Lecture Notes on Integral Calculus - Undergrad Mathematics

1 LectureNotesonIntegralCalculusUBCMath103 LectureNotesby Yue-XianLi (Spring,2004)1 IntroductionandhighlightsDi erentialcalculusyoulearnedin thepasttermwas aboutdi feelembarrassedto ndoutthatyouhave alreadyforgottena number of thingsthatyoulearneddi ,if youstillremember thatdi erentialcalculuswas abouttherateof change,theslope of a graph,andthetangentof a curve, youareprobablyOK. Theessenceof di erentiationis ndingtheratiobetween thedi erence in thevalueoff(x) ,thederivative or theslope of a functionis givenbyf0(x) =dfdx=lim x!0f(x+ x) f(x) x:(1)Integralcalculusthatwe arebeginningto learnnow is willbemostlyaboutaddinganincrementalproc essto arriveat a \total". It willcover threemajoraspectsof integralcalculus:1. Themeaningof integration. We'lllearnthatintegrationanddi erentiationareinverseoperationsof each sidesof thesamecoin(FundamentalTheoremof Caclulus).2. :addinga sequenceof numbersIn essence,integrationis anadvancedformof allstartedlearninghow to addtwo numberssinceas youngas we say \Areyoukidding?

2 AreyoutellingmethatI have to startmy university lifeby learningaddition?".1 Theanswer is ndoutthatdoingadditionis oftenmuch even ndsigmasumis themostdi cultthingto culty is by-passedby usingtheFundamentalTheoremofCaclulus,you shouldNEVER forgetthatyouare actually doinga sigmasumwhenyouarecalculatingan Integral . Thisis onesecretforcorrectlyformulatingtheinteg ralin many appliedproblemswithease!Now,I usea coupleof examplesto show thatyourskillsin :Findthetotalnumber of logsin a triangularpileof fourlayers(see gure).Solution1a:Letthetotalnumber beS4, where`S'standsfor`Sum'andthesubscriptrem indsus thatwe arecalculatingthesumfora pileof 4 |{z}in layer 1+2|{z}in layer 2+3|{z}in layer 3+4|{z}in layer 4= 10:A pieceof cake!Example1b:Now, ndthetotalnumber of logsin a triangularpileof 50layers, ndS50! (Give metheanswer in a fewsecondswithoutusinga calculator).Solution1b:Let'sstartby |{z}in layer 1+2|{z}in layer 2+ +49|{z}in layer 49+50|{z}in layer 50=?

3 Where` ' hadtobe usedtorepresent thenumbersbetween3 'tenoughspaceforwritingallof thereis enoughspace,it is tediousandunnecessaryto writeallof themdownsincetheregularityof thissequencemakes it pieceof cake?Notreallyif youhadnotlearnedGauss' 'llhave to leave itunansweredat :Finally, ndthetotalnumber of logsin a triangularpileofklayers, ndSk(kis any positive integer, 8;888;888is onepossiblechoice)!Solution2:Thisis equivalent to calculatingthesumof the rstkpositive 1 + 2 + + (k 1) +k:Theonlythingwe cansay now is thattheanswer mustbe a functionofkwhich is thetotalnumber of integerswe needto ,we have to leave it unansweredat irregularsequencesAsequenceis a listof numberswrittenin a de sequenceisregularif each termof thesequenceis uniquelydetermined,followinga well-de nedrule,by itsposition/orderinthesequence(oftendeno tedby anintegeri). Veryoften,each termcanbe generatedby anexplicitformulathatis expressedas a functionof thepositioni, (i).

4 We cancallthisformulathesequence generatoror thegeneral example,theith termin thesequenceof integersis identicalto itslocationin thesequence,thus itssequencegeneratorisf(i) =i. Thus,the9thtermis 9 whilethe109thtermis equalto :Thesumof the rsttenoddnumbersisO10= 1 + 3 + 5 + + 19 :Notethattheith oddnumber is equalto theith evennumber minus 1. Theitheven number is simply2i. Thus,theith oddnumber is 2i 1, namelyf(i) = 2i 1. To verify,5 is the3rdoddnumber, 3. Thus,2i 1 = 2 3 1 = 5 which is exactlythenumberwe ,we canwritedownthesumof the rstkoddnumbersforany positive 1 + 3 + 5 + + (2k 1):Example4:Findthesequencegeneratorof thefollowingsumof 100productsof subsequentpairsof 2|{z}1stterm+2 3|{z}2ndterm+3 4|{z}3rdterm+ + 100 101|{z}100thtermSolution4:Sincetheith termis equalto thenumberimultipliedby thesubsequent integerwhich is equaltoi+ 1. Thus,f(i) =i(i+ 1).Knowingthesequencegenerator,we canwritedownthesumofksuch termsforany 1 2 + 2 3 + +k(k+ 1):Example4:Thesequenceof the rst8 digitsof theirrationalnumber = 3:1415926: : :is 8= 3 + 1 + 4 + 1 + 5 + 9 + 2 + 6:We cannot ndthesequencegeneratorsincethesequenceis irregular,we cannotexpressthesumof the rstkdigitsof (forarbitraryk).

5 Orderto short-handthemathematicalexressionof thesumof a regularsequence,a conve-nient notationis nition( sum):Thesumof the rstktermsof a sequencegeneratedby thesequencegeneratorf(i) canbe denotedbySk=f(1)+f(2)+ +f(k) kXi=1f(i)wherethesymbol (theGreekequivalent ofSreads\sigma")means\take thesumof",thegeneralexpressionfortheterm sto be addedor thesequencegeneratorf(i) is calledthesummand,iis calledthesummationindex, 1 andkare,respectively, thestartingandtheendingindicesof ,kXi=1f(i)meanscalculatethesumof all thetermsgenerated by thesequence generatorf(i)forall integersstartingfromi= 1andendingati= thesumis independent of thesummationindexi, henceiis calleda\dummy"variableservingforthesolep urposeof runningthesummationfromthestartingindext o , :ExpressthesumSk=kPi=3i2in :Thesequencegeneratorisf(i) =i2. Notethatthestartingindexis not1 but3!.Thus,the1sttermisf(3)= 32. Thesubsequent termscanbe determinedaccordngly.

6 Thus,Sk=kXi=3i2=f(3)+f(4)+f(5)+ +f(k) = 32+ 42+ 52+ +k2:Aneasycheck fora mistake youstill ndthe\dummy"variableiin anexpandedformor in the nalevaluationof thesum,youranswer mustbe 'sformulaandotherformulasforsimplesumsLe tus returnto Examples1 and2 aboutthetotalnumber of logsin a 'sstartwitha pileof 4 (ina \thought-experiement")putan4identicalpil ewithupsidedownadjacent to theoriginalpile,youobtaina pilethatcontainstwicethenumber of logsthatyouwant to calculate(see gure).Theadvantageof doingthisis that,in thisdouble-sizedpile,each layer containsanequalnumber of is equalto number onthe1st(top)layer plusthenumber onthe4th(bottom) themeantime,theheight of thepileremainsunchanged(4).Thus,thenumbe r in thisdouble-sizedpileis 4 (4 + 1) = justhalfof thisnumberwhich is 'sapplythisideato ndingtheformulain (Original)Sk=1+2+ +k 1 +k:(2)(Inversed)Sk=k+k 1 + +2+1:(3)(Addingthetwo) 2Sk=(k+ 1) + (k+ 1) + + (k+ 1) + (k+ 1)|{z}k termsin total=k(k+ 1):(4)Dividingbothsidesby 2, we obtainGauss'sformulaforthesumof the rstkpositive 1 + 2 + +k=12k(k+ 1):(5)Thisactaullyansweredtheproblemin Example2.

7 Theanswer to Example1bis even 50 51 = 1275:Thefollowingaretwo important simplesumsthatwe thesumof the 12+ 22+ +k2=16k(k+ 1)(2k+ 1):(6)Theotheris thesumof the 13+ 23+ +k3= 12k(k+ 1) 2:(7)We shallnotillustratehow to derive ndit in prove thattheseformulasworkforarbitrarilylarge integersk, we canusea method save time,we' we canprove thingsconcerningarbitrarilylargenumbersi s to guaranteethatthisformulamustbe correctfork=N+ 1 if it is correctfork=N. Thisis like tryingto arrangea string/sequenceof guaranteethatallthedominosin thestring/sequencefalloneaftertheother,w e needto guaranteethatthefallingof each dominowillnecessarilycausethefallingof thesubsequent theessenceof provingthatiftheformulais right fork=N, it mustbe right fork=N+ 1. Thelastthingyouneedto dois to knock downthe rstoneandkeepyour rstdominois equivalent to provingthattheformulais correctfork= 1 which is veryeasyto check in allcases(seeKeshet'snotesfora prove of thesumof the rstkintegerssquared).

8 Rulesof sigmasumsRule1: Summationinvolvingconstant +c+ +c|{z}k termsin total=kc:Notethat:thetotal# of terms=endingindex- startingindex+ : Constant multiplication:multiplyinga sumby a constant is equalto multiplyingeachtermof thesumby (i) =kXi=1cf(i)Rule3: Addingtwo sumswithidenticalstartingandendingindice sis equalto thesumof sumsof (i) +kXi=1g(i) =kXi=1[f(i) +g(i)]:Rule4: Breakonesuminto (i) =nXi=1f(i) +kXi=n+1f(i);(1 n < k):Example7:Let'sgo back to solve thesumof the :Ok= 1 + 3 + + (2k 1) =kX1(2i 1) =kX1(2i) kX11 = 2kX1i k;6whereRules1, 2, and3 areusedin thelasttwo ,we obtainOk=kX1(2i 1) =k(k+ 1) k=k2:Example8:Letus now goback to solve Example4. It is thesumof the rstkproductsofpairsof subsequent :Pk= 1 2 + 2 3 + +k (k+ 1) =kX1i(i+ 1) =kX1[i2+i] =kX1i2+kX1i;whereRule3 was usedin ,we learnedPk=kX1i(i+ 1) =16k(k+ 1)(2k+ 1) +12k(k+ 1) =13k(k+ 1)(k+ 2):Thisis anothersimplesumthatwe :CalculatethesumS=kP1(i+ 2) :Thisproblemcanbe solvedin two di erent rstis to expandthesummandf(i) = (i+ 2)3which yieldS=kX1(i+ 2)3=kX1(i3+ 6i2+ 12i+ 8) =kX1i3+ 6kX1i2+ 12kX1i+ 8k:We cansolve a betterway to solve ndit easierto seehow substitutionworksby (i+ 2)3= 33+ 43+ +k3+ (k+ 1)3+ (k+ 2)3:We seethatthisis simplya sumof notstartat 13andendatk3like in theformulaforthesumof the re-writethesumwitha sigmanotationwithannewindexcalledlwhich startsatl= 3 andendsatl=k+ 2 (thereis noneedto changethesymbol fortheindex,youcankeepcallingitiif youdonotfeelany confusion).

9 Thus,S=kXi=1(i+ 2)3= 33+ 43+ +k3+ (k+ 1)3+ (k+ 2)3=k+2Xl=3l3:7We just nisheddoinga substitutionof is equivalent to replacingibyl=i+ 2. Thisrelationalsoimpliesthati= 1)l= 3 andi=k)l=k+ 2. Thisisactuallyhow youcandeterminethestartingandtheendingva luesof we cansolve thissumusingRule4 (i+2)3=k+2Xl=3l3=k+2Xl=1l3 2Xl=1l3= 12(k+ 2)(k+ 3) 2 12 23= 12(k+ 2)(k+ 3) 2 9 sigmasumTheareaundera curveWe know thattheareaof a rectanglewithlengthlandwidthwisArect=w cancalculatetheareaof a triangleanda triangleanda trapezoidcanbe transformedinto a rectangle(seeFigure).Thus,fora triangleof heighthandbaselengthbAtrig=12hb:Similarl y, fora trapezoidwithbaselengthb, toplengtht, andheighthAtrap=12h(t+b):Followinga verysimilaridea,thesumof a trapezoid-shapedpileof logswithtlogsontoplayer,blogsonthebottom layer,anda height ofh=b t+ 1 layers(see gure)isbXi=ti=t+ (t+ 1) + + (b 1) +b=12h(t+b) =12(b t+ 1)(t+b):(8)Now returningto theproblemof formulais fortheareaof a circleof r2:Now,oncewe learnedsigmaand/orintegration,we cancalculatetheareaunderthecurve ofany functionthatis :Calculatetheareaunderthecurvey=x2betwee n0 and2 (see gure).

10 Solution10:Remember always tryto reducea problemthatyoudonotknow how to solveinto a problemthatyouknow how to Let'sdivideAinto 3 rectanglesof equalwidthw= 2=3. Thus,A w h1+w h2+w h3=(w[x20+x21+x22] =13[02+ 13 2+ 23 2];left endapprox:;w[x21+x22+x23] =13[ 13 2+ 23 2+ 33 2];right endapprox::Byusingtheserectangles,we introducedlargeerrorsin ourestimatesusingtheheights basedonboththeleftandtheright endpoints of increaseaccurary, we needtoincreasethenumber of rectanglesby makingeach divideAintonrectanglesof equalwidthw= 2=n. Thus,usingtheheight basedontheright endpoint of eachsubinterval,we obtainA Sn=A1+A2+ +An=nXi=1w hi=wnXi=1f(xi) =2nnXi=1x2i:It is veryimportant tokeepa clearaccount of theheight of each ,hi=f(xi) =x2i. Thus,thekey is ndingthex-coordinateof theright-endof each rectanglesof of equalwidth,xi=x0+iw=x0+i x=n, where xis thelengthof theintervalx0is theleftendpoint of 0 and x= 2 ,xi=i(2=n).)