Calculus - Weebly1 Calculus Chapter 4A Approximation using the derivative It’s an algebraic take...
Transcript of Calculus - Weebly1 Calculus Chapter 4A Approximation using the derivative It’s an algebraic take...
1
Calculus
Chapter4A Approximationusingthederivative
It’sanalgebraictakeonthelimitrule
𝑓" 𝑥 = lim(→*
𝑓 𝑥 + ℎ − 𝑓(𝑥)ℎ
**that‘formula’isfamiliartoyouisn’tit…youwillneverforgetitwillyouJ**
andthroughthemagicofAlgebraitbecomes
ℎ𝑓" 𝑥 = lim(→*
𝑓 𝑥 + ℎ − 𝑓(𝑥)
lim(→*
𝑓 𝑥 + ℎ = ℎ𝑓" 𝑥 + 𝑓 𝑥
and
𝑓 𝑥 + ℎ ≈ 𝑓 𝑥 + ℎ𝑓′(𝑥)
…noticethelimithasbeenremoved,sothereforehdoesnotequalzero,butasmallactualvalue…thereforethesolutioncanonlybeapproximate,astheoriginalfunctionwasbasedonsettingh=0,buthere,hisNOTzero!**
eg.Approximatethevalueof 37…Notethatthisisverycloseto 36,sowecanapproximate.
Set 𝑓 𝑥 = 𝑥 → 𝑓" 𝑥 = 67 8
Usingtherule 𝑓 𝑥 + ℎ ≈ 𝑓(𝑥) + ℎ×𝑓′ 𝑥
Becomes 𝑓 𝑥 + ℎ ≈ 𝑓 𝑥 + ℎ 67 8
Nowwesay,𝑥 = 36,whichleavesℎ = 1
𝑓 36 + 1 ≈ 𝑓 36 + 1×1
2 36
36 + 1 ≈ 6 +112
∴ 37 ≈ 6112
clearlythebiggerℎis,themoreapproximateyouranswer.
DoExercise4A
**Note,onlydothemarkedquestions…NOneedtodoanyTrigquestionshere**
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Integration
WeshouldhavecoveredsomeofthisinMathsB,sojustasarefresher…
IntegrationistheOppositeofDifferentiation.WecouldsaythatIntegrationisAnti-Differentiation!
WeshalltakemoreofamathematicalapproachtoIntegrationinMathsC,thaninMathsB.
IfweknowourDerivatives,thenwealreadyknowsomeintegrals…it’sjustthereverseprocess!
Ifwehaveaknownderivative,thenweknowtojustworkbackwardstoknowtheIntegral.
TableofknownDerivatives
f(x) f’(x)
etcetc
Letsgetsometerminologyestablished.
isthesignforIntegral.AsaverbwewouldsayweneedtoIntegrate.
() 𝑑𝑥meanscalculatetheIntegralof()withrespectto𝑥
andthethinginsidethebracketsiscalledthe“Integrand”
€
xn
€
nxn−1
€
aeg(x )
€
g'(x)aeg(x )
€
lng(x)
€
g'(x)g(x)
€
sinax
€
acosax
€
cosax
€
−asinax
3
So,firstlyletsset
Integration
f(x) F(x)
or
sin 𝑘𝑥
A
BCD E8E
+
𝐶
cos 𝑘𝑥 DIJ E8E
+
𝐶
𝑠𝑒𝑐7𝑘𝑥 NOJ E8E
+
𝐶
6
PQ
A
8Q 𝑠𝑖𝑛
A
6 8P
+
𝐶
A
6
PQ
A
8Q cos
A
6 8P
+
𝐶
PPQ
T
8Q 𝑡𝑎𝑛
A
6 8P
+
𝐶
Don’tforgettoADDthe“c”Makesureyouunderstandwhythereisa“C”...andNeverforgettoaddit!
€
f (x)∫ = F(x)
€
axn
€
axn+1
n +1+ c
€
f (x) ± g(x)
€
f (x) ± g(x)∫∫
€
kf (x)
€
k f (x)∫
€
(ax + b)n
€
(ax + b)n+1
a(n +1)+ c
€
1ax + b
€
(ax + b)−1
€
1aln(ax + b) + c
€
ekx
€
1kekx + c
€
ln x
€
x ln x − x + c
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TrigIdentities
AdvancedPeriodicFunctionsskillswillbehelpfulwhenIntegratingsomeTrigfunctionsassomemanipulationoftrigonometricIdentitieswillberequired.
Youmaybecalledupontouseanyoftheidentitiesfromthepreviouschapteragainhere…andyoumusthaverecallofanyTrigIdentitymentionedinthetextBookChapters…
IalsoprovidedaHugelistofextraTrigidentitiesthatyoudonotneedtoremember,butitmaybeanideatohaveyoursheetathandtoassisthere.
OK…soletslookatIntegration…
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Chapter4B Integrationbysubstitution
Theykeyis…lookforthederivativeofpartofthefunction,withinanotherpartoftheintegrand…youwillseewhatImeanafterafewpracticequestions.
Thetextbookmakesthislookconfusing…butitreallyisn’t!
IcallitIntegrationby“U”substitution.
Bestwelookatanexample…!
2𝑥 𝑥7 + 1 W 𝑑𝑥
(notetherelationshipbetweeninsideandoutsidethebrackets)
Soweset,𝑢 = 𝑥7 + 1 YZY8= 2𝑥 → 𝑑𝑢 = 2𝑥𝑑𝑥
BacktotheIntegral 2𝑥 𝑥7 + 1 W𝑑𝑥
Rearrangeto 𝑥7 + 1 W2𝑥𝑑𝑥
Bysubstitution(twice)wenowget 𝑢W𝑑𝑢
Whicheasilyevaluatesto 𝑢W = Z[
\
Andagainbysubstitution,wecanfinallysay
2𝑥 𝑥7 + 1 W𝑑𝑥 = 𝟏𝟒𝒙𝟐 + 𝟏 𝟒 + 𝑪
**takenoteoftheIntegralTableonpage132.Iwillnotguaranteeyourcalculatorwillenableyoutosolveyourexamquestionsappropriately.IsuggestyouMemorisethistable.Therereallyareonly4thatyouwon’talreadyknow**
DoExercise4B
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Chapter4C multiplesofaderivativeintheintegrand
HereweusetheLinearpropertyofdifferentialsANDintegralstosolveforintegralquestionswherethederivativewithintheintegrandisnotexactlyintheright‘multiple’.
TheLinearlawofDerivativesandIntegrals:
𝑘𝑔(𝑥) = 𝑘 𝑔(𝑥) YY8𝑘𝑓 𝑥 = 𝑘 Y
Y8𝑓(𝑥)
Seeitinaction: 6𝑥7 𝑥W − 2 c 𝑑𝑥
set 𝑢 = 𝑥W − 2 YZY8= 3𝑥7 → 2YZ
Y8= 6𝑥7 → 2𝑑𝑢 = 6𝑥7𝑑𝑥
backtoIntegral…
6𝑥7 𝑥W − 2 c𝑑𝑥 = 𝑥W − 2 c 6𝑥7𝑑𝑥
= 𝑢c2𝑑𝑢
= 2 𝑢c𝑑𝑢
= 2 Zd
e
= 6W𝑥W − 2 e + 𝐶
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ALTERNATEsolution:
6𝑥7 𝑥W − 2 c 𝑑𝑥
Younoticethatitsnotastraightsubstitution,andthinkthatitwouldbe“nice”ifthefirstfunctionwassimply3𝑥7…thenmanipulateitsothatitis…!
6𝑥7 𝑥W − 2 c𝑑𝑥 = 2 3𝑥7 𝑥W − 2 c 𝑑𝑥
set 𝑢 = 𝑥W − 2 YZY8= 3𝑥7 → 𝑑𝑢 = 3𝑥7𝑑𝑥
2 3𝑥7 𝑥W − 2 c = 2 𝑥W − 2 c 3𝑥7𝑑𝑥
= 2 𝑢c𝑑𝑢
= 2𝑢e
6
=13 𝑥W − 2 e + 𝐶
Doesitmatterwhichwayyoudoit…NO…Justgetitright,andmakesureyourSolutionshowsPerfectcommunication…Don’tskipanysteps,orPanelwillthinkyouarejustusingyourcalculator…J
Seemssimplehey…thealgebradoesgetinterestingintheharderones!
Justlikederivatives,thepreferredformofyoursolutionshouldbe“FactorForm”,andthisgivesyousomeinterestingalgebra!
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TakeQuestion1p)
3𝑥8 − 𝑥
𝑑𝑥
set 𝑢 = 8 − 𝑥 so,clearly 𝑥 = 8 − 𝑢
and YZY8= −1 so, 𝑑𝑢 = −𝑑𝑥
bysubstitution,weget
3 8 − 𝑢
𝑢67
×−𝑑𝑢
= 3𝑢 − 24
𝑢67
𝑑𝑢
= 3𝑢67 − 24𝑢
A67 𝑑𝑢
= 2𝑢W7 − 48𝑢
67
= 2(8 − 𝑥)W7 − 48(8 − 𝑥)
67
= 2(8 − 𝑥)(8 − 𝑥)67 − 24(8 − 𝑥)
67
= 2 8 − 𝑥 − 24 (8 − 𝑥)67
= −2 𝑥 + 16 8 − 𝑥 + 𝐶
AlgebraoperationsstillholdinIntegralequations…SeeWorkedExample8forsomeinnovativealgebraworktoobtainanintegral.Ifgivenaquestion,keeplookingatALLtheequationsyouhave.Manipulatethemandseewhereyoucansubstitutetoobtainthecorrect‘form’.
DoExercise4C
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Chapter4D
IsaboutusingTrigIdentitiestomanipulatetheintegrandintosomethingthatis“Integratable”?
Thisisallaboutpracticeinusingtheidentities…so…bestgettoit!
IseelimitedvalueinenforcingthememorisingofmoreTrigIdentities.Ifitisnotinmylistinmyworksheets,thenIdonotexpectyoutoknowit.
However,doingthisexercisewillgiveyoucrucialpracticeatsolvingintegrals…butIconfirmthereisnoneedforyoutomemorisethesetrigidentities,andsimplysuggestyouwritedownthenewidentitiesinthischaptersomewhereandrefertothemwhenyouworkthroughthischapter.
DoExercise4D
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Chapter4E PartialFractions
Headbackto4C…whatifthereisasimilartermthatisthedifferentialofthesecondterm,butitisnotanevenlinearscalarofit…theanswer…PartialFractions!
Thebooksayspartialfractionsistedious,butIthinkit’sbeautiful!
Asitisalongerprocess,onlydoitwherenecessary,butitwillbenecessarysometimes!
Letsconsider: e88QT78AW
𝑑𝑥
ThisisNot‘u’substitutionasthe6xisnotalinearderivativeofthedenominator!
Whatwewanttodoistosplitthefractionintoasumoftwofractions.
6𝑥𝑥7 + 2𝑥 − 3 =
6𝑥(𝑥 + 3)(𝑥 − 1) =
𝐴𝑥 + 3 +
𝐵𝑥 − 1
clearly
𝐴𝑥 + 3 +
𝐵𝑥 − 1 =
𝐴(𝑥 − 1)(𝑥 + 3)(𝑥 − 1) +
𝐵(𝑥 + 3)(𝑥 − 1)(𝑥 + 3) =
𝐴 + 𝐵 𝑥 + 3𝐵 − 𝐴𝑥7 + 2𝑥 − 3
combiningtheabovetwolines,wegetto:
6𝑥𝑥7 + 2𝑥 − 3 =
𝐴 + 𝐵 𝑥 + 3𝐵 − 𝐴𝑥7 + 2𝑥 − 3
Soby‘equatingcoefficients’,CLEARLY,wehave
𝐴 + 𝐵 = 6 and 3𝐵 − 𝐴 = 0
solvingthesesimultaneouslyweget A=4.5andB=1.5
Nowitiseasy(welleasier)toIntegrate:
4.5𝑥 + 3 +
1.5𝑥 − 1𝑑𝑥
= 4.51
𝑥 + 3 𝑑𝑥 + 1.51
𝑥 − 1𝑑𝑥
= 4.5 ln(𝑥 + 3) + 1.5 ln(𝑥 − 1) + 𝐶
see,mucheasier…J
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Thereisaslightproblemencounteredwhenthereisa“square”inthedenominator,suchas:
6𝑥𝑥7 + 2𝑥 + 1 𝑑𝑥
seeifyoucansolvethisoneaspertheprevioustechnique…youendupwithsimultaneousequationsthatdon’tgetasolutionandyougettoadeadend.
Whenthereisaperfectsquareinthedenominator,yousplitthedenominatorsupdifferentlyandestablishyourPartialFractionsintheform:
𝐴𝑎𝑥 + 𝑏 7 +
𝐵(𝑎𝑥 + 𝑏)
Sointheaboveexamplewewouldset:
6𝑥𝑥7 + 2𝑥 + 1 =
𝐴𝑥 + 1 7 +
𝐵(𝑥 + 1)
6𝑥𝑥7 + 2𝑥 + 1 =
𝐴𝑥 + 1 7 +
𝐵(𝑥 + 1)(𝑥 + 1)(𝑥 + 1)
6𝑥𝑥 + 1 7 =
𝐴 + 𝐵𝑥 + 𝐵𝑥 + 1 7
6𝑥𝑥 + 1 7 =
𝐵𝑥 + 𝐴 + 𝐵𝑥 + 1 7
toget
𝐵 = 6
𝐴 + 𝐵 = 0
so 𝐴 = −6
6𝑥𝑥7 + 2𝑥 + 1 𝑑𝑥 =
−6𝑥7 + 2𝑥 + 1 +
6𝑥 + 1 𝑑𝑥
=−6
𝑥 + 1 7 +6
𝑥 + 1 𝑑𝑥
= −6 𝑥 + 1 A7 𝑑𝑥 + 61
𝑥 + 1 𝑑𝑥
= −6×𝑥 + 1 A7T61−2 + 1 + 6 ln 𝑥 + 1
=6
𝑥 + 1 + 6 ln 𝑥 + 1 + 𝐶
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Thecreativityofmathematiciansissimplyamazing.Canyouimaginehowcreativethefirstpersontosolvethisnextonehadtobe…itssosimple,buthardtoknowwhattodounlessyouhaveseenitbefore:
2𝑥W + 𝑥7 − 5𝑥7 − 1 𝑑𝑥
ConsidertheIntegrand
2𝑥W + 𝑥7 − 5𝑥7 − 1
=2𝑥W − 2𝑥 + 𝑥7 + 2𝑥 − 5
𝑥7 − 1
=2𝑥(𝑥7 − 1) + 𝑥7 + 2𝑥 − 5
𝑥7 − 1
=2𝑥(𝑥7 − 1)𝑥7 − 1 +
𝑥7 + 2𝑥 − 5𝑥7 − 1
= 2𝑥 +𝑥7 − 1 + 2𝑥 − 5 + 1
𝑥7 − 1
= 2𝑥 +𝑥7 − 1𝑥7 − 1 +
2𝑥 − 4𝑥7 − 1
= 2𝑥 + 1 +2𝑥 − 4𝑥7 − 1
andfromhereyouconsiderjustthefractionandset:
2𝑥 − 4𝑥7 − 1 =
𝐴𝑥 + 1 +
𝐵𝑥 − 1
andfromhereyoushouldberight…?
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Chapter4F IntegrationbyParts
It’ssimplyusingalgebrafromourProductDerivativerule…seetheTextbookforhowthealgebraworkstoarriveatthefollowingrule:
𝑢𝑑𝑣𝑑𝑥 = 𝑢𝑣 − 𝑣
𝑑𝑢𝑑𝑥
UsethiswherethereisaPRODUCTintheIntegrand.Sometimesoneofthetermsoftheintegrandmaybesetto“1”…asperWorkedExample18.
Further,choosecarefullythevaluefor𝑢𝑎𝑛𝑑 YoY8…theYo
Y8shouldbesettothepart
oftheIntegrandthatwhenIntegratedpartdoes-NOTgetmoreComplex…!
Youwillsoonknowifyouchosewisely.Ifyouchoosethewrongway,yougobackwardsandthingsgetmorecomplicated…Ifthishappens,thenjuststartagain.
Typically,fortheYoY8part,youwillsetitequaltosomethinglike…sin 𝑥 , 𝑒8
Sometimes,nomatterwhichwayyouchoose,yougetno-where,butpersist,thereWILLbeaway…checkoutworkedexample19…that’ssomeniceMaths!
AndALWAYS…wherepossible,FACTORISEyouanswer…!
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Chapter4G DefiniteIntegrals
Integrationmeasurestheareaunderacurvedfunction.ThatisthevaluecalculatedofF(x)forthevalue“x”,givestheareabetweenthecurveandthexaxisbetweentheoriginpoint“0”andpoint“x”onthexaxis.
***Care…ifthevalueofF(x)isNegative,thisrepresentstheareabeneaththex-axis.***
Tofindtheareaunderacurve,betweentwopointsyoudenoteyouarelookingforadefinitearea,andweusethetermDefiniteIntegral
Sowehavetheareabetweentwopointsaandb:
***Careifthefunctioncrossesthexaxisbetweenthesetwopoints.***
Andifwearelookingfortheareabetweentwofunctions:
again,takecareofintersectingpoints…sketchinggraphsismostuseful,orsimplymathematicallycalculateintersectingpointsandfindtheseparateareas,andthenaddthemtogether.
Propertiesofdefiniteintegrals:
€
f (x)a
b
∫ = F(x)[ ]ab
= F(b) − F(a)
€
f (x)a
b
∫ − g(x)a
b
∫ = f (x) − g(x)[ ]a
b
∫
€
f (x)a
b
∫ = f (x)a
c
∫ + f (x)c
b
∫
€
kf (x)a
b
∫ = k f (x)a
b
∫
€
f (x)a
b
∫ = − f (x)b
a
∫
€
f (x) + g(x)[ ]a
b
∫ = f (x)a
b
∫ + g(x)a
b
∫
€
f (ax + b)∫ =1aF(ax + b) + c
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DefiniteIntegrals…ChangingTerminals:
Whereanintegralisnotstraightforwardandyouneedtosubstitutesomethingfor‘x’tosolve,youalsoneedtoadjusttheterminals,asthevariablewithrespecttowhichtheintegralisbeingcalculatedhaschanged.Asfollows:
𝑥𝑥 + 1 7 𝑑𝑥
7
6
let𝑢 = 𝑥 + 1…so…𝑥 = 𝑢 − 1
YZY8= 1…so…𝑑𝑢 = 𝑑𝑥
So,BEFOREsubstitutingintotheintegral,weneedtoadjusttheterminals.
Where𝑥 = 1,wecanseethat𝑢 = 1 + 1 = 2
Andwhere𝑥 = 2,wecanseethat𝑢 = 2 + 1 = 3
Soourintegralbecomes:
𝑢 − 1𝑢7 𝑑𝑢
W
7
whichsimplifiesto
1𝑢 − 𝑢
A7𝑑𝑢W
7
= ln 𝑢 +1𝑢 7
W
= ln 3 +13 − ln 2 +
12
= 0.238798
Youcouldre-substitutethe𝑥valuesbackinplaceof𝑢andthenusetheinitialterminalvaluesin𝑥…butconvertingterminalsandleavingintermsof𝑢isabitshorter(maybe/maybenot)…butinanycase,conversionofterminalsissomethingwewanttoshowPanel,sodoitthiswaytoshowoffallyourmathematicalability!
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Chapter4H SolidsofRevolution
***Ifandonlyifyoufeeluptoit,skiptopage17andinvestigateIntegrationmorethoroughlybeforeyoudoSolidsofrevolution.Thestuffattheendis
EXTRAanddoesNOTformpartofthecurriculumandisNOTnecessaryfortheexam.Butitmayhelpyouconnectsomemissingpiecesinthissection.The
Curriculumisabitsparseandthereareleapsbetweenstepsherethathavenotbeenexplained.(itisSIXpagesofextraworksomakethedecisionthatisright
foryoubeforeyoucommittodoingthisextralearning)***
OK…sowelcomeback…letsgettoSolidsofRevolution:
It’shardtodrawagooddiagram,sopleasehavealookatthefiguresonpage175asyoureadthroughthesenotes.
Ifweconsiderjusta“disc”where𝑥 = 𝑎,wecouldfinditsareausing𝐴 = 𝜋𝑟7,wheretheradiusis𝑓(𝑥).
Andweknowthat𝑉 = 𝐴uPvw×ℎ
Letsnowconsiderabazillion“discs’atthelimitwhereeachdisc’sℎ → 0frompoint𝑎topoint𝑏.Clearlyifweaddedallthesevolumesup,wehavethevolumeofthesolidofrevolution…J
OK…atthebottomofpage175thetextbook“Conveniently”skipsfrom
𝑉 = limx8→*
𝜋𝑦78z{
8zP
𝛿𝑥
to
𝑉 = 𝜋𝑦7{
P𝑑𝑥
buttheyreallydidn’tsayhowthishappened…!…that’swhyIwentthroughallthatRiemannSeries“guff”…NotthatitmattersbuthereismyverySimplisticwayofthinkingaboutit…
Gobacktothatfirstdiagramandweshallthinkaboutmakingaheapofdiscsallbesideeachotherandwejustaddupalltheirvolumes.
𝑇𝑜𝑡𝑎𝑙𝑉𝑜𝑙𝑢𝑚𝑒 = 𝑉��v� + 𝑉��v� + 𝑉��v� …
becauseeachdiscisa“Cylinder”,wecouldsay
𝑉����Y���wo��Z���� = 𝜋𝑟7ℎ + 𝜋𝑟7ℎ + 𝜋𝑟7ℎ + 𝜋𝑟7ℎ…
17
let’sdosomefactoring…importantlytheradiusofeachdiscisDifferentsoIcan’tcombinetheseastheyareNot“Liketerms”,butmy𝜋andℎareconstants…soweget
𝑉����Y���wo��Z���� = 𝜋 𝑟7 + 𝑟7 + 𝑟7 + 𝑟7 … ℎ
letsconsiderwhat𝑟actuallyis.TheRadiusisthedistancefromthe𝑥-axistothefunction…soisn’tthatjust𝑦?
𝑉����Y���wo��Z���� = 𝜋 𝑦7 + 𝑦7 + 𝑦7 + 𝑦7 … ℎ
theℎofeachdiscisactuallythechangein𝑥-axis…say∆𝑥…oraswewantaninfiniteofdiscswhereℎ → 0wecouldevensay𝑑𝑥…J
𝑉����Y���wo��Z���� = 𝜋 𝑦7 + 𝑦7 + 𝑦7 + 𝑦7 … 𝑑𝑥
Soinsidethatbracketisthesumofabazillionvaluesof𝑦7where𝑦isafunctionof𝑥andchangescontinuouslyandweareaddingthemallup.ThisisverysimilartothefundamentaltheoremofcalculusandiswheretheRiemannSeriescomesin.InsidethatBracketisthesamebasisofhowIntegrationworks,sothatlinkallowsustosay:
𝑉����Y���wo��Z���� = 𝜋 𝑦7 𝑑𝑥
andonelastmodificationisthatclearly𝑦 = 𝑓 𝑥 ,soweget:
𝑽𝑺𝒐𝒍𝒊𝒅𝒐𝒇𝑹𝒆𝒗𝒐𝒍𝒖𝒕𝒊𝒐𝒏 = 𝝅 𝒇(𝒙)𝟐 𝒅𝒙
wecanalsorotateaboutthe𝑦 − 𝑎𝑥𝑖𝑠…andhencethatformulais:
𝑽𝑺𝒐𝒍𝒊𝒅𝒐𝒇𝑹𝒆𝒗𝒐𝒍𝒖𝒕𝒊𝒐𝒏 = 𝝅 𝒇(𝒚)𝟐 𝒅𝒚
andwhenwearetalkingaboutvolumeswherethereisaregionbetweentwofunctionswehave:
𝑽𝑺𝒐𝒍𝒊𝒅𝒐𝒇𝑹𝒆𝒗𝒐𝒍𝒖𝒕𝒊𝒐𝒏 = 𝝅 [𝒇 𝒙 𝟐 − 𝒈 𝒙 𝟐] 𝒅𝒙
TASK:UsingIntegralCalculus,derivetheVolumeofaCylinderandaCone…
𝑉� ���Yw¡ = 𝜋𝑟7ℎ
𝑉���w =13𝜋𝑟
7ℎ
18
***ExtraStuff:Integrationexplained…!
Letstakeastepbackalittle…wellaLOT!ThenextpagesrepresentEXTRAworkthatisNOTtechnicallyneeded.YoudoNOTneedtounderstandthis…youcanstillgetaVHAwithoutthisstuffasitisNOTinthecurriculum…butthereisachancethatitmayconnectafewmissingpiecesandmakeintegrationandCalculusabitmore“real”.
Wehavedonesomeprettyniftymathematicsinthischapter,andyetwehaven’tworkedfromFirstPrinciples…soitispossiblewearen’treallysure‘how’allthisCalculusworks(andthat’sOK).
TolearnthisthoroughlywouldtakeagoodportionofaFullSemesterofUniMaths,soIamjustpickingoutsomebasicconceptsratherthantryingtofullyexplaineverything!
Letsheadbacktodifferentiation:
RecalladerivativehasthenotationY Y8…the‘d’doesn’thaveanythingtodowith
itbeingthefirstletterof‘d’erivative,andisnotreallyamathematicalabbreviationforthewords(withrespectto)…butisshortfor“Delta”.DeltaisaGreekLetterandinCapitaliswrittenas∆.Deltaisusedtonotethechangeinsomething.
YoumayseeY Y8sometimeswrittenas∆
∆8,where∆=Delta(thechangein).
Notethat𝑠𝑙𝑜𝑝𝑒 = ¡�vw¡Z�
= �(P�£w�� �(P�£w��8
= QA ¤8QA8¤
= ∆ ∆8= Y
Y8= lim
(→*
� 8T( A� 8(
…specificallyspeakingwhenwemovefrom∆𝑥to𝑑𝑥,thedifferenceisthatinthe𝑑𝑥,thatchangeinvalueofxapproacheszero,recallwearegettingthegradientofthefunctionatthatinstantaneouspoint,hencethemovetodifferentiationfromfirstprinciples…Nicehey…itallsortofmakessense…?
letsmoveontoAreaunderafunction.
WeknowthatintegrationgivestheAreaunderacurve(thefirsttheoremofcalculusestablishesthis),butletslookattheareaunderacurvebeingthesumoftheareaoflotsofrectanglesunderthecurve:
(maybegetMrFinneytodrawadiagramontheboard?)
19
Considertheline𝑦 = 3onaCartesianPlane.
Nowconsidertherectangleformedbetweenthisfunctionand𝑥 = 1𝑎𝑛𝑑7.
Clearlytheareaequals18!
Letsstarthere…letscutthisrectangleinto3smallerrectanglesbytakingintervalsonthe𝑥-axis.Andthenwecanaddalltheseareasupandtheyshouldtotaltoanareaequalto18…andclearlythisworks!
Howwidearethesechangesin𝑥?Oops,canIsay∆𝑥J
Clearly,thenewwidthofthesesmallerrectanglesare¥A6W= 2
Soourthreerectangles:
𝐴𝑟𝑒𝑎 = 𝑙𝑤 + 𝑙𝑤 + 𝑙𝑤
andalthoughourlengthsare3,clearlywewillsoonbeworkingwithafunctionthatisnotjustparalleltothe𝑥-axis,soletsjustleaveitat𝑙forthemoment
𝐴𝑟𝑒𝑎 = 𝑙2 + 𝑙2 + 𝑙2
𝐴𝑟𝑒𝑎 = 2(𝑙 + 𝑙 + 𝑙)
andnowall𝑙′𝑠 = 3
𝐴𝑟𝑒𝑎 = 2 3 + 3 + 3 = 18
Correct!
WhatifIwantedyoutohavemorerectangles…letssay𝑛rectangles…we’dhaveawidthof¥A6
�…andeventhoughwe’dhavemorerectanglesofsmaller
width,ourlengthoftherectanglewouldremainas3.
𝐴𝑟𝑒𝑎 = 𝐴6 + 𝐴7 + 𝐴W +⋯+ 𝐴�
𝐴𝑟𝑒𝑎 = 𝑙6𝑤 + 𝑙7𝑤 + 𝑙W𝑤 +⋯+ 𝐴�𝑤
𝐴𝑟𝑒𝑎 = (𝑙6 + 𝑙7 + 𝑙W + ⋯+ 𝑙�)𝑤
𝐴𝑟𝑒𝑎 = 𝑙�
�z6
×𝑤
and𝑤 = e�and𝑙6 = 𝑙7 = 𝑙W = 𝑙� = 3
𝐴𝑟𝑒𝑎 =6𝑛× 3
�
�z6
20
weneedtodigressalittle…consider…
3�
�z6
weareadding3toitself,𝑛times…sothesumof3,𝑛timesisequalto3𝑛
nowbackto
𝐴𝑟𝑒𝑎 =6𝑛× 3
�
�z6
nowbecomes
𝐴𝑟𝑒𝑎 =6𝑛×3𝑛
𝐴𝑟𝑒𝑎 =6𝑛×
3𝑛1 = 18
correct…J
Nowletslookattheareaunderageneralfunction…andletsconsideranumberofrectanglesfillingtheareaunderthecurve.Justlikewhenwedrewsecantstofindtheslopeofthecurve,theserectanglesaren’taperfectfit.Wecandowhat’scalledalefthandseries,orarighthandseries…***MrFinneywillneedtodoadiagramonthewhiteboardforthis***
Clearlytheareasofeachindividualrectangleisequalto𝐴 = 𝑙𝑤,andhere𝑙 = 𝑦and𝑤 = ∆𝑥
Maybewecouldwriteitas:
𝐴𝑟𝑒𝑎 = 𝐴6 + 𝐴7 + 𝐴W +⋯+ 𝐴�
or
𝐴𝑟𝑒𝑎 = 𝑦6∆𝑥 + 𝑦7∆𝑥 + 𝑦W∆𝑥 +⋯+ 𝑦�∆𝑥
andfactorise
𝐴𝑟𝑒𝑎 = (𝑦6 + 𝑦7 + 𝑦W + ⋯+ 𝑦�)∆𝑥
Clearly,themorerectangleswehave,themoreaccurateourareabecomes!HowaboutwehaveanInfinitenumberofteentinyrectangles,andhence∆𝑥 → 0 =𝑑𝑥…andwehave:
𝐴𝑟𝑒𝑎 = 𝑦6 + 𝑦7 + 𝑦W + ⋯+ 𝑦� 𝑑𝑥
Takealookinsidethebrackets…wow,thatlookslikeaSeries!
21
𝐴𝑟𝑒𝑎 = 𝑦6 + 𝑦7 + 𝑦W + ⋯+ 𝑦� 𝑑𝑥
letsneatenthatterminologyupalittle:
𝐴𝑟𝑒𝑎 = 𝑓(𝑥)�
�z6
𝑑𝑥
andherewearesayingthereareanInfinitenumberofrectangles,sothereisaninfinitenumberofvaluesof𝑓 𝑥
𝐴𝑟𝑒𝑎 = lim�→¨
𝑓(𝑥)�
�z6
𝑑𝑥
andbywayofthefundamentaltheoremofcalculuswehave:
𝐴𝑟𝑒𝑎 = 𝑓 𝑥 𝑑𝑥
now…letstryandshowyouthe‘link’fromfirstprinciples
lim�→¨
𝑓 𝑥 𝑑𝑥𝑡𝑜 𝑓 𝑥 𝑑𝑥
Ihaven’teverbeenshown,norfound,aGeneralproofthatlinkstheseabovetwoterms…onlysomespecificexamplesshowit…soletsdothat!
Consider…theareaunderthefunction
𝑓 𝑥 = 𝑥 + 1
between𝑥 = 1𝑎𝑛𝑑7
𝐴 = 𝑙𝑤 + 𝑙𝑤 + 𝑙𝑤 +⋯+ 𝑙𝑤
and𝑙 = 𝑦 = 𝑓 𝑥 andarealldifferent
𝐴 = 𝑓 𝑥6 𝑤 + 𝑓 𝑥7 𝑤 + 𝑓 𝑥W 𝑤 +⋯+ 𝑓 𝑥� 𝑤
𝐴 = 𝑓 𝑥6 + 𝑓 𝑥7 + 𝑓 𝑥W + ⋯+ 𝑓 𝑥� 𝑤
thewidthofourrectanglesbecome¥A6�= e
�
𝐴 =6𝑛 𝑓 𝑥6 + 𝑓 𝑥7 + 𝑓 𝑥W + ⋯+ 𝑓 𝑥�
𝐴 =6𝑛 𝑓(𝑥)
�
�z6
𝐴 =6𝑛 (𝑥� + 1)
�
�z6
noteherethat𝑥constantlychanges,andwewantaninfinitenumberofthem,sothevalueofthefunctionisdifferentineach“iteration”.
22
***Canyouseethat𝑥� = ′𝑠𝑡𝑎𝑟𝑡𝑝𝑜𝑖𝑛𝑡′ + ∆𝑥×𝑖…sohere𝑥� = 1 + e��.***
Wewillalsotakethisopportunitytoindicatethatwewantthe“limit”ofourcalculationswhere𝑛 → ∞
𝐴 = lim�→¨
6𝑛 (1 +
6𝑖𝑛 + 1)
�
�z6
𝐴 = lim�→¨
6𝑛 (
6𝑖𝑛 + 2)
�
�z6
𝐴 = lim�→¨
6𝑛 (
6𝑖𝑛 )
�
�z6
+ (2)�
�z6
𝐴 = lim�→¨
6𝑛
6𝑛 (𝑖)
�
�z6
+ (2)�
�z6
youmayrecallfromsequencesandseriesthat:
𝑖�
�z6
=𝑛 𝑛 + 1
2
Andifwesumthenumber2,𝑛times,weget2𝑛:
2�
�z6
= 2𝑛
***theuseofthese“series”shortcutsisessentialinthisprocess,andthese“Series”arecalledReimannSeries.AfteryougotoUni,pleasecomebackandexplainitmorethoroughlytomesoIcanbeclearerwithnextyearsstudents!J
andweget:
𝐴 = lim�→¨
6𝑛
6𝑛𝑛(𝑛 + 1)
2 + 2𝑛
𝐴 = lim�→¨
6𝑛
6(𝑛 + 1)2 + 2𝑛
𝐴 = lim�→¨
6𝑛 3(𝑛 + 1) + 2𝑛
𝐴 = lim�→¨
6𝑛 3𝑛 + 3 + 2𝑛
𝐴 = lim�→¨
6𝑛 5𝑛 + 3
𝐴 = lim�→¨
30 +18𝑛
nowwecanset𝑛 = ∞
23
𝐴 = 30 +18∞
𝐴 = 30 + 0
𝐴𝑟𝑒𝑎 = 30
OK,soletscheckthat.
Youshouldbeabletodrawthefunctionandjustverifythisareathroughinspection.
or
YoushouldbeabletocalculatetheareaunderthefunctionusingYr9mathstrapeziumareaformula.
or
verifybycalculatingthedefiniteintegral
𝑥 + 1¥
6𝑑𝑥
andmaybeyoucandoallthreeverifications…?
now…headbacktoSolidsofRevolution.