2;2].Notethatfiscontinuouson[ 2;2]andthus,wecanapplytheExtremeValueTheorem.174
We rst ndthecriticalnumbersoffintheinterval( 2;2).Wehavef0(x)=2x 1;whichmeansthatweonlyhaveonecriticalnumberin( 2;2):x=12.Thenwecomparethefunctionvalueatthecriticalnumberandtheendpoints.Weseethatf( 2)=6;f(2)=2;f12= 14:Fromthis,weconcludethatfattainsamaximumon[ 2;2]attheleftendpointx= 2.Hence,thenumberwearelookingforis 2..EXAMPLE5:TherangeR(distanceoflaunchsitetopointofimpact)ofaprojectilethatislaunchedatanangle2[0;90]fromthehorizontal,andwitha xedinitialspeedofv0,isgivenbyR()=v20gsin2;wheregistheaccelerationduetogravity.Showthatthisrangeismaximizedwhen=45:v0RangeRSolution.LetR()denotetherangeoftheprojectilethatislaunchedatanangle,measuredfromthehorizontal.WeneedtomaximizeR()=v20gsin2where2[0;=2].NotethatRiscontinuouson[0;=2]andthereforetheExtremeValueTheoremisapplicable.Wenowdi erentiateRto ndthecriticalnumberson[0;=2]:R0()=v20g2cos2=0()2==2:175
Hence==4=45isacriticalnumber.Finally,wecomparethefunctionalvalues:f(0)=0;f(=4)=v20g;f(=2)=0:Thus,weconcludethatfattainsitsmaximumat==4,withvaluev20=g..EXAMPLE6:Anopen-typedrectangularboxistobemadefromapieceofcardboard24cmlongand9cmwidebycuttingoutidenticalsquaresfromthefourcornersandturningupthesides.Findthevolumeofthelargestrectangularboxthatcanbeformed.ssssssss24cm9cmSolution.Letsbethelengthofthesideofthesquarestobecutout,andimaginethe\
aps"beingturneduptoformthebox.Thelength,widthandheightoftheboxwouldthenbe24 2s;9 2s;ands,respectively.Therefore,thevolumeoftheboxisV(s)=(24 2s)(9 2s)s=2(108s 33s2+2s3):WewishtomaximizeV(s)butnotethatsshouldbenonnegativeandshouldnotbemorethanhalfthewidthofthecardboard.Thatis,s2[0;4:5].(Thecases=0ors=4:5doesnotproduceanyboxbecauseoneofthedimensionswouldbecomezero;buttomaketheintervalclosedandbounded,wecanthinkofthosecasesasdegenerateboxeswithzerovolume).SinceVisjustapolynomial,itiscontinuousontheclosedandboundedinterval[0;4:5].Thus,theExtremeValueTheoremapplies.NowV0(s)=216 132s+12s2=4(54 33s+3s2)=4(3s 6)(s 9)andhencetheonlycriticalnumberofVin(0;4:5)is2(s=9isoutsidetheinterval).Wenowcomparethefunctionalvaluesattheendpointsandatthecriticalpoints:s024:5V(s)02000176
Therefore,fromthetable,weseethatVattainsitsmaximumats=2,andthemaximumvolumeisequaltoV(2)=200cm3..EXAMPLE7:Determinethedimensionsoftherightcircularcylinderofgreatestvolumethatcanbeinscribedinarightcircularconeofradius6cmandheight9cm.h6in9inrSolution.Lethandrrespectivelydenotetheheightandradiusofthecylinder.Thevolumeofthecylinderisr2h.Lookingatthecentralcross-sectionofthecylinderandthecone,wecanseesimilartriangles,andso66 r=9h:(2.3)WecannowwriteourobjectivefunctionasV(r)=9r2 32r3=3r23 r2;anditistobemaximized.Clearly,r2[0;6].SinceViscontinuouson[0;6],theExtremeValueTheoremisapplicable.Now,V0(r)=18r 92r2=9r2 r2andhence,ouronlycriticalnumberon(0;6)is4.Wenowcomparethefunctionalvaluesattheendpointsandatthecriticalpoints:r046V(r)0480177
Weseethatthevolumeismaximizedwhenr=4,withvalueV(4)=48.To ndthedimensions,wesolveforhfrom(2.3).Ifr=4,66 4=9h=)h=3:Therefore,thelargestcircularcylinderthatcanbeinscribedinthegivenconehasdimensionsr=4cmandheighth=3cm..EXAMPLE8:Angelo,whoisinarowboat2kilometersfromastraightshoreline,wishestogobacktohishousewhichisontheshorelineand6kilometersfromthepointontheshorelinenearestAngelo.Ifhecanrowat6kphandrunat10kph,howshouldheproceedinordertogettohishouseintheleastamountoftime?shorelineAngelo'shouseStartingPointP6km2kmcSolution.LetcbethedistancebetweenthehouseandthepointPontheshorefromwhichAngelowillstarttorun.UsingthePythagoreanTheorem,weseethatthedistancehewilltravelbyboatisp4+(6 c)2.Notethatspeed=distancetime.Thus,hewillsailforp4+(6 c)26hoursandrunforc10hours.WewishtominimizeT(c)=p4+(6 c)26+c10:Wecanassumethatc2[0;6].Wenowdi erentiateT.Observethatourpreviouslydiscussedrulesofdi erentiationarenotapplicabletothisfunctionbecausewehavenotyetdiscussedhowtodi erentiatecompositionsoffunctions.Weusethede nitionofthederivativeinstead.178
T0(c)=limx!cT(x) T(c)x c=limx!c p4+(6 x)26+x10! p4+(6 c)26+c10!x c=110limx!cx cx c+16limx!c p4+(6 x)2 p4+(6 c)2x c!:Wenowrationalizetheexpressioninthe rstlimitbymultiplyingthenumeratoranddenomi-natorbyp4+(6 x)2+p4+(6 c)2.Thisyields:T0(c)=110+16limx!c(4+(6 x)2) (4+(6 c)2)(x c)(p4+(6 x)2+p4+(6 c)2=110+16limx!c 12(x c)+(x c)(x+c)(x c)(p4+(6 x)2+p4+(6 c)2)=110+16limx!c 12+x+cp4+(6 x)2+p4+(6 c)2=110+16 12+2c2p4+(6 c)2:Solvingforthecriticalnumbersoffon(0;6),wesolveT0(c)=110+16 12+2c2p4+(6 c)2=3p4+(6 c)2 30+5c30p4+(6 c)2=0;andwegetc=92.Comparingfunctionvaluesattheendpointsandthecriticalnumber,T(0)=p406;T92=1315;T(6)=1415;weseethattheminimumofTisattainedatc=92.Thus,AngelomustrowuptothepointPontheshore92kilometersfromhishouseand32kilometersfromthepointontheshorenearesthim.Thenhemustrunstraighttohishouse..SolvedExamplesEXAMPLE1:Foreachofthefollowing,provideamodelthatdepictsthe rstquantitybeingexpressedasafunctionoftheotherquantities.179
1.ThequotientQofanumberxandthesumofitsabsolutevalueand1.2.TheproductPofanumberxandthenumberequalto2lessthanitssquare.3.ThevolumeVofaconeofradiusrandheight2r.4.ThevolumeVofarectangularprismwithlengthl,height2l,andwidth3l.5.ThelengthlofasquareofareaA.6.TheareaAofa4mby6mcardboardsheetwhenidenticalsquareswithsidexarecuto fromthefourcornersofthesheet.Solution.1.Q(x)=xjxj+12.P(x)=x x2 2=x3 2x3.V(r)=13r2(2r)=23r34.V(l)=l2l3l=6l35.l(A)=pA6.A(x)=24 4x2.EXAMPLE2:Findallthecriticalpointsofthegivenfunctions.1.f(x)=4x2 3x=52.f(x)=x33+x2+x+53.f(x)=sinx4.f(x)=ex x5.f(x)=lnx180
Solution.1.Togetthecriticalpointsoff,wedi erentiatefandlookforthezeroesoff0.Indeed,f0(x)=8x 3.Thus,thecriticalpointsarethesolutionsof8x 3=0.Hence,thecriticalpointisx=38.2.f0(x)=x2+2x+1=(x+1)2.Hence,thecriticalpointisx= 1.3.f0(x)=cosx.Thisiszerowhenx=k2,wherekisanoddinteger.4.f0(x)=ex 1.Thus,wewanttolookforxsuchthatex=1.Thecriticalpointisx=0.5.f0(x)=1x.Thisfunctionhasnozeroes.Also,eventhoughf0isunde nedattheorigin,x=0isnotclassi edasacriticialpointsinceitisnotinthedomainoff..EXAMPLE3:Findtheextremaofthegivenfunctionsintheinterval[0;10].1.f(x)=x3 12x+32.f(x)=4x2+5x 13.f(x)=5x3+3x2+4x+54.f(x)=exSolution.1.f0(x)=3x2 12.Equatingthistozero,wehavethat3x2=12,orthatx2=4.Thus,thecriticalpointsarex=2.Wenowevaluatef(0),f(10),andf(2)andcomparethesevalues.Notethatwedidnotevaluatef( 2)since 2=2[0;10].f(0)=3,f(10)=1000 120+3=883andf(2)=8 24+3= 13.Thus,theminimumis(2; 13)andthemaximumis(10;883).2.f0(x)=8x+5.Thus,thecriticalpointisx= 58.Weevaluatethefollowing:f(0)= 1,f(10)=400+50 1=449,andf 58=4 582+5 58 1= 4116.So,theminimumis 58; 4116andthemaximumis(10;449).181
3.f0(x)=15x2+6x+4.Applyingthequadraticformula,weobtainx= 6p36 4(15)(4)30= 6p 20430= 62ip5130:Sincethesearecomplexnumberswithnonzeroimaginaryparts,fhasnocriticalpoints.Wenowevaluatef(0)=5andf(10)=5345.Thus,theminimumis(0;5)andthemaximumis(10;5345).4.f0(x)=ex.Thisfunctionhasnozeroesandsofhasnocriticalpoints.However,f0(x)existsforanyx2R.Comparingf(0)=1andf(10)=e10,weseethatf(10)=e10>1=f(0).Therefore,theminimumis(0;1)andthemaximumis 10;e10..EXAMPLE4:Findtwonumberswhoseproductis25andwhosesumisamaximum.Solution.Wewanttomaximizethesumoftwonumberswhoseproductis25.Inordertowritethisasafunctionofasinglevariable,wehavetowriteoneofthenumbersintermsoftheother.Sincetheirproductis25,wecandenoteasxoneofthenumbers,x6=0,and25xastheotherone.Therefore,weareinterestedinmaximizingthefunctionf(x)=x+25x.Di erentiatingthis,weobtainf0(x)=1 25×2.Equatingthistozero,weobtainx2=25,andsox=5.Toobtainthemaximum,wesubstitutetheseintof.So,f(5)=5+255=10andf( 5)= 5+25 5= 10.Since10> 10;wechoosethepair5and5..EXAMPLE5:If1000cm2ofcardboardisavailabletomakeaboxwithasquarebase, ndthelargestpossiblevolumeofthebox.Solution.Ifweletxbethelengthofasideofthesquarebaseandhbetheheightofthebox,thetotalsurfaceareaofthebox,SisgivenbyS=2×2+4xh.Sincetheboxistobemadefrom1000cm2ofcardboard,wehavethat1000=2×2+4xh,orthatxh=500 x22.Thevolumeofthisbox,V,isgivenbyV=x2h=x(xh)=x500 x22=500x x32.Tomaximizethis,weobtainV0(x)=500 3×22.Togetthecriticalpoint,wesolve500 3×2=0.182
Thusx=10p153.Sincexdenotesthelengthofasideofthebase,itmustbeapositivequantity.Thus,themaximumvolumeoccursatV 10p153!=10p1530BBBBB@500 10p153!231CCCCCA=10p1530B@500 500331CA=10p15310009=10000p1527cm3:.EXAMPLE6:Findthepointontheliney=2x+3thatisclosesttotheorigin.Solution.Anarbitrarypointontheliney=2x+3is(x;y)=(x;2x+3).Thus,thedistance,d,ofthispointfromtheorigincanbewrittenasd(x)=q(x 0)2+(2x+3 0)2:Observethatsincethesquarerootfunctionf(x)=pxisincreasingon[0;+1),disminimizedifandonlyifd2isminimized.Wedothistoevaluatesimplerderivatives.Thus,weminimized2(x)=x2+(2x+3)2.Wegetthecriticalpointsofd2. d2(x)0=2x+2(2x+3)(2)=2x+8x+12=10x+12:Equatingthistozero,wegetacriticalpointx= 65.Clearly,d2becomesarbitrarilylargeasyouinputlargervaluesofx.Thus,x= 65wouldgivetheminimum.Gettingthecorrespondingy-coordinate,wehavey=2 65+3=35.Thus,thepointweareinterestedinis 65;35..183
SupplementaryProblems1.Findtheextremaofthegivenfunctionsinthespeci edinterval.(a)f(x)=x2+2on[1;10](b)f(x)=x2 5x+6on[ 20;20](c)f(x)=sinx+cosxonh0;2i(d)f(x)=ex xon[0;1](e)f(x)=23×3 52×2 3x+1on[ 2;5](f)f(x)=34×4 13×3 32×2+x 1on[ 1;0](g)f(x)=3×4+4×3 12×2+24on[ 1;2](h)f(x)=x4 18×2+4on[ 4;5](i)f(x)=esinxonh2;i(j)f(x)=ecosxonh 2;2i2.Answerthefollowingoptimizationproblemssystematically.(a)Findtwonumberswhosesumis20andwhoseproductisamaximum.(b)Findtwonumberswhosedi erenceis50andwhoseproductisaminimum.(c)Findapositivenumbersuchthatthesumofthenumberandtwiceitsreciprocalisassmallaspossible.(d)Findapositivenumbersuchthatthedi erenceofthenumberanditsreciprocalisaslargeaspossible.(e)Findthedimensionsofarectanglewithperimeter200mandwhoseareaisaslargeaspossible.(f)Findthedimensionsofarectanglewithareais625m2whoseperimeterisassmallaspossible.(g)Findthepointontheliney=3x 1closesttothepoint( 1;1).(h)Findthepointontheparabola,y=x2thatisclosesttothepoint(0;0).(i)Findthemaximumpro tgiventhatthepro tfunctionisP(x)= x4+2×2+1.(j)Findanequationofthetangentlinetoy=x3 4×2+5x+1havingtheleastslope.184
LESSON8:Higher-OrderDerivativesandtheChainRuleLEARNINGOUTCOMES:Attheendofthelesson,thelearnershallbeableto:1.Computehigher-orderderivativesoffunctions;2.IllustratetheChainRuleofdi erentiation;and3.SolveproblemsusingtheChainRule.185
TOPIC8.1:Higher-OrderDerivativesofFunctionsConsiderthefunctiony=f(x).Thederivativey0;f0(x);Dxyordydxiscalledthe rstderiva-tiveoffwithrespecttox.Thederivativeofthe rstderivativeiscalledthesecondderivativeoffwithrespecttoxandisdenotedbyanyofthefollowingsymbols:y00;f00(x);D2xy;d2ydx2Thethirdderivativeoffwithrespecttoxisthederivativeofthesecondderivativeandisdenotedbyanyofthefollowingsymbols:y000;f000(x);D3xy;d3ydx3Ingeneral,thenthderivativeoffwithrespecttoxisthederivativeofthe(n 1)stderivativeandisdenotedbyanyofthefollowingsymbols:y(n);f(n)(x);Dnxy;dnydxnFormally,wehavethefollowingde nition.De nition4.Thenthderivativeofthefunctionfisde nedrecursivelybyf0(x)=limx!0f(x+x) f(x)xforn=1,andf(n)(x)=limx!0f(n 1)(x+x) f(n 1)(x)xforn>1;providedthattheselimitsexist.Thus,thenthderivativeoffisjustthederivativeofthe(n 1)stderivativeoff.Remark1:1.Thefunctionfcanbewrittenasf(0)(x).2.Inthenotationf(n)(x),niscalledtheorderofthederivative.EXAMPLE1:1.Findthefourthderivativeofthefunctionf(x)=x5 3×4+2×3 x2+4x 10.186
Solution.Wedi erentiatethefunctionrepeatedlyandobtainf0(x)=5×4 12×3+6×2 2x+4f00(x)=20×3 36×2+12x 2f000(x)=60×2 72x+12f(4)(x)=120x 72:.2.Findthe rstandsecondderivativesofthefunctionde nedbyy=(3×2 4)(x2 3x):Solution.UsingProductRule,wecomputethe rstderivative:y0=(3×2 4)Dx(x2 3x)+(x2 3x)Dx(3×2 4)=(3×2 4)(2x 3)+(x2 3x)(6x)=6×3 9×2 8x+12+6×3 18×2=12×3 27×2 8x+12:Similarly,weobtainthesecondderivative:y00=Dx(12×3 27×2 8x+12)=36×2 54x 8:.3.Lety=3x+5×2+4.Findd2ydx2.Solution.UsingQuotientRuletwice,wehavedydx=(x2+4)Dx(3x+5) (3x+5)Dx(x2+4)(x2+4)2=(x2+4)(3) (3x+5)(2x)(x2+4)2=3×2+12 6×2 10x(x2+4)2=12 10x 3×2(x2+4)2;187
andd2ydx2=ddx12 10x 3×2(x2+4)2=ddx12 10x 3x2x4+8×2+16=(x4+8×2+16)ddx(12 10x 3×2) (12 10x 3×2)ddx(x4+8×2+16)(x4+8×2+16)2=(x4+8×2+16)( 10 6x) (12 10x 3×2)(4×3+16x)(x4+8×2+16)2=6×5+30×4 48×3+80×2 288x 160(x4+8×2+16)2:.4.Findthethirdderivativeofthefunctionde nedbyg(x)= 4x2ex+5xex 10ex.Solution.Wedi erentiaterepeatedly(applyingtheProductRule)andobtaing(1)(x)=[( 4×2)(ex)+(ex)( 8x)]+[(5x)(ex)+(ex)(5) 10ex]= 4x2ex 3xex 5ex=ex( 4×2 3x 5):g(2)(x)=(ex)( 8x 3)+( 4×2 3x 5)(ex)=ex( 4×2 11x 8):g(3)(x)=(ex)( 8x 11)+( 4×2 11x 8)(ex)=ex( 4×2 19x 19):.5.Iff(x)=exsinx 3excosx, ndf(5)(x).Solution.Wedi erentiaterepeatedly(applyingProductRule)andobtain188
f0(x)=[(ex)(cosx)+(sinx)(ex)] 3[(ex)( sinx)+(cosx)(ex)]=excosx+exsinx+3exsinx 3excosx=ex(4sinx 2cosx):f00(x)=ex[4(cosx) 2( sinx)]+(4sinx 2cosx)(ex)=ex(2cosx+6sinx):f000(x)=ex[2( sinx)+6(cosx)]+(2cosx+6sinx)(ex)=ex(8cosx+4sinx):f(4)(x)=ex[8( sinx)+4(cosx)]+(8cosx+4sinx)(ex)=ex(12cosx 4sinx):f(5)(x)=ex[12( sinx) 4(cosx)]+(12cosx 4sinx)(ex)=ex(8cosx 16sinx):.SolvedExamplesEXAMPLE1:Findthethirdderivativeofthefunctionf(x)=x4 3×3+2×2+2x+1.Solution.Wedi erentiatethefunctionrepeatedlytogetthethirdderivative.f0(x)=4×3 9×2+4x+2f00(x)=12×2 18x+4f000(x)=24x 18:.EXAMPLE2:Findthe rstandsecondderivativesoff(x)= 2×2 3x+3 x3 1.Solution.Togetthe rstderivative,weapplyProductRule.f0(x)=(4x 3) x3 1+ 2×2 3x+3 3×2= 4×4 3×3 4x+3+ 6×4 9×3+9×2=10×4 12×3+9×2 4x+3:189
Wedi erentiatethistogetthesecondderivative.f00(x)=40×3 36×2+18x 4:.EXAMPLE3:Lety=5 6x2x2 1.Findd2ydx2.Solution.Togetdydx,weapplyQuotientRule.dydx= x2 1( 12x) 5 6×2(2x)(x2 1)2= 12×3+12x 10x 12×3(x2 1)2=2x(x2 1)2:ApplyingQuotientRuleagain,weobtaind2ydx2= x4 2×2+1(2) (2x) 4×3 4x[(x2 1)2]2=2×4 4×2+2 8×4+8×2(x2 1)4= 6×4+6×2+2(x2 1)4:.EXAMPLE4:Findthethirdderivativeofg(x)=xex+sinx.Solution.Wedi erentiaterepeatedlytogetg0(x)=ex+xex+cosxg00(x)=ex+ex+xex sinx=2ex+xex sinxg000(x)=2ex+ex+xex cosx=3ex+xex cosx:.190
EXAMPLE5:Iff(x)=exsinx,whatisf(5)(x)?Solution.f0(x)=exsinx+excosx=ex(sinx+cosx)f00(x)=ex(sinx+cosx)+ex(cosx sinx)=2excosxf000(x)=2excosx 2exsinx=2ex(cosx sinx)f(4)(x)=2ex(cosx sinx)+2ex( sinx cosx)= 4exsinxf(5)(x)=( 4exsinx)+( 4excosx)= 4ex(sinx+cosx):.EXAMPLE6:Lety=xexsinx.Findd2ydx2.Solution.Observethatiff(x)=r(x)s(x)t(x),thenwecanregroupfasf(x)=[r(x)s(x)][t(x)].Applyingnowtheproductrule,weobtain,f0(x)=r0(x)s(x)+r(x)s0(x)[t(x)]+[r(x)s(x)]t0(x)=r0(x)s(x)t(x)+r(x)s0(x)t(x)+r(x)s(x)t0(x):Weletr(x)=x,s(x)=ex,andt(x)=sinx.Applyingtheaboveformula,wegetthatdydx=exsinx+xexsinx+xexcosx:Repeatingtheprocess,weobtaind2ydx2=exsinx+excosx+exsinx+xexsinx+xexcosx+excosx+xexcosx xexsinx=2exsinx+2excosx+2xexcosx=2ex(sinx+(1+x)cosx):191
.SupplementaryProblems1.Findy0;y00;y000andy(4)forthefollowingexpressions.(a)y=2×4(b)y=x 3(c)y=x3(d)y=x23(e)y=x1:2(f)y=xe2(g)y=xp2.Findthe rstandsecondderivativesforthefollowing:(a)f(x)=(sinx)px(b)g(x)=x2ex 2cosx(c)h(x)=x3+3×2+xex(d)f(x)=tanx secx+cscx(e)f(x)=xsecx 4x 2(f)f(x)=2×5 4×3+12x 6.(g)Letf(x)= cosx.Findf04,f004,andf0004.(h)Findf(5)(1)iff(x)=x5 4×3+2x 1.(i)FindD2016x(cosx+sinx).(j)FindD2016x(ex).(k)Findf00()iff(x)=sinxcosx.(l)Findf000(4)iff(x)=px 2×2.(m)Considerf(x)=x+1,g(x)=x2+x+1,h(x)=x3 2x+1.Whatisf00(x);g000(x),andh(4)(x)?Ingeneral,iff(x)=anxn+an 1xn 1++a1x+a0,whereaj2Rforj=1;:::;n,whatisf(n+1)(x)?192
TOPIC8.2:TheChainRuleTheChainRulebelowprovidesforaformulaforthederivativeofacompositionoffunctions.Theorem9(ChainRule).Letfbeafunctiondi erentiableatcandletgbeafunctiondi erentiableatf(c).Thenthecompositiongfisdi erentiableatcandDx(gf)(c)=g0(f(c))f0(c):Remark1:AnotherwaytostatetheChainRuleisthefollowing:Ifyisadi erentiablefunctionofude nedbyy=f(u)anduisadi erentiablefunctionofxde nedbyu=g(x),thenyisadi erentiablefunctionofx,andthederivativeofywithrespecttoxisgivenbydydx=dydududx:Inwords,thederivativeofacompositionoffunctionsisthederivativeoftheouterfunctionevaluatedattheinnerfunction,timesthederivativeoftheinnerfunction.EXAMPLE1:1.Recallour rstillustrationf(x)=(3×2 2x+4)2.Findf0(x)usingChainRule.Solution.Wecanrewritey=f(x)=(3×2 2x+4)2asy=f(u)=u2whereu=3×2 2x+4,adi erentiablefunctionofx.UsingtheChainRule,wehavef0(x)=y0=dydududx=(2u)(6x 2)=2(3×2 2x+4)(6x 2)=36×3 36×2+56x 16:.2.Forthesecondillustration,wehavey=sin(2x).Findy0usingChainRule.Solution.Wecanrewritey=sin(2x)asy=f(u)wheref(u)=sinuandu=2x.Hence,y0=dydududx=cosu2=2cos(2x):193
.3.Finddzdyifz=4(a2 y2)2,whereaisarealnumber.Solution.Noticethatwecanwritez=4(a2 y2)2asz=4(a2 y2) 2.ApplyingtheChainRule,wehavedzdy=4 2(a2 y2) 2 1ddy(a2 y2)= 8(a2 y2) 3 2y=16y(a2 y2) 3=16y(a2 y2)3:.Now,supposewewantto ndthederivativeofapowerfunctionofx.Thatis,weareinterestedinDx[f(x)n].Toderiveaformulaforthis,welety=unwhereuisadi erentiablefunctionofxgivenbyu=f(x).ThenbytheChainRule,dydx=dydududx=nun 1f0(x)=n[f(x)]n 1f0(x)Thus,Dx[f(x)]n=n[f(x)]n 1f0(x).ThisiscalledtheGENERALIZEDPOWERRULE.EXAMPLE2:1.Whatisthederivativeofy=(3×2+4x 5)5?Solution.Dx[(3×2+4x 5)5]=5(3×2+4x 5)5 1Dx(3×2+4x 5)=5(3×2+4x 5)4(6x+4):.194
2.Finddydxwherey=p3x3+4x+1.Solution.dydx=12(3×3+4x+1)12 1Dx(3×3+4x+1)=12(3x34x+1) 12(9×2+4)=9×2+42p3x34x+1:.3.Finddydxwherey=(sin3x)2.Solution.dydx=2(sin3x)2 1ddx(sin3x)=2(sin3x)cos3xddx(3x)=2(sin3x)(cos3x)3=6sin3xcos3x:.4.Di erentiate(3×2 5)3.Solution.d(3×2 5)3dx=3(3×2 5)3 1Dx(3×2 5)=3(3×2 5)26x=18x(3×2 5)2:.5.Considerthefunctionsy=3u2+4uandu=x2+5.Finddydx.Solution.ByChainRule,wehavedydx=dydududxwheredydu=6u+4anddudx=2x.Thus,dydx=dydududx=(6u+4)(2x)=6(x2+5)+4(2x)=(6×2+34)(2x)=12×3+68x:195
.SolvedExamplesEXAMPLE1:Letf(x)=px2+3x+1.Findf0(x).Solution.Notethatwecanviewfasf(x)=(gh)(x),whereg(x)=pxandh(x)=x2+3x+1.ApplyingChainRule,wegetf0(x)=g0(h(x))h0(x)=12px2+3x+1(2x+3)=2x+32px2+3x+1:.EXAMPLE2:Supposey=3 x2 12+sin x2 1.Finddydx.Solution.Weletu=x2 1.Thus,wecanwriteyasy=3u2+sinu.ApplyingChainRule,weobtaindydx=dydududx=(6u+cosu)(2x)= 6 x2 1+cos x2 1(2x)=12 x2 1+2xcos x2 1:.EXAMPLE3:Letg(x)=sin(cosx).Whatisg0(x)?Solution.Wenotethatgcanbeexpressedasg(x)=f(h(x)),wheref(x)=sinxandh(x)=cosx.ByChainRule,thederivativeisg0(x)=f0(h(x))h0(x)=cos(cosx)( sinx)= (sinx)(cos(cosx)):.196
EXAMPLE4:Finddydxify=3ptan2(x3 1).Solution.Inthisproblem,wemakerepeateduseofChainRule,startingfromtheoutermostfunction.dydx=13 tan2 x3 113 1ddx tan2 x3 1=13 tan2 x3 1 232tan x3 1ddx tan x3 1=13 tan2 x3 1 232tan x3 1sec2 x3 1ddx x3 1=13 tan2 x3 1 232tan x3 1sec2 x3 1 3×2=2x2tan x3 1sec2 x3 13q(tan2(x3 1))2=2x2tan x3 1sec2 x3 13ptan4(x3 1):.EXAMPLE5:Supposef(x)=(sin(2x)+1)7.Findf02.Solution.We rstobtainf0.f0(x)=7(sin(2x)+1)6ddx(sin(2x)+1)=7(sin(2x)+1)6(2cos(2x))=14cos(2x)(sin(2x)+1)6:Wenowevaluatethisexpressionatx=2.f02=14cos22sin22+16=14( 1)(0+1)6= 14:.197
EXAMPLE6:Letf(x)=tanx+1x 1.Findf0(x).Solution.f0(x)=sec2x+1x 1ddxx+1x 1=sec2x+1x 1(x 1)(1) (x+1)(1)(x 1)2=sec2x+1x 1 2(x 1)2= 2(x 1)2sec2x+1x 1:.SupplementaryProblems1.Finddydx.(a)y= 3×2 4x+53(b)y=psinx+cosx(c)y=2ex2+5 cos e2x(d)y=q1+p1+p1+x(e)y=(tanpx)13(f)y=esecx(g)y=eex(h)y=sintanex3 1(i)y=sin x4 1p1+2×2(j)y=ex3px2 12.Findthederivativesofthefollowingfunctions:(a)f(x)= csc x2+1 cot x2+1(b)g(x)= 1 x2 2(c)h(x)= x3 4×2+152(d)f(x)=px2 ex(e)g(x)=e2x+1 px3+2198
3.Evaluatethederivativesofthegivenfunctionattheprescribedfollowingvalues:(a)f(x)=cos(ex 1),x=0(b)f(x)=ex2+2x+1,x=1(c)g(x)=tan cot x2,x=p4(d)h(x)=sin(ex),x=0(e)f(x)=p3x2+5x 1,x=2199
LESSON9:ImplicitDi erentiationLEARNINGOUTCOMES:Attheendofthelesson,thelearnershallbeableto:1.Illustrateimplicitdi erentiation;2.Applythederivativesofthenaturallogarithmicandinversetangentfunctions;and3.Useimplicitdi erentiationtosolveproblems.200
TOPIC9.1:WhatisImplicitDi erentiation?Wehaveseenthatfunctionsarenotalwaysgivenintheformy=f(x)butinamorecomplicatedformthatmakesitdicultorimpossibletoexpressyexplicitlyintermsofx.Suchfunctionsarecalledimplicitfunctions,andyissaidtobede nedimplicitly.Inthislesson,weexplainhowthesecanbedi erentiatedusingamethodcalledimplicitdi erentiation.Di erentiatingquantitiesinvolvingonlythevariablexwithrespecttoxisnotaproblem;forinstance,thederivativeofxisjust1.Butifafunctionyisde nedimplicitly,thenweneedtoapplytheChainRuleingettingitsderivative.So,whilethederivativeofx2is2x,thederivativeofy2is2ydydx:Moregenerally,ifwehavetheexpressionf(y),whereyisafunctionofx,thenddx(f(y))=ddy(f(y))dydx:Inordertomasterimplicitdi erentiation,studentsneedtoreviewandmastertheapplicationoftheChainRule.Considerasimpleexpressionsuchasy2=4x.Itsgraphisaparabolawithvertexattheoriginandopeningtotheright. 4 22468 8 6 4 224680Ifweconsideronlytheupperbranchoftheparabola,thenybecomesafunctionofx.Wecanobtainthederivativedy=dxbyapplyingtheChainRule.Whendi erentiatingtermsinvolvingy,weareactuallyapplyingtheChainRule,thatis,we rstdi erentiatewithrespecttoy,and201
thenmultiplybydy=dx.Di erentiatingbothsideswithrespecttox,wehavey2=4x;=)ddx(y2)=ddx(4x)=)2ydydx=4:Solvingfordy=dx,weobtaindydx=42y=2y:Noticethatthederivativecontainsy.Thisistypicalinimplicitdi erentiation.Letusnowuseimplicitdi erentiationto ndthederivativesdy=dxinthefollowingexamples.Letusstartwithouroriginalprobleminvolvingthecircle.EXAMPLE1:Findtheslopeofthetangentlinetothecirclex2+y2=5atthepoint( 2;1). 4 22 4 2240(-2,1)x2+y2=5=)ddx(x2+y2)=ddx(5)=)ddx(x2)+ddx(y2)=0=)2x+2ydydx=0:Solution.Solvingfordy=dx,weobtaindydx= 2x2y= xy:Substitutingx= 2andy=1,we ndthattheslopeisdydx=2:202
Noticethatthisisafasterandeasierwaytoobtainthederivative..EXAMPLE2:Finddydxfory3+4y2+3x2y+10=0.Solution.Di erentiatingbothsidesoftheequationgivesddx(y3+4y2+3x2y+10)=ddx(0)=)ddx(y3)+ddx(4y2)+ddx(3x2y)+ddx(10)=0=)3y2dydx+8ydydx+3x2dydx+6xy+0=0:Wecollectthetermsinvolvingdy=dxandrearrangetogetdydx(3y2+8y+3×2)+6xy=0:Thus,dydx= 6xy3y2+8y+3×2:Notethatthederivativeoftheterm3x2yisobtainedbyapplyingProductRule.Weconsider3x2asonefunctionandyasanotherfunction..Implicitdi erentiationcanbeappliedtoanykindoffunction,whethertheyarepolynomialfunctions,orfunctionsthatinvolvetrigonometricandexponentialquantitites.EXAMPLE3:Finddydxforx+y3=exy4.Solution.Di erentiatingbothsideswithrespecttoxgivesddx(x+y3)=ddx(exy4)=)1+3y2dydx=exy4ddx(xy4)=)1+3y2dydx=exy44xy3dydx+y4Collectingalltermswithdy=dxgivesdydx3y2 exy44xy3= 1+exy4y4=)dydx= 1+exy4y43y2 exy44xy3:.203
DERIVATIVESOFTHENATURALLOGARITHMICANDINVERSETANGENTFUNC-TIONSDerivativesoftheNaturalLogarithmicandInverseTangentFunctionsSupposeuisafunctionofx.Thenddx(lnu)=1ududxddx(tan 1u)=11+u2dudxEXAMPLE4:Finddy=dx.1.y=ln(7×2 3x+1)2.y=tan 1(2x 3cosx)3.y=ln(4x+tan 1(lnx))Solution..1.dydx=17×2 3x+1(14x 3)2.dydx=11+(2x 3cosx)2(2+3sinx)3.dydx=14x+tan 1(lnx)4+11+(lnx)21x:.EXAMPLE5:Finddydxforcos(y2 3)=tan 1(x3)+lny.Solution.Di erentiatingbothsidesgivesddx(cos(y2 3))=ddx(tan 1(x3)+lny)=) sin(y2 3)2ydydx=11+(x3)23×2+1ydydx:204
Collectingtermswithdy=dx:dydx 2ysin(y2 3) 1y=3×21+x6=)dydx=3×21+x6 2ysin(y2 3) 1y:.EXAMPLE6:Finddydxfortan 1y=3x2y pln(x y2):Solution.Di erentiatingbothsideswithrespecttoxgivesddx tan 1y=ddx3x2y (ln(x y2))1=2=)11+y2dydx=3x2dydx+6xy 12 ln(x y2) 1=21x y21 2ydydx=)11+y2dydx=3x2dydx+6xy (ln(x y2)) 1=22(x y2)+y(ln(x y2)) 1=2x y2dydx:Wenowisolatedy=dx:dydx 11+y2 3×2 y(ln(x y2)) 1=2x y2!=6xy (ln(x y2)) 1=22(x y2)=)dydx=6xy (ln(x y2)) 1=22(x y2)11+y2 3×2 y(ln(x y2)) 1=2x y2:.SolvedExamplesEXAMPLE1:Findtheequationofthetangentlinetothecurveof4x2+9y2=36atthepoint(0;2).205
Solution.4×2+9y2=36ddx 4×2+9y2=ddx(36)ddx 4×2+ddx 9y2=08x+18ydydx=018ydydx= 8xdydx= 4x9y:Togettheslopeofthetangentline,weevaluatethederivativeatthepoint(0;2).Thus,dydx= 4(0)9(2)=0:Usingthepointslopeform,wegety 2=0(x 0)y=2:.EXAMPLE2:Finddydxforxy2+4y3 6x=10.Solution.ddx xy2+4y3 6x=ddx(10)ddx xy2+ddx 4y3 ddx(6x)=0y2+2xydydx+12y2dydx 6=0 2xy+12y2dydx=6 y2dydx=6 y22xy+12y2:.EXAMPLE3:Finddydxforsin(xy) cos(x+y)=ln(x y).206
Solution.ddx(sin(xy) cos(x+y))=ddx(ln(x y))ddx(sin(xy)) ddx(cos(x+y))=1x y1 dydxcos(xy)y+xdydx+sin(x+y)1+dydx=1x y1 dydxxcos(xy)+sin(x+y)+1x ydydx=1x y ycos(xy) sin(x+y)Therefore,dydx=1x y ycos(xy) sin(x+y)xcos(xy)+sin(x+y)+1x y:.EXAMPLE4:Finddydxify2 3xexy=sin(cos(xy)).Solution.ddx y2 3xexy=ddx(sin(cos(xy)))ddx y2 3ddx(xexy)=cos(cos(xy))ddx(cos(xy))2ydydx 3exy+xddx(exy)=cos(cos(xy))( sin(xy))y+xdydx2ydydx 3exy+xexyy+xdydx= cos(cos(xy))(sin(xy))y+xdydx 2y 3x2exy+xsin(xy)cos(cos(xy))dydx=3exy+3xyexy ycos(cos(xy))sin(xy)Therefore,dydx=3exy+3xyexy ycos(cos(xy))sin(xy)2y 3x2exy+xsin(xy)cos(cos(xy)):.EXAMPLE5:Finddydxforxtany py2+2y=ey.207
Solution.ddxxtany py2+2y=ddy(ey)ddx(xtany) ddxpy2+2y=eydydxtany+xsec2ydydx 12py2+2y(2y+2)dydx=eydydx xsec2x y+1py2+2y ey!dydx= tanyTherefore,dydx= tanyxsec2x y+1py2+2y ey:.EXAMPLE6:Findtheequationofthenormallinetothecurveofy2 xey 1+x3=7atthepoint(2;1).Solution.Togettheequationofthenormalline,wemust rstobtainitsslope.Wenotethatthetangentandnormallinesareperpendicular.Hence,theirslopesarenegativereciprocalsofeachother.Togettheslopeofthetangentline,weapplyimplicitdi erentiation.ddx y2 xey 1+x3=ddx(7)2ydydx ey 1+xey 1dydx+3×2=0 2y xey 1dydx=ey 1 3x2dydx=ey 1 3x22y xey 1:Thus,theexpressionwhichwouldgiveustheslopeofthenormallineatanypointonthecurveisgivenbymNL= 1dydx= 2y xey 1ey 1 3×2=xey 1 2yey 1 3×2:208
Atthepoint(2;1),wegetthattheslopeofthenormallineismNL=2e1 1 2e1 1 3(2)2=01 12=0:Observethat(2;1)mustalsobeonthenormalline.Thus,theequationofthenormallineisy 1=0(x 2)y=1:.EXAMPLE7:Finddy=dx.1.y=ln 3×2 4x+22.y=tan 1(3x 2sinx)3.y=ln 2x tan 1(lnx)Solution.1.dydx=13×2 4x+2(6x 4)=6x 43×2 4x+2:2.dydx=11+(3x 2sinx)2(3 2cosx)=3 2cosx1+(3x 2sinx)2:3.dydx=12x tan 1(lnx)2 11+ln2x1x..EXAMPLE8:Finddydxforsin 2 y3=tan 1 x2 lny.Solution.cos 2 y3 3y2dydx=11+(x2)2(2x) 1ydydx1y 3y2cos 2 y3dydx=2×1+x4dydx=2x(1+x4)1y 3y2cos(2 y3):.209
EXAMPLE9:Finddydxforlny 3xy2=ptan 1y.Solution.1ydydx 3y2+2xydydx=12ptan 1y11+y2dydx 1y 6xy 12ptan 1y11+y2!dydx=3y2dydx=3y2 1y 6xy 12ptan 1y11+y2!:.EXAMPLE10:Findthederivativewithrespecttoxoftan 1 y2+ln(y+1)=sin3(x+1)Solution.11+(y2)22ydydx+1y+1dydx=3sin2(x+1)cos(x+1)2y1+y4+1y+1dydx=3sin2(x+1)cos(x+1)dydx=3sin2(x+1)cos(x+1)2y1+y4+1y+1:.SupplementaryProblems1.Finddydxforthefollowingcurves:(a)x2y py=17 tanx(b)sin(x+y) ex y=secy(c)y2cosx2 x2siny2=1(d)(x 2)2+(y 3)2=25(e)(y 3)27 (x+1)24=1(f)yex+xey=12(g)yex xey=ln(x+y)(h)x+yx y=1(i)xsecy=x y(j)y32ex=2(k)px+py=1210
2. ndtheequationofthetangentandnormallinesatthegivenpoints.(a)x+y=2at(1;1)(b)y3 xy+x2=1at(1;0)(c)cot(x+y) sin(x y)=1at0;2(d)e x y=0at( 1;0)(e)3×2 4y3=8at( 2;1)211
LESSON10:RelatedRatesLEARNINGOUTCOME:Attheendofthelesson,thelearnershallbeabletosolvesitu-ationalproblemsinvolvingrelatedrates.212
Suggestionsinsolvingproblemsinvolvingrelatedrates:1.Ifpossible,provideanillustrationfortheproblemthatisvalidforanytimet.2.Identifythosequantitiesthatchangewithrespecttotime,andrepresentthemwithvariables.(Avoidassigningvariablestoquantitieswhichareconstant,thatis,whichdonotchangewithrespecttotime.Labelthemrightawaywiththevaluesprovidedintheproblem.)3.Writedownanynumericalfactsknownaboutthevariables.Interpreteachrateofchangeasthederivativeofavariablewithrespecttotime.Rememberthatifaquantitydecreasesovertime,thenitsrateofchangeisnegative.4.Identifywhichrateofchangeisbeingasked,andunderwhatparticularconditionsthisrateisbeingcomputed.5.Writeanequationshowingtherelationshipofallthevariablesbyanequationthatisvalidforanytimet.6.Di erentiatetheequationin(5)implicitlywithrespecttot.7.Substituteintotheequation,obtainedin(6),allvaluesthatarevalidattheparticulartimeofinterest.Sometimes,somequantitiesstillneedtobesolvedbysubstitutingtheparticularconditionswrittenin(4)totheequationin(6).Then,solveforwhatisbeingaskedintheproblem.8.Writeaconclusionthatanswersthequestionoftheproblem.Donotforgettoincludethecorrectunitsofmeasurement.EXAMPLE7:Awaterdropletfallsontoastillpondandcreatesconcentriccircularripplesthatpropagateawayfromthecenter.Assumingthattheareaofarippleisincreasingattherateof2cm2/s, ndtherateatwhichtheradiusisincreasingattheinstantwhentheradiusis10cm.Solution.Wesolvethisstep-by-stepusingtheaboveguidelines.(1)Illustrationr213
(2)LetrandAbetheradiusandarea,respectively,ofacircularrippleatanytimet.(3)ThegivenrateofchangeisdAdt=2.(4)Weareaskedto nddrdtattheinstantwhenr=10.(5)TherelationshipbetweenAandrisgivenbytheformulafortheareaofacircle:A=r2:(6)Wenowdi erentiateimplicitlywithrespecttotime.(Bemindfulthatallquantitiesheredependontime,soweshouldalwaysapplyChainRule.)dAdt=(2r)drdt:(7)SubstitutingdAdt=2andr=10gives2=2(10)drdt=)drdt=110:(8)Conclusion:Theradiusofacircularrippleisincreasingattherateof110cm/s..EXAMPLE8:Aladder10meterslongisleaningagainstawall.Ifthebottomoftheladderisbeingpushedhorizontallytowardsthewallat2m/s,howfastisthetopoftheladdermovingwhenthebottomis6metersfromthewall?Solution.We rstillustratetheproblem.ladder10wallxypushLetxbethedistancebetweenthebottomoftheladderandthewall.Letybethedistancebetweenthetopoftheladderandtheground(asshown).Notethatthelengthoftheladderisnotrepresentedbyavariableasitisconstant.Wearegiventhatdxdt= 2.(Observethatthisrateisnegativesincethequantityxdecreaseswithtime.)214
Wewantto nddydtattheinstantwhenx=6.Observethatthewall,thegroundandtheladderdeterminearighttriangle.Hence,therelationshipbetweenxandyisgivenbythethePythagoreanTheorem:x2+y2=100:(2.4)Di erentiatingbothsideswithrespecttotimetgives2xdxdt+2ydydt=0:(2.5)Beforeweproceedtothenextstep,weaskourselvesifwealreadyhaveeverythingweneed.So,dx=dtisgiven,dy=dtisthequantityrequired,xisgiven,BUT,westilldonothavey.Thisiseasytosolvebysubstitutingthegivenconditionx=6intotheequationin(2.4).So,62+y2=100=)y=p100 36=p64=8:Finally,wesubstituteallthegivenvaluesintoequation(2.5):2(6)( 2)+2(8)dydt=0=)dydt=2416=32:Thus,thedistancebetweenthetopoftheladderandthegroundisincreasingattherateof1:5m/s.Equivalently,wecanalsosaythatthetopoftheladderismovingattherateof1:5m/s..EXAMPLE9:Anautomobiletravelingattherateof20m/sisapproachinganintersection.Whentheautomobileis100metersfromtheintersection,atrucktravelingattherateof40m/scrossestheintersection.Theautomobileandthetruckareonperpendicularroads.Howfastisthedistancebetweenthetruckandtheautomobilechangingtwosecondsafterthetruckleavestheintersection?Solution.Letusassumethattheautomobileistravellingwestwhilethetruckistravellingsouthasillustratedbelow.Letxdenotethedistanceoftheautomobilefromtheintersection,ydenotethedistanceofthetruckfromtheintersection,andzdenotethedistancebetweenthetruckandtheautomobile.ytruckxautomobilez215
Thenwehavedxdt= 20(thenegativerateisduetothefactthatxdecreaseswithtime)anddydt=40.Wewantto nddzdtwhent=2.Theequationrelatingx,yandzisgivenbythePythagoreanTheorem.Wehavex2+y2=z2:(2.6)Di erentiatingbothsideswithrespecttot,ddtx2+y2=ddtz2=)2xdxdt+2ydydt=2zdzdt=)xdxdt+ydydt=zdzdt:Beforesubstitutingthegivenvalues,westillneedto ndthevaluesofx,yandzwhent=2.Thisisfoundbythedistance-rate-timerelationship:distance=ratetime:Fortheautomobile,after2seconds,ithastravelledadistanceequalto(rate)(time)=20(2)=40fromthe100metermark.Therefore,x=100 40=60.Ontheotherhand,forthetruck,ithastravelledy=(rate)(time)=40(2)=80:Thevalueofzisfoundfrom(2.6):z=px2+y2=p602+802=p102(36+64)=100:Finally,40( 20)+80(40)=100dzdt=)dzdt=20:Thus,thedistancebetweentheautomobileandthetruckisincreasingattherateof20meterspersecond..EXAMPLE10:Waterispouringintoaninvertedconeattherateof8cubicmetersperminute.Iftheheightoftheconeis12metersandtheradiusofitsbaseis6meters,howfastisthewaterlevelrisingwhenthewateris4-meterdeep?Solution.We rstillustratetheproblem.216
LetVbethevolumeofthewaterinsidetheconeatanytimet.Leth;rbetheheightandradius,respectively,oftheconeformedbythevolumeofwateratanytimet.WearegivendVdt=8andwewishto nddhdtwhenh=4.6rh12Now,therelationshipbetweenthethreede nedvariablesisgivenbythevolumeofthecone:V=3r2h:Observethattherateofchangeofrisneithergivennorasked.Thispromptsusto ndarelationshipbetweenrandh.Fromtheillustration,weseethatbytheproportionalityrelationsinsimilartriangles,weobtainrh=612orr=h2.Thus,V=3r2h=3h22h=12h3:Di erentiatingbothsideswithrespecttot,ddt(V)=ddt12h3=)dVdt=4h2dhdt:Thus,aftersubstitutingallgivenvalues,weobtain8=4(4)2dhdt=)dhdt=3216=2:Finally,weconcludethatthewaterlevelinsidetheconeisrisingattherateof2meters/minute..217
SolvedExamplesEXAMPLE11:Theradiusofacircleisincreasingattherateof5cm/s.Howfastistheareaofthecirclechangingwhentheradiusis6cmlong?Solution.LetrandAbetheradiusandtheareaofthecirclerespectively.rGiven:drdt=5Asked:FinddAdtwhenr=6.TheareaofthecircleisA=r2:Di erentiatebothsidesoftheequationwithrespecttottoobtaindAdt=2rdrdt:Substituter=6anddrdt=5:dAdt=2(6)(5)=60:So,theareaofthecircleisincreasingattherateof60cm2/s..EXAMPLE12:Aladder10meterslongleansagainstawall.Ifthebottomoftheladderisbeingpulledawayfromthewallattherateof2m/min,howfastistheladderslidingdownthewallwhenthetopoftheladderis3metersfromtheground?Solution.Letxandybethedistanceofthebottomandtopoftheladderfromthewallandthegroundrespectively.Given:dxdt=2.Find:dydt,wheny=3.218
ByPythagoreanTheorem,x2+y2=100.Di erentiatingwithrespecttot:2xdxdt+2ydydt=0=)dydt= xydxdt:Substitutex=6,y=3anddxdt=2:dydt= 6(2)3= 4:y10xTherefore,theladderisslidingdownthewallattherateof4m/minwhenthetopoftheladderis3metersfromtheground..EXAMPLE13:Newtonistravelingeastwardat20km/hrwhileLeibnizistravelingnorthwardat10km/hr.At1pm,Leibnizislocated50kmsouthofNewton.AtwhatrateisthedistancebetweenNewtonandLeibnizchangingat3pm?Solution.LetxandybethedistancetraveledbyNewtonandLeibniz,respectively,andletzbethedistancefromNewtontoLebniz.Given:dxdt=20.dydt=10.Find:dzdtat3pm.z2=x2+(50 y)2=)zdzdt=xdxdt+(50 y)dydt:xNewtonLeibniz50 yyzAt3pm,x=40,y=20,z=px2+y2=p402+302=50.Substitutethesevaluestoourequationtoobtain50dzdt=40(20)+(30)(10)dzdt=22:Hence,at3pm,thedistancebetweenNewtonandLeibnizisincreasingattherateof22km/hr..219
EXAMPLE14:Awatertankintheshapeofaninvertedrightcircularconeis9metershigh.Thetoprimofthetankisacirclewitharadiusof3meters.Ifthetankisbeing lledwithwaterattherateof2cubicmetersperminute,whatistherateofchangeofthewaterdepth,inmetersperminute,whenthedepthis5meters?Solution.Letr,handVbetheradius,heightandvolumeoftheconeformedbythewaterinsidethetank,respectively.h123rGiven:dr=dt=2,r=3,h=5.Find:dh=dtwhenh=5.ThevolumeoftheconeisV=13r2h.WewillexpressVisintermsofhonly.Usingsimilartrian-gles,wehavehr=93=)r=h3:SubstitutethisinVtogetV=13r2h=13h32h=127h3:Di erentiatebothsidesoftheequationwithrespecttot:dVdt=19h2dhdt:SubstitutedV=dt=2,h=5andsolvefordh=dt:2=19 52dhdt=)dhdt=1825:Hence,therateofchangeofthewaterdepthwhenthedepthis5metersis1825m3=min..SupplementaryProblems1.Letxandybedi erentiablefunctionsoftandsupposethattheyarerelatedbytheequationxy 1=y2.Finddx=dtwhenx=2,anddy=dt=1.220
2.Letr,h,andAbedi erentiablefunctionsoftandA=2r2+2rh.FinddA=dtwhenr=2,h=5,dr=dt=2anddh=dt=1.3.Startingfromthesamepoint,Roydenstartswalkingeastwardat12m/swhileNilostartsrunningtowardssouthwardat2m/s.HowfastisthedistancebetweenRoydenandNiloincreasingafter3s?4.Aladder20meterslongleansagainstawall.Ifthebottomoftheladderisbeingpushedtowardsthewallattherateof20m/min,howfastisthetopoftheladdermovingupthewallwhenthetopoftheladderis6metersfromtheground?5.CarAleavesaparkinglotat9amandtravelseastwardattherateof60kph.Anothervehicle,carB,leavesthesameparkinglotat10amandtravelssouthwardat40kph.Howfastisthedistancebetweenthesetwocarschangingat11am?6.Asphericalballoonisbeingin
atedinsuchawaythattheradiusisincreasingattherateof1m/s.Whatistherateofchangeofitsvolumewhenitsradiusis5m?7.Asphericalballshrinksattheconstantrateof5cm3/s.Howfastistheradiuschangingwhenitsvolumeis20cm3?8.Asphericalballisbeingpumpedwithairattherateof10cm3/s.Howfastisitssurfaceareachangingwhentheradiusis30cm?9.Standingontopofabuildingandthroughatelescope,JonaobservesMarcorunningtowardsthebuilding.Ifthetelescopeis400mabovethegroundandMarcoisrunningataspeedof20m/s,atwhatrateisthemeasureoftheacuteangleformedbythetelescopeandtheverticalchangingwhenMarcois100mfromthebuilding?10.Standingonacli ,Danielaiswatchingamotorboatthroughatelescopeastheboatapproachestheshorelinedirectlybelowher.Ifthetelescopeis300metersabovethewaterlevelandiftheboatisapproachingthecli at30m/s,atwhatrateistheacuteangle,madebythetelescopewiththecli ,changingwhentheboatis100p3metersfromtheshore?221
CHAPTER2REVIEWI.TRUEorFALSE.WriteTRUEifthestatementisalwaystrue.Otherwise,writeFALSE.1.Iffiscontinuousatx,thenfisdi erentiableatx.2.limx!x0(x+x0)20 x20x x0=20×19.3.Iffisdiscontinuousatx=0,thenfisnotdi erentiableatx=0.II.Solvefordydx.DONOTSIMPLIFY.1.y=x2+3x+12.y=x2+1x 13.xy=sin(x+y)III.Solvethefollowingcompletely.1.Findtwonumbersin[2,10]whosesumis10andwhoseproductismaximum.2.Startingfromthesamepoint,Lexstartedwalkingtowardstheeastatthespeedof4m/sandRaldrantowardsthesouthatthespeedof3m/s.DeterminehowfastisthedistancebetweenLexandRaldchangingafter1second.3.Findf0( 1)giventhatf(x)=1x.4.Findtheslopeofthetangentlineonthecurvegivenbyy=3×2+x+1atx=0.222
Chapter3Integration223
LESSON11:IntegrationLEARNINGOUTCOMES:Attheendofthelesson,thelearnershallbeableto:1.Illustratetheantiderivativeofafunction;2.Computethegeneralantiderivativeofpolynomialfunctions;3.Computethegeneralantiderivativeofrootfunctions;4.Computethegeneralantiderivativeofexponentialfunctions;and,5.Computethegeneralantiderivativeoftrigonometricfunctions.224
TOPIC11.1:IllustrationofanAntiderivativeofaFunctionDe nition5.AfunctionFisanantiderivativeofthefunctionfonanintervalIifF0(x)=f(x)foreveryvalueofxinI.EXAMPLE1:1.Anantiderivativeoff(x)=12×2+2xisF(x)=4×3+x2.Aswecansee,thederivativeofFisgivenbyF0(x)=12×2+2x=f(x).2.Anantiderivativeofg(x)=cosxisG(x)=sinxbecauseG0(x)=cosx=g(x).EXAMPLE2:1.Otherantiderivativesoff(x)=12×2+2xareF1(x)=4×3+x2 1andF2(x)=4×3+x2+1:Infact,anyfunctionoftheformF(x)=4×3+x2+C,whereC2Risanantiderivativeoff(x).ObservethatF0(x)=12×2+2x+0=12×2+2x=f(x):2.Otherantiderivativesofg(x)=cosxareG1(x)=sinx+andG2(x)=sinx 1.Infact,anyfunctionG(x)=sinx+C,whereC2Risanantiderivativeofg(x).Theorem10.IfFisanantiderivativeoffonanintervalI,theneveryantiderivativeoffonIisgivenbyF(x)+C,whereCisanarbitraryconstant.TerminologiesandNotations:Antidi erentationistheprocessof ndingtheantiderivative.ThesymbolZ,alsocalledtheintegralsign,denotestheoperationofan-tidi erentiation.Thefunctionfiscalledtheintegrand.IfFisanantiderivativeoff,wewriteZf(x)dx=F(x)+C.ThesymbolsZanddxgohand-in-handanddxhelpsusidentifythevariableofintegration.TheexpressionF(x)+Ciscalledthegeneralantiderivativeoff.Mean-while,eachantiderivativeoffiscalledaparticularantiderivativeoff.225
SolvedExamplesEXAMPLE1:Letf(x)= 8×3 10x+5andF(x)= 2×4 5×2+5x+3.ShowthatF(x)isanantiderivativeoff(x).Solution.F0(x)= 8×3 10x+5=f(x)..EXAMPLE2:DetermineifF(x)=x3+x+1,G(x)=x3+2x+1orH(x)=x3+x+3is/areantiderivativesoff(x)=3×2+1.Solution.F0(x)=3×2+1=f(x).Thus,F(x)isanantiderivativeoff(x).G0(x)=3×2+26=f(x).Therefore,G(x)isnotanantiderivativeoff(x).H0(x)=3×2+1=f(x).Hence,H(x)isanantiderivativeoff(x).BothF(x)andH(x)areantiderivativesoff(x).Inthisexample,weareabletoillustratethatanantiderivativeofafunctionisnotunique..SupplementaryProblems1.Determinetheantiderivativesofthefollowingfunctions.(a)f(x)=8×7+2×3 1(b)f(x)= 7(c)g(x)=2×3 2x 1(d)h(x)=secxtanx 2sinx2.MatchthefunctionsinColumnAwiththeircorrespondingantiderivativesinColumnB.ColumnA1.f(x)= 4×2 9x 12.f(x)=11×2 1213.f(x)=2x+34.f(x)=x3+15.f(x)=x(x2 3)6.f(x)=(x 3)(2x+1)ColumnBa.F(x)=14×4+x 32b.F(x)= 43×3 92×2 x+35c.F(x)=23×3 52×2 4x 4d.F(x)=113×3 121x 2e.F(x)=14×4 32x2f.F(x)=x2+3x 1226
TOPIC11.2:AntiderivativesofAlgebraicFunctionsTheorem11.(TheoremsonAntidi erentiation)1.Zdx=x+C.2.Ifnisanyrealnumberandn6= 1,thenZxndx=xn+1n+1+C:3.Ifaisanyconstantandfisafunction,thenZaf(x)dx=aZf(x)dx:4.Iffandgarefunctionsde nedonthesameinterval,Z[f(x)g(x)]dx=Zf(x)dxZg(x)dx:EXAMPLE1:Determinethefollowingantiderivatives:1.Z3dx2.Zx6dx3.Z1x6dx4.Z4pudu5.Z 12×2+2xdx6.Zt 2t 3ptdt7.Zx2+1x2dxSolution.WewillusetheTheoremsonAntidi erentiationtodeterminetheantiderivatives.1.Using(a)and(c)ofthetheorem,wehaveZ3dx=3Zdx=3x+C:2.Using(b)ofthetheorem,wehaveZx6dx=x6+16+1+C=x77+C:3.Using(b)ofthetheorem,wehaveZ1x6dx=Zx 6dx=x 6+1 6+1+C=x 5 5+C= 15×5+C:4.Using(b)and(c)ofthetheorem,wehaveZ4pudu=4Zu12du=4u12+112+1+C=4u3232+C=8u323+C:227
5.Using(b),(c)and(d)ofthetheorem,wehaveZ 12×2+2xdx=12Zx2dx+2Zxdx=12×33+2×22+C=4×3+x2+C:6.Using(b),(c)and(d),wehaveZt2t 3ptdt=Z2t2 3t32dt=2Zt2dt 3Zt32dt=2t33 6t525+C:7.Using(a),(b)and(d),wehaveZx2+1x2dx=Z 1+x 2dx=x+x 1 1+C=x 1x+C:.SolvedExamplesEvaluatethefollowingintegrals.EXAMPLE1:Z4dxSolution.Z4dx=4Zdx=4x+C:.EXAMPLE2:Z( 4)dxSolution.Z( 4)dx=( 4)Zdx= 4x+C:.EXAMPLE3:Zx8dxSolution.Zx8dx=x8+18+1+C=x99+C:.EXAMPLE4:Z1x9dxSolution.Z1x9dx=Zx 9dx=x 9+1 9+1+C=x 8 8+C= 18×8+C:.228
EXAMPLE5:ZpxdxSolution.Zpxdx=Zx12dx=x12+112+1+C=x3232+C=2×323+C:.EXAMPLE6:Z5pudxSolution.Z5pudx=5Zu12du=5u12+112+1+C=5u3232+C=10u323+C:.EXAMPLE7:Z 10×4+2×3dxSolution.Z 10×4+2×3dx=Z10x4dx+Z2x3dx=10×55+2×44+C=2×5+x42+C:.EXAMPLE8:Z x2 3 px+xdxSolution..Z x2 3 px+xdx=Zx52+x3 3×12 3xdx=x7272+x44 3×3232 3×22+C=2×727 2×32+x44 3×22+C:.EXAMPLE9:Z 16×3 3×2 1dxSolution..Z 16×3 3×2 1dx=Z16x3dx Z3x2dx Zdx=16×44 3×33 x+C=4×4 x3 x+C:.229
EXAMPLE10:Z2x4 xx3dxSolution.Z2x4 xx3dx=Z 2x x 2dx=2×22 x 1 1+C=x2+1x+C:.SupplementaryProblemsFindtheintegralsofthefollowing.1.Z3dx2.Z( 2)dx3.Zx4dx4.Zx20dx5.Z2pxdx6.Z3pxdx7.Z75px2dx8.Z114px13dx9.Z x5 16×7dx10.Z 2×4+x3dx11.Z 54×9+10×4 23dx12.Z 23x 17×16+10dx13.Z1x5dx14.Z1x23dx15.Z1×2 4x 5dx16.Zx 4 3×2 1dx17.Zpx x2 3x+1dx18.Zx2 2px+24x 1dx19.Z(3w w)2dw20.Z(1 5x)2dx21.Zt4 t3 1t2dt22.Zx6 x2+1x8dx23.Zy3 4y2 3y+14y 2dy24.Z2x4+3×3 2×2 x+32x+3dx230
TOPIC11.3:AntiderivativesofFunctionsYieldingExponentialFunctionsandLogarithmicFunctionsInthislesson,wewillpresentthebasicformulasforintegratingfunctionsthatyieldexponentialandlogarithmicfunctions.Letus rstrecallthefollowingdi erentiationformulas:1.Dx(ex)=ex.2.Dx(ax)=axlna.3.Dx(lnx)=1x.Becauseantidi erentiationistheinverseoperationofdi erentiation,thefollowingtheoremshouldbeimmediate.Theorem12.(Theoremsonintegralsyieldingtheexponentialandlogarithmicfunc-tions)1.Zexdx=ex+C.2.Zaxdx=axlna+C:Here,a>0witha6=1:3.Zx 1dx=Z1xdx=lnjxj+C:EXAMPLE1:Findtheintegralsofthefollowingfunctions.1.Z(ex+2x)dx2.Z3xdx3.Z3x+1dx4.Z2xdxSolution.Wewillusethetheoremtodeterminetheintegrals.1.Using(1)and(2)oftheorem,wehaveZ(ex+2x)dx=Z(ex)dx+Z(2x)dx=ex+2xln2+C:2.Using(2)ofthetheorem,wehaveZ3xdx=3xln3+C:231
3.Using(2)ofthetheorem,wehaveZ3x+1dx=Z(3x)(31)dx=3Z(3x)dx=33xln3+C:4.Using(3)ofthetheorem,wegetZ2xdx=2Z1xdx=2lnjxj+C:.SolvedExamplesIntegratethefollowing.EXAMPLE1:Z2exdxSolution.Z2exdx=2Zexdx=2ex+C:.EXAMPLE2:Z5xdxSolution.Z5xdx=5xln5+C:.EXAMPLE3:Z7x+2dxSolution.Z7x+2dx=Z727xdx=72Z7xdx=49ln77x+C:.EXAMPLE4:Z3xdxSolution.Z3xdx=3Z1xdx=3lnjxj+C:.232
EXAMPLE5:Z3ex 2x+1×2dxSolution.Z3ex 2x+1×2dx=Z3ex 2x x 2dx=3Zexdx 2Z1xdx+Zx 2dx=3ex 2lnjxj x 1+C=3ex 2lnjxj 1x+C:.SupplementaryProblemsEvaluatethefollowingantiderivatives.1.Z4exdx2.Z( 3)exdx3.Z2xdx4.Z3xdx5.Zex+2dx6.Z2x+4dx7.Z4xdx8.Z97xdx9.Z2ex x 18×2dx10.Z92x+x2 34×2dx11.Z73x+1 3×5 x3 12xdx12.Z3×2 27x43x3 3exdx13.Z (2x+1)(x 3)x3 ex2+x 1ex2+1!dx14.Z 2×3 x2+x 32×3 x2+1 x x2 1x4!dx233
TOPIC11.4:AntiderivativesofTrigonometricFunctionsTheorem13.(AntiderivativesofTrigonometricFunctions)1.Zsinxdx= cosx+C2.Zcosxdx=sinx+C3.Zsec2xdx=tanx+C4.Zcsc2xdx= cotx+C5.Zsecxtanxdx=secx+C6.Zcscxcotxdx= cscx+CEXAMPLE1:Determinetheantiderivativesofthefollowing:1.Z(cosx sinx)dx2.Zcot2xdx3.Ztan2vdv4.Zsinxcos2xdxSolution.Wewillusethetheoremonantiderivativesoftrigonometricfunctions.1.Using(a)and(b)ofthetheorem,wehaveZ(cosx sinx)dx=Zcosxdx Zsinxdx=sinx ( cosx)+C=sinx+cosx+C:2.Sinceweknowthatcot2x=csc2 1,thenZcot2xdx=Z csc2x 1dx=Zcsc2x Zdx= cot2 x+C:3.Sincetan2v=sec2v 1,wehaveZtan2vdv=Z sec2v 1dv=Zsec2vdv Zdv=tanv v+C:4.Zsinxcos2xdx=Zsinxcosx1cosxdx=Ztanxsecxdx=secx+C:.SolvedExamplesEvaluatethefollowingantiderivatives.EXAMPLE1:EvaluateZ4sinxdx.234
Solution.Z4sinxdx=4Zsinxdx= 4cosx+C:.EXAMPLE2:EvaluateZcosx+3dx.Solution.Zcosx+3dx=Zcosxcos3 sinxsin3dx=Z12cosxdx Zp32sinxdx=12Zcosxdx p32Zsinxdx=12sinx+p32cosx+C:.EXAMPLE3:Zcosx1 cos2xdxSolution.Zcosx1 cos2xdx=Zcosxsin2xdx=Z1sinxcosxsinxdx=Zcscxcotxdx= cscx+C:.SupplementaryProblemsEvaluatethefollowingintegrals.1.Z7csc2xdx2.Z2secxtanxdx235
3.Z 1+tan2xdx4.Z 1+cot2xdx5.Zsinx 6dx6.Zcosx 6dx7.Z(1 cosx)(1+cosx)(1 sinx)(1+sinx)cscxdx8.Z(1 sinx)(1+sinx)(1 cosx)(1+cosx)secxdx236
LESSON12:TechniquesofAntidi erentiationTIMEFRAME:7hoursLEARNINGOUTCOMES:Attheendofthelesson,thelearnershallbeableto:1.Computetheantiderivativeofafunctionusingthesubstitutionrule;and2.Computetheantiderivativeofafunctionusingatableofintegrals(includingthosewhoseantiderivativesinvolvelogarithmicandinversetrigonometricfunctions).237
TOPIC12.1:Antidi erentiationbySubstitutionandbyTableofIntegralsAntidi erentiationismorechallengingthandi erentiation.To ndthederivativeofagivenfunction,therearewell-establishedrulesthatarealwaysapplicabletodi erentiablefunctions.Forantidi erentiation,theantiderivativesgiveninthepreviouslessonmaynotsucetointe-grateagivenfunction.Aprerequisiteisknowledgeofthebasicantidi erentiationformulas.Someformulasareeasilyderived,butmostofthemneedtobememorized.Nohardandfastrulescanbegivenastowhichmethodappliesinagivensituation.Incollege,severaltechniquessuchasintegrationbyparts,partialfractions,trigonometricsubtitutionwillbeintroduced.Thislessonfocusesonthemostbasictechnique-antidi erentiationbysubstitution-whichistheinverseoftheChainRuleindi erentiation.Thereareoccasionswhenitispossibletoperformadicultpieceofintegrationby rstmakingasubstitution.Thishasthee ectofchangingthevariableandtheintegrand.Theabilitytocarryoutintegrationbysubstitutionisaskillthatdevelopswithpracticeandexperience,butsometimesasensiblesubstitutionmaynotleadtoanintegralthatcanbeevaluated.Wemustthenbepreparedtotryoutalternativesubstitutions.SupposewearegivenanintegraloftheformZf(g(x))g0(x)dx.Wecantransformthisintoanotherformbychangingtheindependentvariablextouusingthesubstitutionu=g(x).Inthiscase,dudx=g0(x)dx.Therefore,Zf(g(x))g0(x)dx=Zf(u)du:Thischangeofvariableisoneofthemostimportanttoolsavailabletous.Thistechniqueiscalledintegrationbysubstitution.Itisoftenimportanttoguesswhatwillbetheusefulsubstitution.Usually,wemakeasubstitutionforafunctionwhosederivativealsooccursintheintegrand.238
EXAMPLE1:EvaluateZ(x+4)5dx.Solution.Noticethattheintegrandisinthe fthpoweroftheexpression(x+4).Totacklethisproblem,wemakeasubstitution.Weletu=x+4.Thepointofdoingthisistochangetheintegrandintoamuchsimpleru5.However,wemusttakecaretosubstituteappropriatelyforthetermdxtoo.Now,sinceu=x+4itfollowsthatdudx=1andsodu=dx.So,substituting(x+4)anddx,wehaveZ(x+4)5dx=Zu5du:Theresultingintegralcanbeevaluatedimmediatelytogiveu66+C.Recallingthatu=x+4,wehaveZ(x+4)5dx=Zu5du=u66+C=(x+4)66+C:.Analternativewayof ndingtheantiderivativeaboveistoexpandtheexpressionintheinte-grandandantidi erentiatetheresultingpolynomial(ofdegree5)termbyterm.WewillNOTdothis.Obviously,thesolutionaboveissimplerthanthementionedalternative.EXAMPLE2:EvaluateZ(x5+2)95x4dx.Solution.Ifweletu=x5+2,thendu=5x4dx,whichispreciselytheotherfactorintheintegrand.Thus,intermsofthevariableu,thisisessentiallyjustapowerruleintegration.Thatis,Z(x5+2)95x4dx=Zu9du;whereu=x5+2=u1010+C=(x5+2)1010+C:239
.Again,thealternativewayistoexpandouttheexpressionintheintegrand,andintegratetheresultingpolynomial(ofdegree49)termbyterm.Again,wewouldratherNOTdothis.EXAMPLE3:EvaluateZz2p1+z3dz.Solution.Inthisexample,weletu=1+z3sothatdudz=3z2.Ifu=1+z3,thenweneedtoexpressz2dzintheintegrandintermsofduoraconstantmultipleofdu.Fromdudz=3z2itfollowsthatdu=3z2dzandz2dz=13du.Thus,Zz2p1+z3dz=Z1p1+z3z2dz=Z1pu13du=13Zu 12du=13 u1212!+C=23u12+C=23(1+z3)12+C.EXAMPLE4:EvaluateZxpx2 1dx.Solution.Noticethatifu=x2 1,thendu=2xdx.Thisimpliesthatxdx=12du,sowehaveZxpx2 1dx=Z1u1212du=12Zu 12du=12 u1212!+C=(x2 1)12+C:240
.EXAMPLE5:EvaluateZ3x5px3+5dx.Solution.Letu=x3+5.Thenx3=u 5anddu=3x2dx.Thus,Zx3px3+5dx=Zu 5u12dx=Zuu12 5u12dx=Z(u12 5u 12)dx=u3232 5u1212+C=23u32 10u12+C=23(x3+5)32 10(x3+5)12+C:.EXAMPLE6:EvaluateZdxx3plnx.Solution.Wesubstituteu=lnxsothatdu=1xdx,whichoccursintheintegrand.Thus,Zdxx3plnx=Z13pudu=Zu 13du=u2323+C=32(lnx)23+C.Werecallthetheoremwestatedinthepreviouslesson.Theorem14.(Theoremsonintegralsyieldingtheexponentialandlogarithmicfunctions)1.Zexdx=ex+C:241
2.Zaxdx=axlna+C:Here,a>0witha6=1.3.Zx 1dx=Z1xdx=lnjxj+C:EXAMPLE7:Evaluatethefollowingintegrals.1.Ze3xdx2.Z24xdx3.Z12x 1dxSolution..1.Weletu=3x.Thendu=3dx.Hence,dx=du3.So,Ze3xdx=Zeudu3=13Zeudu=13eu+C=13e3x+C:2.Here,weletu=4xandsodu=4dx.Thus,dx=du4.Hence,wehaveZ24xdx=Z2udu4=14Z2udu=142uln2+C=14ln224x+C:3.Supposeweletu=2x 1.Thendu=2dx.Hence,dx=du2.WehaveZ12x 1dx=Z1udu2=12Z1udu=12lnjuj+C=12lnj2x 1j+C:.EXAMPLE8:EvaluateZcos(4x+3)dx.Solution.Observethatifwemakethesubstitutionu=4x+3,theintegrandwillcontainamuchsimplerform,cosu,whichwecaneasilyintegrate.So,ifu=4x+3,thendu=4dxanddx=14du.So,242
Zcos(4x+3)dx=Zcosu14du=14Zcosudu=14sinu+C=14sin(4x+3)+C:.EXAMPLE9:EvaluatetheintegralZsinxcosxdx.Solution.Notethatifweletu=sinx,itsderivativeisdu=cosxdxwhichistheotherfactorintheintegrandandourintegralbecomesZsinxcosxdx=Zudu=u22+C1=sin2x2+C1:.Alternativesolutiontotheproblem:Ifweletu=cosx,thendu= sinxdxwhichisalsotheotherfactorintheintegrand.Doingasimilartechniqueasabove,wegetZsinxcosxdx= cos2x2+C2:Evenifbothanswersaredi erent,thepresenceoftheconstantsavoidsanycontradiction,forexample,lettingC2=12+C1showsthattheanswersareequivalent.243
EXAMPLE10:EvaluatetheintegralZesinxcosxdx.Solution.Weletu=sinxsothattheotherfactorintheintegrandcosxdx=du.Thus,theintegralbecomesZesinxcosxdx=Zeudu=eu+C=esinx+C:.RecallthatwehadearlierpresentedtheintegralsZsinxdxandZcosxdx.Nowthatweal-readyknowintegrationbysubstitution,wecannowpresenttheintegralsofothertrigonometricfunctions:tanx,cotx,secx,andcscx.First,letususesubstitutiontechniqueto ndZtanxdx.Notethattanx=sinxcosx.Hence,Ztanxdx=Zsinxcosxdx:Now,ifweletu=cosx,thendu= sinxdx.Hence,wehaveZtanxdx=Zsinxcosxdx=Z1u( du)= Z1udu= lnjuj+C= lnjcosxj+C:Equivalently, lnjcosxj+C=lnjcosxj 1=lnjsecxj+C:Similarly,wecanusesubstitutiontechniquetoshowZcotxdx=lnjsinxj+C.Here,weusecotx=cosxsinxandchooseu=sinx.Letusnow ndZsecxdx.Theusualtrickistomultiplythenumeratorandthedenominatorbysecx+tanx.Zsecxdx=Z(secx)secx+tanxsecx+tanxdx=Z1secx+tanx(secxtanx+sec2x)dx:244
Now,ifweletu=secx+tanx,thendu=secxtanx+sec2x.Thus,wehaveZsecxdx=Z1secx+tanx(secxtanx+sec2x)dx:=Z1udu=lnjuj+C=lnjsecx+tanxj+C:Similarly,wecanshowZcscxdx=lnjcscx cotxj+C.Hence,wehavethefollowingformulas:1.Ztanxdx= lnjcosxj+C=lnjsecuj+C.2.Zcotxdx=lnjsinxj+C.3.Zsecxdx=lnjsecx+tanxj+C.4.Zcscxdx=lnjcscx cotxj+C.EXAMPLE11:EvaluateZx4sec(x5)dx:Solution.Weletu=x5.Thendu=5x4dx.Thus,x4dx=du5.WehaveZx4sec(x5)dx=Zsecudu5=15Zsecudu=15lnjsecu+tanuj+C=15lnjsecx5+tanx5j+C:.EXAMPLE12:EvaluateZ4+cosx4sinx4dx:245
Solution.Letu=x4.Thendu=14dxandso4du=dx.Thus,wehaveZ4+cosx4sinx4dx=Z4+cosusinu4du=4Z4sinu+cosusinudu=4Z4cscu+cotudu=4Zcotudu+16Zcscudu=4lnjsinuj+16lnjcscu cotuj+C=4ln sinx4 +16ln cscx4 cotx4 +C:.INTEGRALSOFINVERSECIRCULARFUNCTIONS1.Zdup1 u2=sin 1u+C2.Zdu1+u2=tan 1u+C3.Zduupu2 1=sec 1u+CIftheconstant1intheseintegralsisreplacedbysomeotherpositivenumber,onecanusethefollowinggeneralizations:Leta>0.ThenZdxpa2 x2=sin 1xa+C(3.1)Zdxa2+x2=1atan 1xa+C(3.2)Zdxxpx2 a2=1asec 1xa+C(3.3)246
EXAMPLE13:EvaluateZ1p9 x2dx.Solution.FromFormula(3.1)witha=3,wewritethisintoZ1p9 x2dx=Z1p32 x2dx=sin 1x3+C:.EXAMPLE14:Zdx9x2+36Solution.Letu=3x,du=3dx.ThenfromFormula(3.2),Zdx9x2+36=13Zduu2+36=1316tan 1u6+C=118tan 13×6+C=118tan 1x2+C:.EXAMPLE15:EvaluateZdxxlnxp(lnx)2 9:Solution.Letu=lnx,du=1xdx.ThenfromFormula(3.3),Zdxxlnxp(lnx)2 9=Zduupu2 9=13sec 1u3+C=13sec 1lnx3+C:.247
EXAMPLE16:Zdxp9+8x x2.Solution.Observethatbycompletingthesquares,andFormula(3.1),Zdxp9+8x x2=Zdxp9 (x2 8x)=Zdxp9 (x2 8x+16 16)=Zdxp25 (x 4)2:Letu=x 4,du=dx.ThenZdxp9+8x x2=Zdup25 u2=sin 1u5+C=sin 1x 45+C:.EXAMPLE17:Z18x+39×2+6x+2dx.Solution.Letu=9×2+6x+2,du=(18x+6)dx.ThenZ18x+39×2+6x+2dx=Z18x+69×2+6x+2dx Z39x2+6x+2dx=Zduu 3Zdx9x2+6x+1+1=lnjuj 3Zdx(3x+1)2+1:Letv=3x+1,dv=3dx.ThenbyFormula(3.2),Z18x+39×2+6x+2dx=lnjuj 33Z1v2+1dv=lnjuj tan 1v+C=lnj9x2+6x+2j tan 1(3x+1)+C:248
.SUMMARY/TABLEOFINTEGRALS1.Zdx=x+C2.Zxndx=xn+1n+1+C,ifn6= 13.Zaf(x)dx=aZf(x)dx4.Z[f(x)g(x)]dx=Zf(x)dxZg(x)dx5.Zexdx=ex+C6.Zaxdx=axlna+C7.Zx 1dx=Z1xdx=lnjxj+C8.Zsinxdx= cosx+C9.Zcosxdx=sinx+C10.Zsec2xdx=tanx+C11.Zcsc2xdx= cotx+C12.Zsecxtanxdx=secx+C13.Zcscxcotxdx= cscx+C14.Ztanxdx= lnjcosxj+C15.Zcotxdx=lnjsinxj+C16.Zsecxdx=lnjsecx+tanxj+C17.Zcscxdx=lnjcscx cotxj+C18.Zdxpa2 x2=sin 1xa+C19.Zdxa2+x2=1atan 1xa+C20.Zdxxpx2 a2=1asec 1xa+CSolvedExamplesEvaluatethefollowingantiderivatives.EXAMPLE1:Z(3x+1)15dxSolution.Letu=3x+1.Thendu=3dx.Z(3x+1)15dx=13Zu15du=13u1616+C=u1648+C=(3x+1)1648+C:.249
EXAMPLE2:Z(8×3+2×2 1)7(6×2+x)dxSolution.Letu=8×3+2×2 1.Thendu=(24×2+4x)dx=4(6×2+x)dx.Z(8×3+2×2 1)7(6×2+x)dx=14Zu7du=14u88+C= 8×3+2x 1832+C:.EXAMPLE3:Zx2p50 5x3dxSolution.Letu=50 5×3.Thendu= 15x2dxwhichimpliesx2dx=du 15.Therefore,Zx2p50 5x3dx=Z1pudu 15= 115Zu 12du= 115u1212+C= 215p50 5×3+C:.EXAMPLE4:Ze 9z+11dzSolution.Letu= 9z+11.Thendu= 9dz.So,Ze 9z+11dz= 19Zeudu= 19eu+C= 19e 9z+11+C:.250
EXAMPLE5:Zx2x2 4dxSolution.Letu=x2 4.Thendu=2xdx.Zx2x2 4dx=12Z2udu=122uln2+C=2×2 42ln2+C:.EXAMPLE6:Zdx3x+7Solution.Letu=3x+7.Thendu=3dx.Zdx3x+7=13Zduu=13lnjuj+C=13lnj3x+7j+C:.EXAMPLE7:Ze2xe2x+3dxSolution.Letu=e2x+3.Thendu=2e2xdx.Ze2xe2x+3dx=12Zduu=12lnjuj+C=12lnje2x+3j+C:.EXAMPLE8:Zt(t2+1)pln(t2+1)dt251
Solution.Letu=ln t2+1.Thendu=2tt2+1dt.Zt(t2+1)pln(t2+1)dt=Z12pudu=12Zu 12du=12u 12+1 12+1+C=u12+C=pln(t2+1)+C:.EXAMPLE9:Ztan2(2x)dxSolution.Letu=2xthendu=2dx.Ztan2(2x)dx=12Ztan2udu=12Z sec2u 1du=12Zsec2udu 12Zdu=tanu2 u2+C=tan(2x)2 x+C:.EXAMPLE10:Zsin4xcosxdxSolution.Letu=sinx.Thendu=cosxdx.Zsin4xcosxdx=Zu4du=u55+C=sin5x5+C:.252
EXAMPLE11:Zsec2(tanx)sec2xdxSolution.Letu=tanx.Thendu=sec2xdx.Zsec2(tanx)sec2xdx=Zsec2udu=tanu+C=tan(tanx)+C:.EXAMPLE12:Z3+cossindSolution.Letu=+cos.Thendu= sind.Z3+cossindx= Z3udu= 3uln3+C= 3+cosln3+C:.EXAMPLE13:Ztan2dSolution.Letu=2.Thendu=12d:Ztan2d=2Ztanud:Fromthetableofintegrals,Ztan2d= 2ln cosu +C= 2ln cos2 +C:253
.EXAMPLE14:Zcot(3x)csc2(3x)(1 cot(3x))3dxSolution.Letu=1 cot(3x).Thendu=3csc2(3x)dx.Zcot(3x)csc2(3x)(1 cot(3x))3dx=13Z(1 u)u3du=13Z u 3 u 2du=13u 2 2 u 1 1+C= 16u2+13u+C= 16(1 cot(3x))2+13(1 cot(3x))+C:.EXAMPLE15:Z1p4 x2dxSolution.SinceZdupa2 u2=sin 1ua+C,takinga=2gives,Z1p4 x2dx=Z1p22 x2dx=sin 1x2+C:.EXAMPLE16:Zdx9x2+25Solution.Letu=3x.Thendu=3dx.Usingformula(19)oftheTableofIntegrals,Zdx9x2+25=13Zduu2+52=1315tan 1u5+C=115tan 13×5+C:.EXAMPLE17:Zdx(2x 5)2+9254
Solution.Letu=2x 5.Thendu=2dx.Formula(19)impliesZdx(2x 5)2+9=12Zdxu2+32=1213tan 1u3+C=16tan 12x 53+C:.EXAMPLE18:Zdxp20+8x x2Solution.Completingthesquares,Zdxp20+8x x2=Zdxp20 (x2 8x)=Zdxp20 (x2 8x+16 16)=Zdxq36 (x 4)2Letu=x 4.Thendu=dx.Formula(18)impliesZdxp20 8x+x2=Zdup36 u2=sin 1u6+C=sin 1x 46+C:.EXAMPLE19:Zx+2×2+2x+5dxSolution.Letu=x2+2x+5.Thendu=(2x+2)dx=2(x+1)dx.Zx+2×2+2x+5dx=Zx+1×2+2x+5dx+Zdxx2+2x+5=12Zduu+Zdxx2+2x+5=12Zduu+Zdx(x+1)2+4Letv=x+1.Thendv=dx.255
Zx+2×2+2x+5dx=12lnjuj+Zdvv2+4+C=12lnjuj+12tan 1v2+C=12lnjx2+2x+5j+12tan 1x+12+C:.EXAMPLE20:Zdxp36x 36Solution.ObservethatZdxp36x 36=Zdxq(6x)2 62:Letu=6x.Thendu=6xln6dx.Hencedx=du6xln6=duuln6.So,Zdxp36x 36=1ln6Zduupu2 62=1ln616sec 1u6+C=16ln6sec 16×6+C=16ln6sec 1 6x 1+C:.EXAMPLE21:Z1ex+e xdxSolution.Z1ex+e xdx=Z1ex+e xexexdx=Zex(ex)2+1dx:Letu=ex.Thendu=exdx.So,Z1ex+e xdx=Zduu2+1=tan 1u+C=tan 1(ex)+C:256
.SupplementaryProblems1.Z(5x 3)100dx2.Z(x 3)1=4dx3.Zx2+1223×3+x5dx4.Z(8x 3) 4×2 3x34dx5.Zxp1 4x2dx6.Zxp3 5x2dx7.Rxpx 1dx8.Z(x)px2 1dx9.Z15x+1dx10.Zdx3x+911.Z4x 6(x 1)(x 2)dx12.Z4x 4x(x 2)dx13.Z2xq(ln(x))3dx14.Z3(x 1)q(ln(x+1))7dx15.Z15e2 5xdx16.Z(2x 3)ex2 3xdx17.Ze4x+1e2x 1dx18.Ze 2xe 2x 1dx19.Z27x+3dx20.Z77x 7dx21.Z3x3 x+1(12×2 4)dx22.Z11x7 3x+cosx 7×6 3 sinxdx23.Rcos4xsinxdx24.Zcsc4xcot3xdx25.Zxsec3(x2)tan(x2)dx26.Zsin2xdx Hint:cos(2x)=1 2sin2x27.Z1p4 (x 1)2dx28.Z19x2+25dx29.Zdxxlnxp(lnx)2 3630.Z1p4x 4dx31.Zdxp7+6x x232.Z8x+14×2+12x+10dx257
LESSON13:ApplicationofAntidi erentiationtoDi erentialEquationsLEARNINGOUTCOME:Attheendofthelesson,thelearnershallbeabletosolvesepa-rabledi erentialequationsusingantidi erentiation.258
TOPIC13.1:SeparableDi erentialEquationsAdi erentialequation(DE)isanequationthatinvolvesx,yandthederivativesofy.Thefollowingareexamplesofdi erentialequations:1.dydx=2x+52.dydx= xy3.y00+y=0Theorderofadi erentialequationpertainstothehighestorderofthederivativethatappearsinthedi erentialequation.The rsttwoexamplesaboveare rst-orderDEsbecausetheyinvolveonlythe rstderivative,whilethelastexampleisasecond-orderDEbecausey00appearsintheequation.Asolutiontoadi erentialequationisafunctiony=f(x)orarelationf(x;y)=0thatsatis estheequation.Forexample,y=x2+5x+1isasolutiontodydx=2x+5sinceddx(y)=ddx(x2+5x+1)=2x+5:Therelationx2+y2=1isasolutiontodydx= xybecauseifwedi erentiatetherelationimplicitly,weget2x+2ydydx=0=)dydx= xy:Finally,y=sinxsolvesthedi erentialequationy00+y=0sincey0=cosxandy00= sinx,andthereforey00+y=( sinx)+sinx=0:Solvingadi erentialequationmeans ndingallpossiblesolutionstotheDE.Adi erentialequationissaidtobeseparableifitcanbeexpressedasf(x)dx=g(y)dy;wherefandgarefunctionsofxandy,respectively.Observethatwehaveseparatedthevariablesinthesensethattheleft-handsideonlyinvolvesxwhiletheright-handsideispurely259
intermsofy.Ifitispossibletoseparatethevariables,thenwecan ndthesolutionofthedi erentialequationbysimplyintegrating:Zf(x)dx=Zg(y)dy;andapplyingappropriatetechniquesofintegration.Notethattheleft-handsideyieldsafunc-tionofx,sayF(x)+C1,whiletheright-handsideyieldsafunctionofy,sayG(y)+C2.WethusobtainF(x)=G(y)+C(Here,C=C2 C1)whichwecanthenexpressintoasolutionoftheformy=H(x)+C,ifpossible.Wewillnowlookatsomeexamplesofhowtosolveseparabledi erentialequations.EXAMPLE1:Solvethedi erentialequationdydt=14y:Solution.Observethaty=0isasolutiontothedi erentialequation.Supposethaty6=0.Wedividebothsidesoftheequationbyytoseparatethevariables:dyy=14dt:Integratingtheleft-handsidewithrespecttoyandintegratingtheright-handsidewithrespecttotyieldZdyy=Z14dt=)lnjyj=14t+C:Takingtheexponentialofbothsides,weobtainjyj=e14t+C=eCe14t:Therefore,y=Ae14t;whereA=eCisanypositiveconstant.Therefore,alongwiththesolutiony=0thatwehavefoundatthestart,thesolutiontothegivendi erentialequationisy=Ae14t,whereAisanyrealnumber..EXAMPLE2:Solvethedi erentialequation2ydx 3xdy=0:Solution.Ify=0,thendydx=0whichmeansthatthefunctiony=0isasolutiontothedi erentialequation2y 3xdydx=0.Assumethaty6=0.Separatingthevariables,weget2ydx=3xdy260
2xdx=3ydy:Integratingbothsides,weobtainZ2xdx=Z3ydy2lnjxj=3lnjyj+Clnjxj2 C=lnjyj3jyj3=e Celnjxj2=e Cjxj2:Therefore,thesolutionsarey=3pAx2=3whereA=e Cisanypositiveconstant.Alongwiththetrivialsolutiony=0,thesetofsolutionstothedi erentialequationisy=Bx2=3whereBisanyrealnumber.Note:Theabsolutevaluebarsaredroppedsincex2isalreadynonnegative..EXAMPLE3:Solvetheequationxy3dx+ex2dy=0:Solution.First,weseparatethevariables.Wehave xy3dx=ex2dy xe x2dx=1y3dy:Integratingbothsidesoftheequationwithrespecttotheirvariables,wehave Zxe x2dx=Zy 3dy:Meanwhile,ifu= x2;thendu= 2xdxsothat du2=xdx.Hence, Zxe x2dx=12Zeudu=12eu+C=12e x2+C:Therefore, Zxe x2dx=Zy 3dy12e x2= y 22+C:.Ifwesolveforyintermsofx,wegety= A e x2 1=2,whereA=2Cisanyconstant.EXAMPLE4:Solvetheequation3(y+2)dx xydy=0:261
Solution.Separatingthevariablesofthedi erentialequationgivesus3(y+2)dx=xydy3xdx=yy+2dy3xdx=(y+2) 2y+2dy3xdx=1 2y+2dy:Nowthatwehaveseparatedthevariables,wenowintegratetheequationtermbyterm:Z3xdx=Z1 2y+2dy3lnjxj=y 2lnjy+2j+C:.Notethatinthepreviousexamples,aconstantofintegrationisalwayspresent.Ifthereareinitialconditions,orifweknowthatthesolutionpassesthroughapoint,wecansolvethisconstantandgetaparticularsolutiontothedi erentialequation.EXAMPLE5:Findtheparticularsolutionsofthefollowinggiventheircorrespondinginitialconditions:1.dydt=14ywheny=100andt=02.2ydx 3xdy=0whenx=1andy=13.xy3dx+ex2dy=0whenx=0andy=14.3(y+2)dx xydy=0whenx=1andy= 1Solution.Wewillusethegeneralsolutionsfromthepreviousexamples.1.ThesolutiontoExample1isy=Ae1=4t.Usingtheconditionsy=100andt=0,weget100=Ae0.Hence,A=100andthereforetheparticularsolutionisy=100e14t.2.ThesolutiontoExample2isy=Bx2=3.Substituting(x;y)=(1;1)gives1=B12=3=B.Hence,theparticularsolutionisy=3px2:262
3.FromtheExample3,thegeneralsolutionisy=1pA e x2.Substituting(x;y)=(1;0)yields1=1pA 1.Sincethesquarerootofarealnumberisnevernegative,pA 1=+1andsoA=2.Finally,theparticularsolutionisy=1p2 e x2:4.FromExample4,thegeneralsolutionis3lnjxj=y 2lnjy+2j+C:Substitutingthegivenvalues(x;y)=(1; 1),weobtain3lnj1j= 1 2lnj 1+2j+C:SimplifyingthisgivesC=1:Hence,theparticularsolutionistherelation3lnjxj=y 2lnjy+2j+1:.SolvedExamplesEXAMPLE1:Findthegeneralsolutiontothedi erentialequationdydx=x2y+3.Solution.dydx=x2y+3(y+3)dy=x2dxZ(y+3)dy=Zx2dxy22+3y=x33+C:.EXAMPLE2:Findthegeneralsolutiontothedi erentialequationdydx=1(ey+1)(1+x2).Solution.dydx=1(ey+1)(1+x2)(ey+1)dy=11+x2dxZ(ey+1)dy=Z11+x2dxey+y=tan 1x+C:.263
EXAMPLE3:Findthegeneralsolutiontothedi erentialequationdydx=y x3+xy 4:.Solution.dydx=y x3+xy 4dydx=yy 4x3+x1y 4ydy= x3+xdxZy 4ydy=Z x3+xdxZ1 4ydy=Z x3+xdxy 4lnjyj=x44+x22+C:.EXAMPLE4:Solvetheinitialvalueproblemxy+2x+dydx=x2(y+2)+y+2andy(0)=4:Solution.xy+2x+dydx=x2(y+2)+y+2x(y+2)+dydx=x2(y+2)+(y+2)dydx=x2(y+2) x(y+2)+(y+2)dydx=(y+2) x2 x+11y+2dy= x2 x+1dxZ1y+2dy=Z x2 x+1dxlnjy+2j=x33 x22+x+C:Sincey=4whenx=0,wehaveln(4+2)=Cln6=C:264
Thesolutiontotheinitialvalueproblemislnjy+2j=x33 x22+x+ln6:.EXAMPLE5:Solvetheinitialvalueproblemdydx=6e3x+1andy(0)=0:Solution.dydx=6e3x+1dy= 6e3x+1dxZdy=Z 6e3x+1dxy=6e3x3+x+Cy=2e3x+x+C:Substitutingx=0andy=0,weget0=2e0+0+C 2=CHencethesolutiontotheinitialvalueproblemisy=2e3x+x 2:.SupplementaryProblems1.Determinewhethereachofthefollowingdi erentialequationsisseparableornot,ifitisseparable,rewritetheequationintheformg(y)dy=f(x)dx.(a)dydx=y+3(b)dydx=x2y xy+3y(c)dydx=cos(3x y)(d)dydxx 3y+2(e)1sinxdydx=cos 3y2 y 1(f) xy3+3x+dydx=3 y3(g)dydx=p4+y x2.Findthegeneralsolutionofthefollowingdi erentialequations.(a)dydx=33y(b)dydx=y2(1+ex)(c)dydx=16×3(y 1)(d)dydx= x+12(y+3)(e)(tany)dydx=csc2(x+3)265
3.Solvethefollowinginitialvalueproblems.(a)dydx=x2 2andy(3)=7(b)dydx=xy 4andy(2)=9(c)dydx=2x3y2andy(2)=1(d)Findtheparticularsolutionofthedi erentialequationd2ydx2= 1x5determinedbytheconditionsy= 112anddydx=54whenx=1.266
LESSON14:ApplicationofDi erentialEquationsinLifeSci-encesLEARNINGOUTCOME:Attheendofthelesson,thelearnershallbeabletosolvesitua-tionalproblemsinvolving:exponentialgrowthanddecay,boundedgrowth,andlogisticgrowth.267
TOPIC14.1:SituationalProblemsInvolvingGrowthandDecayProblemsEXPONENTIALGROWTHANDDECAYLety=f(t)bethesizeofacertainpopulationattimet,andletthebirthanddeathratesbethepositiveconstantsbandd,respectively.Therateofchangedydtinthepopulationywithrespecttotimetisgivenbydydt=ky,wherek=b d:Ifkispositive,thatiswhenb>d,thentherearemorebirthsthandeathsanddydtdenotesgrowth.Ifkisnegative,thatiswhenb
Withthevalueswehaveenumeratedandsolved,theboundedequationisnowoftheformy=5000 4000(0:975)t:Wecannow ndtherequiredpopulation,y5.y=5000 4000(0:975)55000 4000(0:881)=5000 3524=1476:Therefore,therewillbeapproximately1476pawikansinthebreedingsite..Thenextexampleillustratesasortof\decay.”Rememberwesaidearlierthatthereareoccasionswheny>K?Thisisoneinstance.EXAMPLE4:Supposethatnewly-bakedcupcakesaretakenoutoftheovenwhichissetat100degrees.Roomtemperatureisfoundtobe25degrees,andin15minutesthecupcakesarefoundtohaveatemperatureof50degrees.Determinetheapproximatetemperatureofthecupcakesafter30minutes.Solution.Newton’sLawofCoolingstatesthattherateofchangeofthetemperatureofanobjectisequaltothedi erencebetweentheobject’stemperatureandthatofthesurroundingair.Thisgivesthedi erentialequationdydt= k(y 25):Sincethesituationanticipatesthatthetemperatureofanobject,y,willdecreasetowardsthatofthesurroundingair,ya.Thus,yisassumedtobegreaterthanya.Furthermore,todenotethedecrease,theconstantofproportionalityiswrittenas k,withk>0.tinthisproblemismeasuredinminutes.Byseparationofvariables,thisbecomesy=25+Ce kt:271
y0=100,wegetC=75andtheequationbecomesy=25+75e kt:The50-degreetemperatureafter15minutesgivese k=131=15;andtheequationchangesfurthertoy=25+7513t=15:Wecannowproceedtoapproximatethetemperatureafter30minutes:y=25+751330=15=25+75132=25+751925+8:33=33:33:Hence,after30minutes,thecupcakes’temperaturewillbeapproximately33degrees..LOGISTICGROWTHLety=f(t)bethesizeofacertainpopulationattimet.Wesaythatthetypeofgrowthyhasiscalledlogisticgrowthifysatis esthedi erentialequationdydt=ky(K y),whereyisthesizeofthepopulation.EXAMPLE5:TenPhilippineeagleswereintroducedtoanationalpark10yearsago.Therearenow23eaglesinthepark.Theparkcansupportamaximumof100eagles.Assumingalogisticgrowthmodel,whenwilltheeaglepopulationreach50?Solution.Tosolvetheproblem,we rstrecognizehowthegiveninformationwill tintoandimproveourequation.272
SinceK=100,wehavey=1001+Ce 100ktSincey0=23,wecansolveforC.10=1001+Ce010=1001+C10+10C=10010C=90orC=9:Hence,theequationbecomesy=1001+9e 100kt:Thecurrentpopulationof23eaglesisequaltothepopulationafter10years,ory10=23.Thispieceofinformationallowsustosolvefortheexponentialterm.23=1001+9e 100k1023=1001+9e 1000k1+9e 1000k=100239e 1000k=10023 19e 1000k=7723e 1000k=77239ore 1000k0:37:Insteadofsolvingfork,itwillsuceto ndasubstitutefore Kk=e 100k.Clearly,ife 1000k0:37,thene 100k(0:37)1=10.So,y=1001+9(0:37)t=10:Wearenowreadytoanswerthequestion,\Whenwilltheeaglepopulationreach50?”Giventhemostrecentversionofourlogisticequation,wejustsubstituteywith50andsolvefort,thetimerequiredtohave50eaglesinthepopulation.273
50=1001+9(0:37)t=1050(1+9(0:37)t=10)=10050+450(0:37)t=10=100(0:37)t=10=100 50450(0:37)t=10=19(0:37)t=1910ln(0:37)t=ln1910tln(0:37)=10ln19t=10ln(1=9)ln(0:37)t10(2:2)=22:Theeaglepopulationinthesaidnationalparkwillreach50inapproximately22years..SolvedExamplesSolveeachitemcompletely.EXAMPLE1:Itisobservedthatthenumberofbacteriapresentintheportionofthesmallintestineofananimalgrowsexponentially.Initially,thereare10,000bacteriapresent.Threehourslater,thebacteriagrowto500,000.Assumingnobacteriadieintheprocess,howmanybacteriawilltherebeafteroneday?Afterhowmanyhourswillthebacteriadoubleitsnumber?Solution.Lety(t)bethenumberofbacteriaintheintestineafterthours.Sincethenumberofbacteriafollowsanexponentialgrowthmodeltheny(t)=y0ektforsomeconstanty0andk.Usingtheassumptionsthatthereare10,000bacteriapresentinitiallyand500,000bacteriaarepresentafter3hoursthen10;000=y(0)=y0ek0=)y0=10;000and500;000=y(3)=10;000e3k=)e3k=50=)e3k=501=3:274
Hencey(t)=10;000(50)(1=3)t:Soafter1day=24hours,therewillbey(24)=10;000(50)(1=3)(24)=3:91017cellsintheintestineandthecellswilldoubleitsnumberafter2y(0)=y(t)20000=10000(50)(1=3)t50(1=3)t=2ln50(1=3)t=ln2tln50(1=3)t=ln2t=ln2ln 501=3t=0:53hours:.EXAMPLE2:Thehalf-lifeofaradioactivesubstanceisde nedtobetheamountoftimeittakesforthesubstacetodecay50%ofitsamount.IfsubstanceXhasahalf-lifeof3,600years,whatpartofsubstanceXwillremainafter4,500years?Solution.Lety(t)betheamountofsubstanceXthatremainsaftertyears.y(t)followsanexponentialdecaymodel.Lety0betheamountofsubstanceXatt=0.SincesubstanceXhasahalflifeof3,600years,theny02=y(3600)=y0e3600k=)e3600k=12=)ek=12(1=3600):Hence,y(t)=y012t(1=3600):andwhent=4;500,y(4500)=y0124500(1=3600)0:42y0:Approximately42%ofthesubstancewillremainafter4,500years..EXAMPLE3:Inacommunitywith300families,itwasfoundoutthat50familieswereinfectedbyanincurabledisease.Afteroneweek,thenumberoffamiliesthatwereinfectedbythediseasewasincreasedby10.Assumingthatthespreadofdiseasefollowsaboundedgrowthmodel,whenwill90%ofthecommunitybeinfected?275
Solution.Lety(t)bethenumberofinfectedfamiliesatweekt.Sincethespreadofdiseasefollowsaboundedgrowthmodel,thenK=300y0=50y(1)=60C=K y0=300 50=250So,y(t)=300 250e kt:Since60familieswereinfectedafteroneweek,theny(1)=60.Wecansolveforkusingthisinformation.300 250e k=60 250e k= 240e k=2425:Thenourmodelbecomesy(t)=300 2502425t:Wewantto ndtsuchthaty(t)=300(0:9).300 2502425t=300(0:9)2502425t=300(0:1)2425t=325ln2425t=ln325tln2425=ln325t=ln325ln2425t51:94Hence,90%ofthecommunitywillbeinfectedbythediseasein51.94weeks..276
EXAMPLE4:Athermometerreads18Cinsideanairconditionedhouse.Itisplacedout-sidethehousewheretheairtemperatureis30C.Threeminuteslater,itisfoundthatthethermometerreadingis21C.Findthethermometerreadingafter6minutes.Solution.Lety(t)bethetemperaturereadingofthethermometerattimetminutesafteritwasplacedoutside.ThenbyNewton’sLawofCooling,dydt= k(y 30):Solvingbyseparationofvariableswehave 1kdyy 30=dt 1klnjy 30j=t+Cy 30=e k(t+C)y=Ke kt+30;whereK=e kC.Sincey(0)=18,18=Ke k0+30K= 12;andy(3)=21implies21= 12e 3k+30 9= 12e 3ke 3k=0:75e k=0:751=3:Hencey(t)= 12(0:75)(1=3)t+30:In6minutes,thetemperaturereadingofthethermometerwillbey(6)= 12(0:75)(1=3)(6)+30=23:25C:.EXAMPLE5:Scientistsstockaverylargeaquariumwith500shrimpsforanexperimentandtheaquariumcanaccommodateupto10,000shrimps.Thenumberofshrimpstripledduringthe rstquarteroftheyear.277
1.Assumingthatthesizeoftheshrimppopulationsatis esthelogisticequation, ndanexpressionforthesizeofthepopulationaftertquarters.2.Howlongwillittakethepopulationtoreach3,000?Solution.Lety(t)betheshrimppopulationaftertquarters.Sincethepopulationsatis esthelogisticequation,theny(t)=K Ce ktwhereK=10;000,y(0)=500andy(1)=1500.Wesubstitutethesevaluestoobtainthefollowing500=10;000 C=)C=9;500and1;500=10;000 9;500e k=)9;500e k=8;500=)e k=1719:Hencey(t)=10;000 9;5001719t:Wenowsolvethetimewhenthepopulationbecomes3,000.y(t)=3;00010;000 9;5001719t=3;0009;5001719t=7;0001719t=1419ln1719t=ln1419tln1719=ln1419t=ln1419ln1719t=2:7456quarters.278
SupplementaryProblems1.Apetridishhas10;000bacteriainit.After4hoursthebacteria’spopulationincreasedto20,000.Ifthenumberofcoloniesgrowsexponentially,writeaformulaforthenumberofbacteriainthedishatanytimet,wheretisinhours.2.Inafarm,thereinitallywere10rabbits.Afterthreemonths,therearealready20rabbits.Assumingthatthegrowthisexponentialandnorabbitdiesintheprocess,atwhattimewilltherebe100rabbits?3.SupposethatthevalueofaP100,000assetdecreasesataconstantpercentagerateof10%peryear.(a)Finditsworthafter20years.(b)Whenwilltheassetbehalfofitsoriginalvalue?4.Acertaincultureisabletosustain100coloniesofbacteria.Initially,therewere10coloniesintheculture.After5hours,thecoloniesincreasedto25.Howmanycolonieswilltherebein24hours?Assumethatthebacteriafollowthelimitedgrowthmodel.5.Abowlofsoupat95C(toohot)isplacedina22Croom.Oneminutelater,thesouphascooledto83C.Whenwillthetemperaturebe49C(justright)?6.Abacterialpopulationisknowntohavealogisticgrowthpatternwithinitialpopulation1000andanequilibriumpopulationof10,000.Acountshowsthatattheendof1hourthereare2000bacteriapresent.Determinethepopulationasafunctionoftime.7.TheDepartmentofAnimalWelfarereleased100deerintoagamepreserve.Duringthe rst5years,thepopulationincreasesto450deer.Findamodelforthepopulationgrowthassuminglogisticgrowthwithalimitof5000deer.Whatwillbethepredictedpopulationofthedeerin10years?In20years?279
LESSON15:RiemannSumsandtheDe niteIntegralLEARNINGOUTCOMES:Attheendofthelesson,thelearnershallbeableto:1.ApproximatetheareaofaregionunderacurveusingRiemannsums:(a)left,(b)right,and(c)midpoint;and2.De nethede niteintegralasthelimitoftheRiemannsums.280
TOPIC15.1:ApproximationofAreausingRiemannSumsMETHODOFEXHAUSTIONNoticethatgeometryprovidesformulasfortheareaofaregionboundedbystraightlines.However,itdoesnotprovideformulastocomputetheareaofageneralregion.Forexample,itisquiteimpossibletocomputefortheareaoftheregionbelowtheparabolay=x2usinggeometryalone.1y=x2Area?xyEventheformulafortheareaofthecircleA=r2usesalimitingprocess.Before,sincepeopleonlyknewhowto ndtheareaofpolygons,theytriedtocovertheareaofacirclebyinscribingn-gonsuntiltheerrorwasverysmall.ThisiscalledtheMethodofExhaustion.ThemethodofexhaustionisattributedtotheancientGreekmathematicianAntiphonofAthens(ca.5thcenturyBCE),whothoughtofinscribingasequenceofregularpolygons,eachwithdoublethenumberofsidesthanthepreviousone,toapproximatetheareaofacircle.Ourmethodforapproximatingtheareaofaregionusesthesametechnique.However,insteadofinscribingregularn-gons,weusethesimplestpolygon{rectangles.RIEMANNSUMS281
Throughoutthislesson,wewillassumethatfunctionfispositive(thatis,thegraphisabovethex-axis),andcontinuousontheclosedandboundedinterval[a;b].ThegoalofthislessonistoapproximatetheareaoftheregionRboundedbyy=f(x),x=a;x=b;andthex-axis.aby=f(x)RxyWe rstpartition[a;b]regularly,thatis,intocongurentsubintervals.Similartothemethodofexhaustion,we llRwithrectanglesofequalwidths.TheRiemannsumoffreferstothenumberequaltothecombinedareaoftheserectangles.Noticethatasthenumberofrectanglesincreases,theRiemannsumapproximationoftheexactareaofRbecomesbetterandbetter.Ofcourse,theRiemannsumdependsonhowweconstructtherectanglesandwithhowmanyrectangleswe lltheregion.WewilldiscussthreebasictypesofRiemannsums:Left,Right,andMidpoint.PARTITIONPOINTSFirst,wediscusshowtodivideequallytheinterval[a;b]intonsubintervals.Todothis,wecomputethestepsizex,thelengthofeachsubinterval:x=b an:Next,weletx0=a,andforeachi=1;2;:::;n,wesettheithintermediatepointtobexi=a+ix:Clearly,thelastpointisxn=a+nx=a+n b an=b.Pleaserefertothefollowingtable:x0x1x2x3xixn 1xnaa+xa+2xa+3xa+ixa+(n 1)xbWecallthecollectionofpointsPn=fx0;x1;:::;xngasetofpartitionpointsof[a;b].Notethattodivideanintervalintonsubintervals,weneedn+1partitionpoints.282
EXAMPLE1:Findthestepsizexandthepartitionpointsneededtodividethegivenintervalintothegivennumberofsubintervals.Example:[0;1];6)x=16;P6=0;16;13;12;23;56;1 .[0;1];7Answer:x=17;P7=f0;17;27;37;47;57;67;1g[2;5];6Answer:x=12;P6=f2;52;3;72;492;5g[ 3;4];4Answer:x=74;P4=f 3; 54;12;94;4g[ 5; 1];5Answer:x=45;P5=f 5; 215; 175; 135; 95; 1g[ 10; 3];8Answer:x=78;P8=f 10; 738; 334; 598; 132; 458; 194; 318; 3gAssumethattheinterval[a;b]isalreadydividedintonsubintervals.Wethencovertheregionwithrectangleswhosebasescorrespondtoasubinterval.ThethreetypesofRiemannsumdependontheheightsoftherectangleswearecoveringtheregionwith.LEFTRIEMANNSUMThenthleftRiemannsumLnisthesumoftheareasoftherectangleswhoseheightsarethefunctionalvaluesoftheleftendpointsofeachsubinterval.Forexample,weconsiderthefollowingillustration.Wesubdividetheintervalintothreesubin-tervalscorrespondingtothreerectangles.Sinceweareconsideringleftendpoints,theheightofthe rstrectangleisf(x0),theheightofthesecondrectangleisf(x1),andtheheightofthethirdrectangleisf(x2).”””x0x1x2x3y=f(x)xy283
Therefore,inthisexample,the3rdleftRiemannsumequalsL3=f(x0)(x1 x0)+f(x1)(x2 x1)+f(x2)(x3 x2)=(f(x0)+f(x1)+f(x2))x:Ingeneral,if[a;b]issubdividedintonintervalswithpartitionpointsfx0;x1;:::;xng,thenthenthleftRiemannsumequalsLn=(f(x0)+f(x1)+:::+f(xn 1))x=nXk=1f(xk 1)x:Wede netherightandmidpointRiemannsumsinasimilarmanner.RIGHTRIEMANNSUMThenthrightRiemannsumRnisthesumoftheareasoftherectangleswhoseheightsarethefunctionalvaluesoftherightendpointsofeachsubinterval.Forexample,weconsiderthefollowingillustration.Wesubdividetheintervalintothreesubin-tervalscorrespondingtothreerectangles.Sinceweareconsideringrightendpoints,theheightofthe rstrectangleisf(x1),theheightofthesecondrectangleisf(x2),andtheheightofthethirdrectangleisf(x3).”””x0x1x2x3y=f(x)xyTherefore,inthisexample,the3rdrightRiemannsumequalsR3=f(x1)(x1 x0)+f(x2)(x2 x1)+f(x3)(x3 x2)=(f(x1)+f(x2)+f(x3))x:284
Ingeneral,if[a;b]issubdividedintonintervalswithpartitionpointsfx0;x1;:::;xng,thenthenthrightRiemannsumequalsRn=(f(x1)+f(x2)+:::+f(xn))x=nXk=1f(xk)x:MIDPOINTRIEMANNSUMThenthmidpointRiemannsumMnisthesumoftheareasoftherectangleswhoseheightsarethefunctionalvaluesofthemidpointsoftheendpointsofeachsubinterval.Forthesakeofnotation,wedenotebymkthemidpointoftwoconsecutivepartitionpointsxk 1andxk;thatis,mk=xk 1+xk2:Wenowconsiderthefollowingillustration.Wesubdividetheintervalintothreesubintervalscorrespondingtothreerectangles.Sinceweareconsideringmidpointsoftheendpoints,theheightofthe rstrectangleisf(m1),theheightofthesecondrectangleisf(m2),andtheheightofthethirdrectangleisf(m3).”””x0m1x1m2x2m3x3y=f(x)xyTherefore,inthisexample,the3rdmidpointRiemannsumequalsM3=f(m1)(x1 x0)+f(m2)(x2 x1)+f(m3)(x3 x2)=(f(m1)+f(m2)+f(m3))x:Ingeneral,if[a;b]issubdividedintonintervalswithpartitionpointsfx0;x1;:::;xng,thenthenthmidpointRiemannsumequalsLn=(f(m1)+f(m2)+:::+f(mn))x=nXk=1f(mk)x:285
EXAMPLE2:Findthe4thleft,right,andmidpointRiemannsumsofthefollowingfunctionswithrespecttoaregularpartitioningofthegivenintervals.1.f(x)=x2on[0;1]2.f(x)=sinxon[0;].Solution..1.First,notethatx=1 04=14:Hence,P4=0;14;12;34;1 .Wethencomputethemidpointsofthepartitionpoints:18;38;58;78 :1412341y=x2xy1412341xy181438125834781xyThe4thleftRiemannsumequalsL4=f(0)+f14+f12+f34x=0+116+14+91614=0:21875:The4thrightRiemannsumequalsR4=f14+f12+f34+f(1)x=116+14+916+114=0:46875:Lastly,the4thmidpointRiemannsumequalsM4=f18+f38+f58+f78x=164+964+2564+496414=0:328125:286
2.First,notethatx= 04=4:Hence,P4=0;4;2;34;1 .Wethencomputethemidpointsofthepartitionpoints:8;38;58;78 :4234y=sinxxy4234xy843825834781xyThe4thleftRiemannsumequalsL4=f(0)+f4+f2+f34x=sin0+sin4+sin2+sin344=1:896:::The4thrightRiemannsumequalsR4=f4+f2+f34+f()x=sin4+sin2+sin34+sin4=1:896:::Finally,the4thmidpointRiemannsumequalsM4=f8+f38+f58+f78x=sin8+sin38+sin58+sin784=2:052:::.Thefollowingexampleshowsthatarbitrarilyincreasingthenumberofpartitionpointsdoesnotnecessarilygiveabetterapproximationofthetrueareaoftheregion.EXAMPLE3:Letf(x)=1 jx 1jandconsidertheclosedregionRboundedbyy=f(x)andthex-axisontheinterval[0;2].287
121y=1 jx 1jxyRShowthatrelativetoregularpartitioning,thesecondleftRiemannsumL2isabetterapprox-imationoftheareaofRthanthethirdleftRiemannsumL3offon[0;2].Solution.First,observethattheexactareaofRis12(1)(2)=1:121xy234321xyNow,thestepsizeandpartitionpoints,repectively,arex=1andP2=f0;1;2gforL2andx=23andP3=f0;23;43;2gforL3.UsingtheformulafortheleftRiemannsum,wehavethefollowingcomputations:L2=(f(0)+f(1))x=(0+1)1=1whileL3=f(0)+f23+f43x=0+23+2323=4323=89:Clearly,L2iscloser(andinfact,equal)totheexactvalueof1,thanL3..REFINEMENTTodealwiththequestionabove,wede netheconceptofare nementofapartition:De nition6.ApartitionQofanintervalIisare nementofanotherpartitionPofIifPQ,meaning,QcontainsallpartitionpointsofPandmore.288
EXAMPLE4:FortheintervalI=[0;2],P3=f0;23;43;2gisnotare nementofP2=f0;1;2g.However,P4=f0;12;1;32;2gisare nementofP2whileP6=f0;13;23;1;43;53;2gisare nementofP3:EXAMPLE5:P2nisalwaysare nementofPn.InfactP2ncontainsallpartitionpointsofPnandallthemidpointstherein.Theorem15.SupposeQisare nementofP.Then,any(left,midpoint,right)RiemannsumapproximationofaregionRrelativetoQisequaltoorisbetterthanthesamekindofRiemannsumapproximationrelativetoP.Itisnothardtoconvinceourselvesaboutthevalidityoftheabovetheorem.Considerthediagramsbelow.TheoneontheleftillustratesL3whiletherightoneillustratesL6relativetotheregularpartitionofthegiveninterval.”””x0x1x2x3y=f(x)xy””””””x0x1x2x3x4x5x6y=f(x)xyRemark1:Thetheoremalsoconformstotheprocedureintheclassicalmethodofexhaustionwhereintheyusetheareasofinscribedregularn-gonstoapproximatetheareaofacircle.Thesequenceofn-gonstheGreeksconsideredwassuchthatthenextn-gonwouldhavetwicethenumberofsidesasthepreviousone.Remark2:Aconsequenceofthetheoremisthatforanypositiveintegern,thesequenceLn;L2n;L4n;L8n;L16n;:::isamonotonesequenceconvergingtotheexactareaoftheregion.ThismeansthatifAistheexactareaoftheregionandLnA,thenLnL2nL4nL8nL16n:::A:289
Forinstance,inExample3,A=1andL3=89.TheaboveinequalitiesimplythatL3L6A.Indeed,L6=f(0)+f13+f23+f(1)+f43+f53x=0+13+23+1+23+1313=1:IRREGULARPARTITIONSometimes,thepartitionofanintervalisirregular,thatis,thelengthsofthesubintervalsarenotequal.Thiskindofpartitioningisusuallyusedwhenyouwanttoobtainare nementofapartition(andtherebygetabetterapproximation)withoutcomputingforalotmorepoints.Forexample,supposethatyouthinkthataRiemannsumrelativetothepartitionP=f0;23;43;2gisalreadyclosetotheexactvalue,thenyoucansimplyinsertonemorepoint,say1,togetthepartitionP0=f0;23;1;43;2g:SinceP0isare nementofP,thenaRiemannsumrelativetoitshouldhaveabettervaluethanthatofPandyoujusthavetocomputeforonemorefunctionalvalue,f(1),ratherthan3morevalues.TogetRiemannsumsrelativetoirregularpartitions,theideaisthesame,youjusthavetobecarefulaboutthevariablestepsizes.ConsideranirregularpartitionP=fx0;x1;x2;x3;:::;xngofaninterval.Ingeneralitmaynotbethecasethatx1 x0=x2 x1.So,wede nethestepsizesxk.Foreachk2f1;2;:::;ng,de nethekthstepsizexktobethelengthofthekthsubinterval,i.e.xk=xk xk 1:Withthisnotation,theleftRiemannsumwithrespecttothepartitionPisLP=f(x0)x1+f(x1)x2+:::+f(xn 2)xn 1+f(xn 1)xn=nXk=1f(xk 1)xk:290
Now,supposewearegiveny=f(x)andapartitionP=fx0;x1;x2;x3g.ThentheleftRiemannsumwithre-specttothepartitionPisLP=f(x0)x1+f(x1)x2+f(x2)x3:”””x1x2x3x0x1x2x3y=f(x)xyVerysimilarly,therightRiemannsumisgivenbyRP=f(x1)x1+f(x2)x2+:::+f(xn 1)xn 1+f(xn)xn=nXk=1f(xk)xk;andthemidpointRiemannsumisgivenbyMP=f(m1)x1+f(m2)x2+:::+f(mn 1)xn 1+f(mn)xn=nXk=1f(mk)xk;withthesameconventionthatmk=xk 1+xk2,themidpointofthekthinterval.EXAMPLE6:RelativetothepartitionP=f0;12;23;34;1g, ndtheleft,right,andmidpointRiemannsumsoff(x)=x2ontheinterval[0;1].Solution.ObservethatPpartitions[0;1]into4irregularsubintervals:[0;12];[12;23];[23;34];[34;1].Thestepsizesarex1=12;x2=16;x3=112;x4=14:ThisimpliesthattheRiemannsumsare291
LP=f(0)x1+f12x2+f23x3+f34x4=012+1416+49112+91614=0:2193:RP=f12x1+f23x2+f34x3+f(1)x4=1412+4916+916112+114=0:4959:MP=f14x1+f712x2+f1724x3+f78x4=11612+4914416+289576112+496414=0:3212:1223341y=x2xy1223341xy1223341xy.SolvedExamplesEXAMPLE1:Findthestepsizexandthepartitioningneededtodividethegivenintervalintogivennumberofregularsubintervals.a.[0;2];8b.[0;]:4c.[ 1;4]:5Solution..a.First,wesolveforx.Fora=0,b=2,andn=8,x=b an=2 08=28=14:Next,wecomputeforthepartitionP6:P6=0;14;12;34;1;54;32;74;2:292
b.Similarto(a),wecomputeforthevaluesofxandP4.Forxwehave,x= 04=4;andforthepartition,wehaveP4=0;4;2;34;:c.Notethat,x=1.Hence,P5=f 1;0;1;2;3;4g:.EXAMPLE2:Findthe5thleft,right,andmidpointRiemannsumsofthefollowinggivenfunctionswithrespecttotheregularpartitioningofthegivenintervals.a.f(x)=2xon[0;1]b.g(x)=(x+1)2on[ 1;4]Solution..a.First,notethatx=1 05=15andP5=0;15;25;35;45;1.Also,wecomputeforthemidpointsofthepartitionpoints:110;310;12;710;910.The5thleftRiemannsumisL5=f(0)+f15+f25+f35+f45x=0+25+45+65+8515=45=0:80:Next,the5thrightRiemannsumisR5=f15+f25+f35+f45+f(1)x=25+45+65+85+215=1:20:Lastly,the5thmidpointRiemannsumisM5=f110+f310+f12+f710+f910x=210+35+1+75+9515=1:293
b.Notethatx=4 ( 1)5=55=1.Then,P5=f 1;0;1;2;3;4g:Furthermore,themidpointsofthepartitionpointsare 12;12;32;52;72:The5thleftRiemannsumisL5=[g( 1)+g(0)+g(1)+g(2)+g(3)]x=(0+1+4+9+16)1=30:The5thrightRiemannsumisR5=[g(0)+g(1)+g(2)+g(3)+g(4)]x=(1+4+9+16+25)1=55:Finally,the5thmidpointRiemannsumisM5=g 12+g12+g32+g52+g72x=14+94+254+494+8141=1654=41:25:.EXAMPLE3:Findthe6thleft,right,andmidpointRiemannsumsofthefollowinggivenfunctionswithrespecttotheregularpartitioningofthegivenintervals.a.h(x)=cosxon[0;2]b.k(x)=1xon[1;4]Solution..a.Computingforx=2 06=12.Hence,P6=0;12;6;4;3;512;2.Wefurthercomputeforthemidpointsofthethepartitionpoints:24;8;524;724;38;1124:The6thleftRiemannsumisL6=h(0)+h12+h6+h4+h3+h512x=cos0+cos12+cos6+cos4+cos3+cos51212=1:125:::294
The6thrightRiemannsumisR6=h12+h6+h4+h3+h512+h2x=cos12+cos6+cos4+cos3+cos512+cos212=0:863:::Lastly,the6thmidpointRiemannsumisM6=h24+h8+h524+h724+h38+h1124x=cos24+cos8+cos524+cos724+cos38+cos112412=1:002:::b.Notethatx=4 16=12.Hence,P6=1;32;2;52;3;72;4andthemidpointsofthispartitionare54;74;94;114;134;154.The6thleftRiemannsumisL6=k(1)+k32+k(2)+k52+k(3)+k72x=1+23+12+25+13+2712=223140=1:592:::The6thrightRiemannsumisR6=k32+k(2)+k52+k(3)+k72+k(4)x=23+12+25+13+27+1412=341280=1:217:::The6thmidpointRiemannsumisM6=k54+k74+k94+k114+k134+k154x=45+47+49+411+413+41512=1:376:::.295
EXAMPLE4:Letf(x)=x+1andconsidertheclosedregionboundedbyy=f(x)andthex-axisontheinterval[0;3]:12314y=x+1xyShowthatrelativetotheregularpartitioning,thethirdrightRiemannsumR3isabetterapproximationtotheareaoftheclosedregionthanthesecondrightRiemannsumR2offon[0;3].Solution.Theclosedregionisatrapezoidwithbases1and4andheight3.Therefore,theareaisequalto1+423=152=7:5:Thestepsizeandthepartitionarex=1:5andP2=f0;1:5;3g,respectively,forR2andx=1andP3=f0;1;2;3g,respectively,forR3.WecannowcomputeforthevaluesofR2andR3.R2=(f(1:5)+f(3))x=(2:5+4)1:5=9:75;whileR3=(f(1)+f(2)+f(3))x=(2+3+4)1=9:ComparingthevaluesofR2andR3totheoriginalareaofthetrapezoid,R3isindeedabetterapproximationthanR2..EXAMPLE5:LetP1=f0;1;2;3g,P2=f0;14;12;34;1;32;2;3gandP3=f0:8;1;2;3gbepartitionsoninterval[0;3].ExplainwhyP2isare nementofP1butnotP3.Solution.P2isare nementofP1becauseP1P2.Ontheotherhand,P3isnotare nementofP1sinceP3*P1:.EXAMPLE6:GiventheirregularpartitionP=f0;34;1;32;2g, ndtheleft,right,andmidpointRiemannsumsoff(x)=pxontheinterval[0;2]:296
Solution.NotethatPpartitions[0;2]into4subintervals:[0;34],[34;1],[1;32],and[32;2]:Thestepsizesofeachsubarex1=34,x2=14,x3=12,andx4=12.Therefore,theRiemannsumsrelativetothepartitionParegiven:LP=f(0)x1+f34x2+f(1)x3+f32x4=034+r3414+112+r3212=2:078:::RP=f34x1+f(1)x2+f32x3+f(2)x4=r3434+114+r3212+p212=2:218:::MP=f38x1+f78x2+f54x3+f74x4=r3834+r7814+r5412+r7412=1:913:::.SupplementaryProblems1.Letf(x)=3xbede nedon[0;2].FindtheleftRiemannsumrelativetotheregularpartitionP3:2.Letf(x)=p4 x2bede nedon[ 1;1].FindtherightRiemannsumrelativetotheregularpartitionP4:3.Letf(x)=3x+1bede nedon[1;3].FindtherightRiemannsumrelativetotheregularpartitionP3:4.Letf(x)=x3+2bede nedon[ 1;1].FindtheleftRiemannsumrelativetotheregularpartitionP2:5.Letf(x)=x2+2x+4bede nedon[0;1].FindtherightRiemannsumrelativetotheregularpartitionP3:6.Letf(x)=1+px+1bede nedon[ 1;1].FindthemidpointRiemannsumrelativetotheregularpartitionP4:297
7.Letf(x)=1x+1bede nedon[1;3].FindtheleftRiemannsumrelativetotheregularpartitionP4:8.Letf(x)=sin(2x)bede nedon[0;].FindtherightRiemannsumrelativetotheregularpartitionP5:9.Letf(x)=2tan(x)bede nedon[0;4].FindthemidpointRiemannsumrelativetotheregularpartitionP3:10.Letf(x)=x2 4bede nedon[0;5].FindtheleftRiemannsumrelativetothepartitionP=0;14;12;1;43;2 298
TOPIC15.2:TheFormalDe nitionoftheDe niteIntegralDEFINITEINTEGRALWeworkwithacontinuouspositivefunctiony=f(x)de nedonaclosedandboundedinterval[a;b].Theobjectiveofthislessonisto ndtheareaoftheregionRboundedbyy=f(x)fromabove,thex-axisfrombelow,thelinex=afromtheleftandx=bfromtheright.aby=f(x)RxyToavoidcomplications,wejustconsiderthecasewherethepartitionontheintervalisregular.WerecallthatPn=fx0;x1;:::;xn 1;xng(wherex0=aandxk=xk 1+xwithx=b an)partitions[a;b]intoncongruentsubintervals.Foreachsubintervalk=1;2;:::;n,letxkbeanypointinthekthsubinterval[xk 1;xk].Then,theRiemannsum,de nedbythischoiceofpoints,relativetothepartitionPisf(x1)x+f(x2)x+:::+f(xn 1)x+f(xn)x=nXk=1f(xk)x:IfthisisaleftRiemannsum,thenxk=xk 1;IfthisisarightRiemannsum,thenxk=xk;and nally,IfthisisamidpointRiemannsum,thenxk=12(xk 1+xk).Inanycase,weknowthattheaboveRiemannsumisonlyanapproximationoftheexactareaofR.Tomakethisestimateexact,weletnapproachin nity.ThislimitoftheRiemannsumiswhatwecallthede niteintegraloffover[a;b]:Zbaf(x)dx=limn!1nXk=1f(xk)x;299
ifthislimitexists.Thevalueofthisintegraldoesnotdependonthekind(left,right,ormidpoint)ofRiemannsumbeingused.Rememberthatxactuallydependsonn:x=b an.So,thistermcannotbetakenoutfromthelimitoperator.Aftertakingthelimit,thisxbecomesourdi erentialdx.TheintegralsignRandthedi erentialdxactasdelimiters,whichindicatethateverythingbetweenthemistheintegrand-theupperboundaryoftheregionwhoseareaiswhatthisintegralisequalto.Thenumbersaandbarecalledthelowerandupperlimitsofintegration,respectively.RecallthattheintegralsignRwasearlierusedtodenotetheprocessofantidif-ferentiation.Thereisareasonwhythesamesymbol(R)isbeingused{weshallseelaterthatantiderivativesareintimatelyrelatedto ndingareasbelowcurves.GeometricInterpretationoftheDe niteIntegralLetfbeapositivecontinuousfunctionon[a;b].Thenthede niteintegralZbaf(x)dxistheareaoftheregionboundedbyy=f(x),thex-axis,x=a,andx=b.Consequently,thede niteintegraldoesnotdependonthevariablex.Changingthisvariableonlychangesthenameofthex-axisbutnottheareaoftheregion.Therefore,Zbaf(x)dx=Zbaf(t)dt=Zbaf(u)du:COMPUTINGDEFINITEINTEGRALSBYAPPEALINGTOGEOMETRICFORMULASUsingthegeometricinterpretationofthede niteintegral,wecanalwaysthinkofade niteintegralasanareaofaregion.Ifweareluckythattheregionhasanareathatiseasytocomputeusingelementarygeometry,thenweareabletosolvethede niteintegralwithoutresortingtoitsde nition.EXAMPLE1:Findtheexactvaluesofthefollowingde niteintegrals:1.Z213dx2.Z20(1 jx 1j)dx3.Z31(3x+1)dx4.Z1 1p1 x2dxSolution.Usingtheabovede nitionofthede niteintegral,wejustdrawtheregionand nditsareausingelementarygeometry.300
1.Thegraphofy=3isahorizontalline.Sincetheregionisarectangle,itsareaequalsLW=23=6:Therefore,Z313dx=6:133y=3xy2.Thegraphofy=1 jx 1jisasgiven.Theshadedregionisatrianglewithbaseb=2andheighth=1.Itsareaequals12bh=1.Therefore,Z20(1 jx 1j)dx=1:121y=1 jx 1jxy3.Thegraphofy=3x+1isalineslantingtotheright.Theshadedregionisatrapezoidwithbasesb1=4andb2=10andheighth=2.Itsareaequals12(b1+b2)h=12(4+10)2=14.Therefore,Z31(3x+1)dx=14:123410y=3x+1xy4.Thegraphofy=p1 x2isasemicirclecenteredattheoriginwithradius1.Theareaoftheshadedregionis12(1)2:Therefore,Z1 1p1 x2dx=2: 111y=p1 x2xy301
.LIMITSATINFINITYOFRATIONALFUNCTIONSConsidertherationalfunctionf(x)=3x+1x+3.Wedescribethebehaviorofthisfunctionforlargevaluesofxusingatableofvalues:xf(x)=3x+1x+31002:922331;0002:9920210;0002:99920100;0002:99992Clearly,asxgoesverylarge,thevalueofyapproachesthevalueof3.Wesaythatthelimitoff(x)=3x+1x+3asxapproachesin nityis3,andwewritelimx!13x+1x+3=3:Tocomputelimitsatin nityofrationalfunctions,itisveryhelpfultoknowthefollowingtheorem:Theorem16.Ifp>0,thenlimx!11xp=0.Weillustratethetheoremusingthegraphofy=1xp:246820y=1xpObservethatasxtakeslargevalues,thegraphapproachesthex-axis,orthey=0line.Thismeansthatthevaluesofycantakearbitrarilysmallvaluesbymakingxverylarge.Thisiswhatwemeanbylimx!11xp=0.Thetechniqueinsolvingthelimitatin nityofrationalfunctionsistodividebythelargestpowerofxintherationalfunctionandapplytheabovetheorem.302
EXAMPLE2:Computethelimitsofthefollowingrationalfunctions.(a)limx!12x+45x+1(b)limx!14 x+x23x2 2x+7(c)limx!120x+13×3 5x+1(d)limx!13×2+48x 1Solution..(a)Thehighestpowerofxhereis1.So,2x+45x+1=2x+45x+11x1x=2+4×5+1x:UsinglimittheoremsandTheorem16,limx!12x+45x+1=limx!12+4limx!11xlimx!15+limx!11x=2+05+0=25:(b)Thehighestpowerofxhereis2.So,byTheorem16,limx!14 x+x23x2 2x+71x21x2=limx!14×2+1x+13 2x+7×2=0 0+13 0+0=13:(c)Thehighestpowerofxhereis3.Again,usingTheorem16,limx!120x+13×3 5x+11x31x3=limx!120×2+1×33 5×2+1×3=0+03 0+0=0:(d)Thehighestpowerofxhereis2.So,limx!13×2+48x 11x21x2=limx!13+4x28x 1×2:Noticethatthenumeratorapproaches3butthedenominatorapproaches0.Therefore,thelimitdoesnotexist..303
LimitsatIn nityofRationalFunctionsSupposethedegreesofthepolynomialsp(x)andq(x)aremandn,respectively.Thereareonlythreecasesthatcouldhappeninsolvingforlimx!1p(x)q(x).Ifm=n,thenlimx!1p(x)q(x)exists.Infact,thelimitisnonzeroandequalstheratiooftheleadingcoecientofp(x)totheleadingcoecientofq(x).Ifm
Weapplypropertiesofthesummation:distributingthesummationsymboloverthesum,factoringoutthosewhichareindependentoftheindexk,and nally,applyingtheformulas,nXk=11=nandnXk=1k=n(n+1)2:ThisgivesRn=8nnXk=11+12n2nXk=1k=8nn+12n2n(n+1)2=8+61+1n:Finally,byde nitionZ313x+1dx=limn!1Rn=limn!18+61+1n:UsingTheorem16,limn!11n=0.Therefore,itfollowsthatZ313x+1dx=14,asdesired..Forthenextexample,weneedthefollowingformulaforthesumofthe rstnperfectsquares:nXk=1k2=n(n+1)(2n+1)6:(3.4)EXAMPLE4:Usethede nitionofthede niteintegralasalimitofaRiemannsumtoshowthatZ10x2dx=13:Solution.Forconvenience,letusagainusetherightRiemannsumrelativetothepartitionPn=f0;1n;2n;:::;n 1n;1gof[0;1].Clearly,x=1nandxk=x0+kx=0+k1n=kn:BytheformulaoftherightRiemannsum,wehaveRn=nXk=1f(xk)xk=nXk=1kn21n:Simplifying,andusingformula(3.4),weobtainRn=1n3nXk=1k2=1n3n(n+1)(2n+1)6=1n32n3+3n2+n6=13+12n+16n2:Thede niteintegralisjustthelimitoftheaboveexpressionasntendstoin nity.WeuseTheorem16toevaluatethelimitsofthelasttwotermsintheexpression.Therefore,wehaveZ10x2dx=limn!1Rn=limn!113+12n+16n2=13:305
Thisillustratesthepowerofthede niteintegralincomputingareasofnon-polygonalregions.1y=x2xyArea=13.PROPERTIESOFTHEDEFINITEINTEGRALAsalimitofasum,thede niteintegralsharesthecommonpropertiesofthelimitandofthesummation.Theorem17(LinearityoftheDe niteIntegral).Letfandgbepositivecontinuousfunctionson[a;b]andletc2R.Then1.Zbacf(x)dx=cZbaf(x)dx2.Zba(f(x)g(x))dx=Zbaf(x)dxZbag(x)dxEXAMPLE5:SupposewearegiventhatZbaf(x)dx=2andZbag(x)dx=7.Findtheexactvalueofthefollowing:1.Zba3f(x)dx2.Zbaf(x) g(x)dx3.Zba2f(x)+g(x)dx4.Zba3f(x) 2g(x)dxSolution.Usingthepropertiesoftheintegral,1.Zba3f(x)dx=3Zbaf(x)dx=3(2)=6:2.Zbaf(x) g(x)dx=Zbaf(x)dx Zbag(x)dx=2 7= 5:306
3.Zba2f(x)+g(x)dx=2Zbaf(x)+Zbag(x)dx=2(2)+7=11:4.Zba3f(x) 2g(x)dx=3Zbaf(x)dx 2Zbag(x)dx=3(2) 2(7)= 8:.Anotherimportantpropertyofthede niteintegraliscalledadditivity.Beforeweproceed,we rstde nethefollowing:De nition7.Letfbeacontinuouspositivefunctionon[a;b].Then1.Zaaf(x)dx=0,and2.Zabf(x)dx= Zbaf(x)dx:The rstoneisveryintuitiveifyouvisualizethede niteintegralasanareaofaregion.Sincetheleftandrightboundariesarethesame(x=a),thenthereisnoregionandtheareathereforeis0.Thesecondonegivesmeaningtoade niteintegralwheneverthelowerlimitofintegrationisbiggerthantheupperlimitofintegration.Wewillseelaterthatthisisneededsothepropertyofadditivitywillbeconsistentwithourintuitivenotion.Theorem18(AdditivityoftheDe niteIntegral).LetfbeapositivecontinuousfunctiononaclosedandboundedintervalIcontainingdistinctnumbersa;bandc.ThenZbaf(x)dx+Zcbf(x)dx=Zcaf(x)dxnomatterhowa,bandcareorderedontheintervalI.EXAMPLE6:SupposethatZ20f(x)dx=IandZ10f(x)dx=J.FindZ21f(x)dx:Solution.Byadditivity,Z20f(x)dx=Z10f(x)dx+Z21f(x)dx:SubstitutingthegivenvaluesyieldsI=Z21f(x)dx+J:ThisimpliesthatZ21f(x)dx=I J..307
EXAMPLE7:SupposewearegiventhatZbaf(x)dx=3;Zcdf(x)=10andZbdf(x)dx=4.FindZacf(x)dx.Solution.Byadditivity,Zacf(x)dx=Zdcf(x)dx+Zbdf(x)dx+Zabf(x)dx= Zcdf(x)dx+Zbdf(x)dx Zbaf(x)dx= 10+4 3= 9:.THEDEFINITEINTEGRALASANETSIGNEDAREAWealwaysassumedthatthefunctionthatweareconsideringisalwayspositive.Whathappensifthefunctionhasanegativepart?Howdoweinterpretthisgeometrically?GeneralGeometricInterpretationofaDe niteIntegralasaNetSignedAreaSupposethatfisacontinuousfunctionon[a;b].(Noticethatwedroppedtheassumptionthatfmustbepositive.)Thenthede niteintegralZbaf(x)dxisthenetsignedareaoftheregionwithboundariesy=f(x),x=a,x=b,andthex-axis.Thenetsignedareaequalsthesumofalltheareasabovethex-axisminusthesumofalltheareasbelowthex-axis.Ine ect,weareassociatingpositiveareasfortheregionsabovethex-axisandnegativeareasfortheregionsbelowthex-axis.308
Forexample,considerthefollow-inggraphofy=f(x)on[a;b].IftheareasoftheshadedregionsareA,BandC,asshown,thenZbaf(x)dx=A+C B:ABCaby=f(x)xyEXAMPLE8:Interpretthefollowingintegralsassignedareastoobtainitsvalue.1.Z502x 2dx2.Z=2 =2sinxdxSolution..1.Thegraphofy=2x 2isshownontheright.From0to1,theareaofthetrian-gle(belowthex-axis)is12(1)(2)=1:From1to5,theareaofthetriangle(abovethex-axis)is12(4)(8)=16:Therefore,thenetsignedareaequalsZ502x 2dx=16 1=15:15-28y=2x 2xy2.Thegraphofy=sinxisshownontheright.Observethatbecausey=sinxissym-metricwithrespecttotheorigin,theregion(belowthex-axis)from =2to0iscongruenttotheregion(abovethex-axis)from0to=2.Therefore,thenetsignedareaequalsZ=2 =2sinxdx=0: 2 2-11xy309
.SolvedExamplesEXAMPLE1:Findtheexactvaluesofthefollowingde niteintegrals.1.Z4 16dx2.Z303xdx3.Z41(2x+1)dx4.Z3 3p9 x2dxSolution..1.Observethattheregionisarectangle.Hence,theareaisequaltoLW=56=30:Therefore,Z4 16dx=30 146y=6xy2.Fromgeometry,theareaoftriangleis12timesthebase,3,andaheight,9.Therefore,Z303xdx=272:39y=3xxy3.Therequiredareaiscoveredbyatrapezoid.NotethattheareaofatrapezoidisgivenbyA=b1+b22h=3+923=18:Therefore,Z41(2x+1)dx=18:1439y=2x+1xy310
4.Observethattheshadedregionisasemicirclecenteredattheoriginwithradius3.Hence,A=(3)2=9:Therefore,Z3 3p9 x2dx=9: 333y=p9 x2xy.EXAMPLE2:Usethede nitionofde niteintegralasalimitoftheRiemannsumtoshowthatZ42(2x 1)dx=10:Solution.First,wegettherightRiemannsum(notethatthechoiceofrighthereisarbitrary)off(x)=2x 1relativetotheregularpartitionPnof[2;4].Thisgivesx=4 2n=2n.Hence,thepartitionpointsarexk=x0+kx=2+2kn:Thus,wehaveRn=nXk=1f(xk)x=nXk=1(2xk 1)x:Wetakenoteofthefollowingformulas:nXk=11=nandnXk=1k=n(n+1)2:Thus,Rn=nXk=122+2kn 12n=nXk=14+4kn 12n=nXk=13+4kn2n=nXk=16n+8kn2=6nnXk=11+8n2nXk=1k=6nn+8n2n(n+1)2=6+4n2+nn2=6+41+1n=10+4n:Byde nition,Z42(2x 1)dx=limn!1Rn=limn!110+4n=10:311
.EXAMPLE3:ShowthatZ40(3×2+3)dx=76usingthede nitionofthede niteintegralasalimitsum.Solution.WeagainusetherightRiemannsumoff(x)=3×2+3relativetotheregularpartitionPnof[0;4].Notethatx=4n.Thus,thepointsarexk=x0+kx=4kn:AnotherusefulformulainevaluatingtheRiemannsumisnXk=1k2=n(n+1)(2n+1)6=2n3+3n2+n6:UsingtheformulafortherightRiemannsum,wehaveRn=nXk=1f(xk)x=nXk=1 34kn2+3!4n=nXk=1316k2n2+34n=nXk=148k2n2+34n=nXk=1192k2n2+12n=192n2nXk=1k2+12nnXk=11=192n2n3+3n2+n6+12n(n)=19213+12n+16n2+12:Therefore,bythede nitionofde niteintegral,Z40(3×2+3)dx=limn!1Rn=limn!119213+12n+16n2+12=19213+12=76:.EXAMPLE4:Usethede nitionofde niteintegraltoshowZ52(4x x2)dx=3:Solution.UsingtherightRiemannsumrelativetothepartitionPnof[2;5],itisclearthatx=3nandxk=2+3kn:312
Thus,Rn=nXk=1(4xk x2k)x=nXk=1 42+3kn 2+3kn2!3n=nXk=18+12kn 4+12kn+9k2n23n=nXk=14 9k2n23n=nXk=112n 27k2n3=12nnXk=11 27n3nXk=1k2=12 2713+12n+16n2:Hence,bythede nitionofde niteintegral,Z52(4x x2)dx=limn!1Rn=limn!112 2713+12n+16n2=12 9=3:.EXAMPLE5:SupposethatZbaf(x)dx=9andZbag(x)dx=4.Findtheexactvalueofthefollowing:1.Zba2f(x)dx2.Zba(f(x) g(x))dx3.Zba(f(x)+2g(x))dx4.Zba(3f(x) 2g(x))dx313
Solution..1.Zba2f(x)dx=2Zbaf(x)dx=2(9)=18:2.Zba(f(x) g(x))dx=Zbaf(x)dx Zbag(x)dx=9 4=5:3.Zba(f(x)+2g(x))dx=Zbaf(x)dx+2Zbag(x)dx=9+2(4)=9+8=17:4.Zba(3f(x) 2g(x))dx=3Zbaf(x)dx 2Zbag(x)dx=3(9) 2(4)=19:.EXAMPLE6:Interpretthefollowingintegralsasnetsignedareastoobtainitsvalues.1.Z41( 2x+4)dx2.Z1 1x3dxSolution..1.Theareaofthetrianglefrom1to2is1,whiletheareaoftrianglefrom2to4is4.Therefore,thenetsignedareaisZ41( 2x+4)dx= 3:124-42y= 2x+4xy2.Noticethatthegraphofy=x3issymmetricwithrespecttotheorigin.Hence,theregionfrom 1to0isthesameastheregionfrom0to1.Thus,thenetsignedareaisZ1 1x3dx=0: 1 1-11y=x3xy.314
SupplementaryProblems1.Evaluateeachintegralbyinterpretingitintermsofsignedareas.(a)Z1 14dx(b)Z305dx(c)Z312dx(d)Z2 2jxjdx(e)Z6 4(4 jxj)dx(f)Z1 3(2 jx+1j)dx(g)Z41(2x+2)dx(h)Z52(3x 4)dx(i)Z 1 4( x+2)dx(j)Z6 6p36 x2dx(k)Z30p9 x2dx(l)Z4 4p16 x2dx2.Given:Z21f(x)dx=5,Z41f(x)dx,Z21g(x)dx,andZg(x)dx=1.Findthefollowing.(a)Z11f(x)dx(b)Z2 3g(x)dx(c)Z21(4g(x) 3f(x))dx(d)Z14f(x)dx(e)Z42f(x)dx3.Usethede nitionofde niteintegralasalimitofRiemannsumtoevaluatethefollowingde niteintegrals.(a)Z504xdx(b)Z30(5×2 2x)dx(c)Z0 2(3×2+2x)dx(d)Z20(x2+10)dx(e)Z0 3(4×2 5x 1)dx(f)Z 1 5(x2+3x+5)dx315
LESSON16:TheFundamentalTheoremofCalculusLEARNINGOUTCOMES:Attheendofthelesson,thelearnershallbeableto:1.IllustratetheFundamentalTheoremofCalculus;and2.Computethede niteintegralofafunctionusingtheFundamentalTheoremofCalculus.316
TOPIC16.1:IllustrationoftheFundamentalTheoremofCal-culusFundamentalTheoremofCalculus(FTOC)Letfbeacontinuousfunctionon[a;b]andletFbeanantiderivativeoff,thatis,F0(x)=f(x).ThenZbaf(x)dx=F(b) F(a):EXAMPLE1:NotethatF(x)=x33isanantiderivativeoff(x)=x2(sinceF0(x)=f(x).)Hence,byFTOC,Z10x2dx=x33 10=F(1) F(0)=13 0=13;aswehaveseenbefore.VerticalBarNotationWeadoptthefollowingnotationforevaluatingF(x)fromx=atox=b:F(x) ba=F(b) F(a):Forexample,(1+x x2) 21=(1+2 22) (1+1 12)=( 1) (1)= 2;andsinx =2=4=sin(=2) sin(=4)=1 p22=2 p22:Usingtheabovenotation,theFTOCnowstates:IfFisanantiderivativeoff,thenZbaf(x)dx=F(x) ba:Theconstantofintegrationthatwasnecessaryforinde niteintegrationwillnowjustcancelout: F(x)+C ba= F(b)+C F(a)+C=F(b) F(a)=F(x) ba:317
ThenextexampleswillvalidatethatFTOCworksbyredoingtheexamplesintheprevioussection.EXAMPLE2:Withoutreferringtothegraphsoftheintegrands, ndtheexactvaluesofthefollowingde niteintegrals:1.Z213dx2.Z20(1 jx 1j)dx3.Z31(3x+1)dx4.Z1 1p1 x2dxSolution.WeintegrateusingtheFundamentalTheoremofCalculus.1.Z213dx=3x 21=3(2 1)=3:2.Thesolutionforthisproblemtakesafewmorestepsbecausetheabsolutevaluefunctionisessentiallyapiecewisefunction.RecallthatifEisanyexpression,thenjEj=EifE0,whilejEj= EifE<0.Withthisinmind,1 jx 1j=8<:1 (x 1)ifx 101 [ (x 1)]ifx 1<0=8<:2 xifx1xifx<1:Therefore,byadditivity,Z20(1 jx 1j)dx=Z212 xdx+Z10xdx=2x x22 21+x22 10=4 42 2 12+12 0=1:3.Z31(3x+1)dx=3x22+x 31=272+3 32+1=14:4.Thereisatechniqueofintegrationneededtointegratep1 x2.Thisiscalledtrigonomet-ricsubstitution,andthestudentwilllearnthisincollege.Fornow,weconvinceourselvesthatZp1 x2dx=12sin 1x+xp1 x2+C318
bydi erentiatingtheright-handsideandobservingthatityieldstheintegrandoftheleft-handside.Hence,byFTOC,Z1 1p1 x2dx=12sin 1x+xp1 x2 1 1=122+0 12 2 0=2:.BecauseFTOCevaluatesthede niteintegralusingantiderivativesandnotby\netsignedarea,"wedonothavetolookatthegraphandlookatthoseregionsbelowthex-axis,thatis,FTOCworksevenifthegraphhasa\negativepart."EXAMPLE3:EvaluatethefollowingintegralsusingFTOC.1.Z50(2x 2)dx2.Z=2 =2sinxdxSolution..1.Z50(2x 2)dx=2x22 2x 50=(25 10) (0 0)=15:2.Z=2 =2sinxdx=( cosx) =2 =2=(0 0)=0:Theseanswersarethesameaswhenweappealedtothegeometricalsolutionoftheintegralintheprevioussection..SolvedExamplesEXAMPLE1:UsingFTOC, ndZ0cosxdx.Solution.NotethatF(x)=sinxisanantiderivativeoff(x)=cosxsinceF0(x)=cosx.Therefore,byFTOC,Z0cosxdx=sinx 0=sin sin0=0:.319
Withoutreferringtothegraphsoftheintegrands, ndtheexactvalueofthefollowingde niteintegrals.EXAMPLE2:Z316dxSolution.Z316dx=6x 31=6(3) 6(1)=12:.EXAMPLE3:Z303xdxSolution.Z303xdx=3x22 30=32(9 0)=272..EXAMPLE4:Z41(2x+1)dxSolution.Z41(2x+1)dx=(x2+x) 41=(16+4) (1+1)=20 2=18:.SupplementaryProblemsEvaluatethefollowingintegralsusingFTOC.1.Z105dx2.Z 1 213dx3.Z21(jx+1j+2)dx4.Z1 1jxjdx5.Z32(2x 1)dx6.Z42(3x+7)dx7.Z1 1 x2+1dx8.Z20 1+x2dx9.Z2010xdx10.Z31(2+2x)dx320
TOPIC16.2:ComputationofDe niteIntegralsusingtheFundamentalTheoremofCalculusIntheprevioussection,weillustratedhowtheFundamentalTheoremofCalculusworks.Iffisacontinuousfunctionon[a;b]andFisanyantiderivativeoff,thenZbaf(x)dx=F(b) ba=F(b) F(a);wheref(x)=F0(x).Forthislesson,westartwitharemarkabouttheapplicabilityoftheFTOCandproceedwithansweringsomeexercises,eitherindividuallyorbygroup.Remark1:Ifthefunctionisnotcontinuousonitsintervalofintegration,theFTOCwillnotguaranteeacorrectanswer.Forexample,weknowthatanantiderivativeoff(x)=x 2isF(x)= x 1:So,ifweapplytheFTOCtofon[ 1;1],wegetZ1 11x2dx= 1x 1 1=( 1 (+1))= 2:ThisisabsurdastheregiondescribedbyZ1 1x 2dxisgivenbelow: 11y=x 2xyClearly,theareashouldbepositive.Thestudyofde niteintegralsoffunctionswhicharedis-continuousonaninterval(whichmaynotbeclosednorbounded)iscalledimproperintegrationandwillbestudiedincollege.321
TableofIntegralsObservethatforFTOCtowork,thestudentmustbeabletoproduceanantiderivativefortheintegrand.Thisiswhythestudentmustbecomfortablewiththe rstfewlessonsinthischapter.AverycommonandindispensableformulaisthePowerRule:Zxndx=xn+1n+1+C;n6= 1:Forothercases,werecallthetableofintegralsforreference.1.Zdx=x+C2.Zxndx=xn+1n+1+C;ifn6= 13.Zaf(x)dx=aZf(x)dx4.Zf(x)g(x)dx=Zf(x)dxZg(x)dx5.Zexdx=ex+C:6.Zaxdx=axlna+C:7.Zx 1dx=Z1xdx=lnjxj+C:8.Zsinxdx= cosx+C9.Zcosxdx=sinx+C10.Zsec2xdx=tanx+C11.Zcsc2xdx= cotx+C12.Zsecxtanxdx=secx+C13.Zcscxcotxdx= cscx+C14.Ztanxdx= lnjcosxj+C15.Zcotxdx=lnjsinxj+C16.Zsecxdx=lnjsecx+tanxj+C17.Zcscxdx=lnjcscx cotxj+C18.Zdxpa2 x2=sin 1xa+C19.Zdxa2+x2=1atan 1xa+C20.Zdxxpx2 a2=1asec 1xa+CEXAMPLE1:UsingFTOC,evaluatethefollowingde niteintegrals:1.Z41pxdx2.Z21x3 2x2+4x 2xdx3.Z=40cosx+tanxdx4.Z1=20dxp1 x2322
Solution..1.Z41pxdx=Z41x1=2dx=x3=23=2 41=23(8 1)=143:2.We rstdividethenumeratorbythedenominatortoexpressthefractionasasum.Z21x3 2x2+4x 2xdx=Z21x2 2x+4 2xdx=x33 x2+4x 2lnjxj 21=83 4+8 2ln2 13 1+4 0=103 2ln2:3.Z=40cosx+tanxdx=(sinx+lnjsecxj) =40=p22+ln(p2)4.Z1=20dxp1 x2=sin 1x 1=20=sin 1(1=2) sin 1(0)=6 0=6:.INTEGRANDSWITHABSOLUTEVALUESSolvingde niteintegralswithabsolutevaluesintheintegrandshasbeendiscussedinpassinginpreviousexamples.Wewillnowgiveamorein-depthdiscussion.Ashasbeensaid,foranycontinuousexpressionE,itsabsolutevaluecanalwaysbewritteninpiecewiseform:jEj=8<:EifE0 EifE<0:Therefore,the rststepinsolvingthiskindofintegralistoeliminatetheabsolutevaluebars.Thisisdonebydividingtheintervalofintegration[a;b]intotwosubintervalsaccordingtothepiecewiseversionofthefunction.Hence,oneintegrandiseitherpurelypositive(orzero)andtheotherispurelynegative.323
EXAMPLE2:Evaluatethefollowingde niteintegrals:1.Z2 1jx 3jdx2.Z73jx 3jdx3.Z41jx 3jdx4.Z204x+j2x 1jdx5.Z2=30jcosxjdx6.Z3 2jx2 1jdxSolution.Foritems1-3,observethatbyde nition,jx 3j=8<:x 3;ifx 30 (x 3);ifx 3<0;=8<:x 3;ifx3; x+3;ifx<3:1.Sincex 3isalwaysnegativeontheintervalofintegration[ 1;2],wereplacejx 3jwith (x 3).Hence,Z2 1jx 3jdx=Z2 1 x+3dx= x22+3x 2 1= 42+6 12 3=152:2.Sincex 3isalwaysnonnegativeon[3;7],wereplacejx 3jwithx 3.So,Z73jx 3jdx=Z73x 3dx=x22 3x 73=492 21 92 9=8:3.Sincex 3isneitherpurelypositivenorpurelynegativeon[1;4],weneedtodividethisintervalinto[1;3]and[3;4].Onthe rst,wereplacejx 3jwith x+3,whileonthesecond,wereplacejx 3jwithx 3.Z41jx 3jdx=Z31 x+3dx+Z43x 3dx= x22+3x 31+x22 3x 43= 92+9 12+3+162 12 92 9=92 52+ 82+92=52:4.Wesplittheintervalofintegrationinto[0;12]and[12;2]sincej2x 1j=8<:2x 1if2x 10 (2x 1)if2x 1<0:324
Hence,4x+j2x 1j=8<:4x+(2x 1)ifx124x (2x 1)ifx<12:Therefore,Z204x+j2x 1jdx=Z1202x+1dx+Z2126x 1dx=(x2+x) 1=20+(3x2 x) 21=2=14+12 (0+0)+(12 2) 34 12=212:5.Thecosinefunctionisnonnegativeon[0;=2]andnegativeon[=2;2=3].Seegraphbelow.223y=cosxxyHence,jcosxj=8<:cosx;ifcosx0; cosx;ifcosx<0=8<:cosx;ifx2[0;=2]; cosx;ifx2[=2;2=3]:Therefore,Z2=30jcosxjdx=Z=20cosxdx+Z2=3=2 cosxdx=sinx =20+( sinx) 2=3=2=(1 0)+[( p3=2) ( 1)]=2 p32:6.Thequadraticfunctiony=x2 1isbelowthex-axisonlywhenx2( 1;1).Thisconclusioncanbededucedbysolvingtheinequalityx2 1<0orbyreferringtothegraphbelow.325
2 113y=x2 1xyThisimpliesthatwehavetodivide[ 2;3]intothethreesubintervals[ 2; 1],[ 1;1];and[1;3].On[ 1;1],wereplacejx2 1jwith x2+1whileontheothersubintervals,wesimplyreplacejx2 1jwithx2 1.Hence,Z3 2jx2 1jdx=Z 1 2x2 1dx+Z1 1 x2+1dx+Z31x2 1dx=x33 x 1 2+ x33+x 1 1+x33 x 31= 13+1 83+2+ 13+1 13 1+273 3 13 1=283:.SolvedExamplesUsingFTOC,evaluatethefollowingde niteintegrals.EXAMPLE1:Z941pxdxSolution.Z941pxdx=Z94x 12dx=x1212 94=2(3 2)=2:.326
EXAMPLE2:Z2012x3+6x2+4x2xdxSolution.Z2012x3+6x2+4x2xdx=Z20(6x2+3x+2)dx=6x33+3x22+2x 20=2(2)3+3(2)22+2(2) 2(0)3+3(0)22+2(0)=26:.EXAMPLE3:Z40(sinx+secxtanx)dxSolution.Z40(sinx+secxtanx)dx=( cosx+secx) 40= cos4+sec4 ( cos0+sec0)= p22+p2! ( 1+1)=p22:.EXAMPLE4:Z21dxxpx2 1Solution.Z21dxxpx2 1=sec 1x 21=sec 1(2) sec 1(1)=3:.327
EXAMPLE5:Z1 1j2x 1jdxSolution.j2x 1j=(2x 1ifx12 2x+1ifx<12Therefore,Z1 1j2x 1jdx=Z12 1( 2x+1)dx+Z112(2x 1)dx= x2+x 12 1+(x2 x) 112= 14+12 ( 1 1)+0 14 12=52:.EXAMPLE6:Z3 2jx+1jdxSolution.jx+1j=(x+1ifx 1 x 1ifx< 1:Therefore,Z3 2jx+1jdx=Z 1 2( x 1)dx+Z3 1(x+1)dx= x22 x 1 2+x22+x 3 1= 12+1 42+2+92+3 12 1=172:.EXAMPLE7:Z0 6j2x+4jdxSolution.j2x+4j=(2x+4ifx 2 2x 4ifx< 2:328
Therefore,Z0 6j2x+4jdx=Z 2 6( 2x 4)dx+Z0 2(2x+4)dx=( x2 4x) 2 6+(x2+4x) 0 2=[( 4+8) ( 36+24)]+[0 (4 8)]=20:.EXAMPLE8:Z1 1(j3x+1j+3x)dxSolution.j3x+1j=(3x+1ifx 13 3x 1ifx< 13:Now,Z1 1(j3x+1j+3x)dx=Z 13 1[( 3x 1)+3x]dx+Z1 13[(3x+1)+3x]dx=Z 13 1 1dx+Z1 13(6x+1)dx=( x) 13 1+(3x2+x) 1 13= 13+1+(3+1) 13 13=103:.EXAMPLE9:Z4 4jsinxjdxSolution.jsinxj=(sinxifx2[0;4] sinxifx2[ 4;0):329
Therefore,Z4 4jsinxjdx=Z0 4 sinxdx+Z40sinxdx=cosx 0 4 cosx 40=hcos0 cos 4i hcos4 cos0i= 1 p22!+ p22+1!=2 p2:.EXAMPLE10:Z40jx2 4jdxSolution.jx2 4j=(x2 4ifx2orx 2 x2+4if 2
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