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**

2

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

4

TrigIdentities

AdvancedPeriodicFunctionsskillswillbehelpfulwhenIntegratingsomeTrigfunctionsassomemanipulationoftrigonometricIdentitieswillberequired.

Youmaybecalledupontouseanyoftheidentitiesfromthepreviouschapteragainhere…andyoumusthaverecallofanyTrigIdentitymentionedinthetextBookChapters…

IalsoprovidedaHugelistofextraTrigidentitiesthatyoudonotneedtoremember,butitmaybeanideatohaveyoursheetathandtoassisthere.

OK…soletslookatIntegration…

5

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

6

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

7

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!

8

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

9

Chapter4D

IsaboutusingTrigIdentitiestomanipulatetheintegrandintosomethingthatis“Integratable”?

Thisisallaboutpracticeinusingtheidentities…so…bestgettoit!

IseelimitedvalueinenforcingthememorisingofmoreTrigIdentities.Ifitisnotinmylistinmyworksheets,thenIdonotexpectyoutoknowit.

However,doingthisexercisewillgiveyoucrucialpracticeatsolvingintegrals…butIconfirmthereisnoneedforyoutomemorisethesetrigidentities,andsimplysuggestyouwritedownthenewidentitiesinthischaptersomewhereandrefertothemwhenyouworkthroughthischapter.

DoExercise4D

10

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

11

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

12

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…?

13

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…!

14

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

15

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!

16

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.