Lesson 28: The Fundamental Theorem of Calculus

. . . . . .

Section5.4TheFundamentalTheoremofCalculus

Math1aIntroductiontoCalculus

April16, 2008

Announcements

◮ Midtermisfinished: x̄ ≈ 43, σ ≈ 6.◮ MidtermIII isWednesday4/30inclass◮ Friday5/2isMovieDay!◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323

. . . . . .

Announcements

◮ Midtermisfinished: x̄ ≈ 43, σ ≈ 6.◮ MidtermIII isWednesday4/30inclass◮ Friday5/2isMovieDay!◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323

. . . . . .

Outline

Lasttime: TheSecondFundamentalTheoremofCalculusMyfirsttableofintegrals

TheFirstFundamentalTheoremofCalculusTheAreaFunctionStatementandproofof1FTCBiographies

Differentiationoffunctionsdefinedbyintegrals“Contrived”examplesErfOtherapplications

Factsabout g from fA problem

. . . . . .

Thedefiniteintegralasalimit

DefinitionIf f isafunctiondefinedon [a,b], the definiteintegralof f from ato b isthenumber∫ b

af(x)dx = lim

∆x→0

n∑i=1

f(ci) ∆x

. . . . . .

Theorem(TheSecondFundamentalTheoremofCalculus)Suppose f isintegrableon [a,b] and f = F′ foranotherfunction F,then ∫ b

af(x)dx = F(b) − F(a).

. . . . . .

TheIntegralasTotalChange

Anotherwaytostatethistheoremis:∫ b

aF′(x)dx = F(b) − F(a),

or theintegralofaderivativealonganintervalisthetotalchangebetweenthesidesofthatinterval. Thishasmanyramifications:

. . . . . .



aF′(x)dx = F(b) − F(a),


TheoremIf v(t) representsthevelocityofaparticlemovingrectilinearly,then ∫ t1

t0v(t)dt = s(t1) − s(t0).

. . . . . .



aF′(x)dx = F(b) − F(a),


TheoremIf MC(x) representsthemarginalcostofmaking x unitsofaproduct, then

C(x) = C(0) +

∫ x

0MC(q)dq.

. . . . . .



aF′(x)dx = F(b) − F(a),


TheoremIf ρ(x) representsthedensityofathinrodatadistanceof x fromitsend, thenthemassoftherodupto x is

m(x) =

∫ x

0ρ(s)ds.

. . . . . .

Myfirsttableofintegrals∫[f(x) + g(x)] dx =

∫f(x)dx +

∫g(x)dx∫

xn dx =xn+1

n + 1+ C (n ̸= −1)∫

ex dx = ex + C∫sin x dx = − cos x + C∫cos x dx = sin x + C∫sec2 x dx = tan x + C∫

sec x tan x dx = sec x + C∫1

1 + x2dx = arctan x + C

∫cf(x)dx = c

∫f(x)dx∫

1xdx = ln |x| + C∫

ax dx =ax

ln a+ C∫

csc2 x dx = − cot x + C∫csc x cot x dx = − csc x + C∫

1√1− x2

dx = arcsin x + C

. . . . . .

Outline





. . . . . .

Anareafunction

Let f(t) = t3 anddefine g(x) =

∫ x

0f(t)dt. Canweevaluatethe

integralin g(x)?

..0 .x

Dividingtheinterval [0, x] into n pieces

gives ∆x =xnand xi = 0 + i∆x =

ixn.

So

Rn =xn· x

3

n3+

xn· (2x)3

n3+ · · · + x

n· (nx)3

n3

=x4

n4(13 + 23 + 33 + · · · + n3

)=

x4

n4[12n(n + 1)

]2=

x4n2(n + 1)2

4n4→ x4

4

as n → ∞.

. . . . . .

Anareafunction

Let f(t) = t3 anddefine g(x) =

∫ x

0f(t)dt. Canweevaluatethe

integralin g(x)?

..0 .x

Dividingtheinterval [0, x] into n pieces

gives ∆x =xnand xi = 0 + i∆x =

ixn.

So

Rn =xn· x

3

n3+

xn· (2x)3

n3+ · · · + x

n· (nx)3

n3

=x4

n4(13 + 23 + 33 + · · · + n3

)=

x4

n4[12n(n + 1)

]2=

x4n2(n + 1)2

4n4→ x4

4

as n → ∞.

. . . . . .

Anareafunction, continued

So

g(x) =x4

4.

Thismeansthatg′(x) = x3.

. . . . . .

Anareafunction, continued

So

g(x) =x4

4.

Thismeansthatg′(x) = x3.

. . . . . .

Theareafunction

Let f beafunctionwhichisintegrable(i.e., continuousorwithfinitelymanyjumpdiscontinuities)on [a,b]. Define

g(x) =

∫ x

af(t)dt.

◮ Whenis g increasing?

◮ Whenis g decreasing?◮ Overasmallinterval, what’stheaveragerateofchangeof g?

. . . . . .

Theareafunction


g(x) =

∫ x

af(t)dt.

◮ Whenis g increasing?◮ Whenis g decreasing?

◮ Overasmallinterval, what’stheaveragerateofchangeof g?

. . . . . .

Theareafunction


g(x) =

∫ x

af(t)dt.

◮ Whenis g increasing?◮ Whenis g decreasing?◮ Overasmallinterval, what’stheaveragerateofchangeof g?

. . . . . .

Theorem(TheFirstFundamentalTheoremofCalculus)Let f beanintegrablefunctionon [a,b] anddefine

g(x) =

∫ x

af(t)dt.

If f iscontinuousat x in (a,b), then g isdifferentiableat x and

g′(x) = f(x).

. . . . . .

Proof.Let h > 0 begivensothat x + h < b. Wehave

g(x + h) − g(x)h

=

1h

∫ x+h

xf(t)dt.

Let Mh bethemaximumvalueof f on [x, x + h], and mh theminimumvalueof f on [x, x + h]. From§5.2wehave

mh · h ≤

∫ x+h

xf(t)dt

≤ Mh · h

So

mh ≤ g(x + h) − g(x)h

≤ Mh.

As h → 0, both mh and Mh tendto f(x). Zappa-dappa.

. . . . . .


g(x + h) − g(x)h

=1h

∫ x+h

xf(t)dt.


mh · h ≤

∫ x+h

xf(t)dt

≤ Mh · h

So

mh ≤ g(x + h) − g(x)h

≤ Mh.


. . . . . .


g(x + h) − g(x)h

=1h

∫ x+h

xf(t)dt.


mh · h ≤

∫ x+h

xf(t)dt

≤ Mh · h

So

mh ≤ g(x + h) − g(x)h

≤ Mh.


. . . . . .


g(x + h) − g(x)h

=1h

∫ x+h

xf(t)dt.


mh · h ≤

∫ x+h

xf(t)dt ≤ Mh · h

So

mh ≤ g(x + h) − g(x)h

≤ Mh.


. . . . . .


g(x + h) − g(x)h

=1h

∫ x+h

xf(t)dt.


mh · h ≤∫ x+h

xf(t)dt ≤ Mh · h

So

mh ≤ g(x + h) − g(x)h

≤ Mh.


. . . . . .


g(x + h) − g(x)h

=1h

∫ x+h

xf(t)dt.


mh · h ≤∫ x+h

xf(t)dt ≤ Mh · h

So

mh ≤ g(x + h) − g(x)h

≤ Mh.


. . . . . .


g(x + h) − g(x)h

=1h

∫ x+h

xf(t)dt.


mh · h ≤∫ x+h

xf(t)dt ≤ Mh · h

So

mh ≤ g(x + h) − g(x)h

≤ Mh.


. . . . . .

MeettheMathematician: JamesGregory

◮ Scottish, 1638-1675◮ AstronomerandGeometer

◮ Conceivedtranscendentalnumbersandfoundevidencethatπ wastranscendental

◮ Provedageometricversionof1FTC asalemmabutdidn’ttakeitfurther

. . . . . .

MeettheMathematician: IsaacBarrow

◮ English, 1630-1677◮ ProfessorofGreek,theology, andmathematicsatCambridge

◮ Hadafamousstudent

. . . . . .

MeettheMathematician: IsaacNewton

◮ English, 1643–1727◮ ProfessoratCambridge(England)

◮ PhilosophiaeNaturalisPrincipiaMathematicapublished1687

. . . . . .

MeettheMathematician: GottfriedLeibniz

◮ German, 1646–1716◮ Eminentphilosopheraswellasmathematician

◮ Contemporarilydisgracedbythecalculusprioritydispute

. . . . . .

DifferentiationandIntegrationasreverseprocesses

Puttingtogether1FTC and2FTC,wegetabeautifulrelationshipbetweenthetwofundamentalconceptsincalculus.

◮ddx

∫ x

af(t)dt = f(x)

◮ ∫ b

aF′(x)dx = F(b) − F(a).

. . . . . .

DifferentiationandIntegrationasreverseprocesses

Puttingtogether1FTC and2FTC,wegetabeautifulrelationshipbetweenthetwofundamentalconceptsincalculus.

◮ddx

∫ x

af(t)dt = f(x)

◮ ∫ b

aF′(x)dx = F(b) − F(a).

. . . . . .

Outline





. . . . . .

Differentiationofareafunctions

Example

Let g(x) =

∫ x

0t3 dt. Weknow g′(x) = x3. Whatifinsteadwehad

h(x) =

∫ 3x

0t3 dt.

Whatis h′(x)?

SolutionWecanthinkof h asthecomposition g ◦ k, where g(u) =

∫ u

0t3 dt

and k(x) = 3x. Then

h′(x) = g′(k(x))k′(x) = 3(k(x))3 = 3(3x)3 = 81x3.

. . . . . .

Differentiationofareafunctions

Example

Let g(x) =

∫ x

0t3 dt. Weknow g′(x) = x3. Whatifinsteadwehad

h(x) =

∫ 3x

0t3 dt.

Whatis h′(x)?

SolutionWecanthinkof h asthecomposition g ◦ k, where g(u) =

∫ u

0t3 dt

and k(x) = 3x. Then

h′(x) = g′(k(x))k′(x) = 3(k(x))3 = 3(3x)3 = 81x3.

. . . . . .

Example

Let h(x) =

∫ sin2 x

0(17t2 + 4t− 4)dt. Whatis h′(x)?

SolutionWehave

ddx

∫ sin2 x

0(17t2 + 4t− 4)dt

=(17(sin2 x)2 + 4(sin2 x) − 4

)· ddx

sin2 x

=(17 sin4 x + 4 sin2 x− 4

)· 2 sin x cos x

. . . . . .

Example

Let h(x) =

∫ sin2 x

0(17t2 + 4t− 4)dt. Whatis h′(x)?

SolutionWehave

ddx

∫ sin2 x

0(17t2 + 4t− 4)dt

=(17(sin2 x)2 + 4(sin2 x) − 4

)· ddx

sin2 x

=(17 sin4 x + 4 sin2 x− 4

)· 2 sin x cos x

. . . . . .

ErfHere’safunctionwithafunnynamebutanimportantrole:

erf(x) =2√π

∫ x

0e−t2 dt.

Itturnsout erf istheshapeofthebellcurve. Wecan’tfind erf(x),explicitly, butwedoknowitsderivative.

erf′(x) =2√πe−x2 .

Example

Findddx

erf(x2).

SolutionBythechainrulewehave

ddx

erf(x2) = erf′(x2)ddx

x2 =2√πe−(x2)22x =

4√πxe−x4 .

. . . . . .


erf(x) =2√π

∫ x

0e−t2 dt.

Itturnsout erf istheshapeofthebellcurve.

Wecan’tfind erf(x),explicitly, butwedoknowitsderivative.

erf′(x) =2√πe−x2 .

Example

Findddx

erf(x2).


ddx


x2 =2√πe−(x2)22x =

4√πxe−x4 .

. . . . . .


erf(x) =2√π

∫ x

0e−t2 dt.


erf′(x) =

2√πe−x2 .

Example

Findddx

erf(x2).


ddx


x2 =2√πe−(x2)22x =

4√πxe−x4 .

. . . . . .


erf(x) =2√π

∫ x

0e−t2 dt.


erf′(x) =2√πe−x2 .

Example

Findddx

erf(x2).


ddx


x2 =2√πe−(x2)22x =

4√πxe−x4 .

. . . . . .


erf(x) =2√π

∫ x

0e−t2 dt.


erf′(x) =2√πe−x2 .

Example

Findddx

erf(x2).


ddx


x2 =2√πe−(x2)22x =

4√πxe−x4 .

. . . . . .


erf(x) =2√π

∫ x

0e−t2 dt.


erf′(x) =2√πe−x2 .

Example

Findddx

erf(x2).


ddx


x2 =2√πe−(x2)22x =

4√πxe−x4 .

. . . . . .

Otherfunctionsdefinedbyintegrals

◮ Thefuturevalueofanasset:

FV(t) =

∫ ∞

tπ(τ)e−rτ dτ

where π(τ) istheprofitabilityattime τ and r isthediscountrate.

◮ Theconsumersurplusofagood:

CS(p∗) =

∫ p∗

0f(p)dp

where f(p) isthedemandfunctionand p∗ istheequilibriumprice(dependsonsupply)

. . . . . .

Outline





. . . . . .

Factsabout g from f

Let f bethefunctionwhosegraphisgivenbelow.Supposethethepositionattime t secondsofaparticlemoving

alongacoordinateaxisis s(t) =

∫ t

0f(x)dx meters. Usethegraph

toanswerthefollowingquestions.

. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)

Whatistheparticle’svelocityattime t = 5?

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)

Whatistheparticle’svelocityattime t = 5?

SolutionRecallthatbytheFTC wehave

s′(t) = f(t).

So s′(5) = f(5) = 2.

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)

Istheaccelerationofthepar-ticleattime t = 5 positiveornegative?

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)

Istheaccelerationofthepar-ticleattime t = 5 positiveornegative?

SolutionWehave s′′(5) = f′(5), whichlooksnegativefromthegraph.

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)

Whatistheparticle’spositionattime t = 3?

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)

Whatistheparticle’spositionattime t = 3?

SolutionSinceon [0,3], f(x) = x, wehave

s(3) =

∫ 3

0x dx =

92.

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)

Atwhattimeduringthefirst9secondsdoes s haveitslargestvalue?

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)


Solution

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)


SolutionThecriticalpointsof s arethezerosof s′ = f.

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)


SolutionBylookingatthegraph, weseethat f ispositivefromt = 0 to t = 6, thennegativefrom t = 6 to t = 9.

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)


SolutionTherefore s isincreasingon[0, 6], thendecreasingon[6, 9]. Soitslargestvalueisatt = 6.

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)

Approximately when is theaccelerationzero?

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)

Approximately when is theaccelerationzero?

Solutions′′ = 0 when f′ = 0, whichhappensat t = 4 and t = 7.5(approximately)

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)

Whenistheparticlemovingtowardtheorigin? Awayfromtheorigin?

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)


SolutionTheparticleismovingawayfromtheoriginwhen s > 0and s′ > 0.

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)


SolutionSince s(0) = 0 and s′ > 0 on(0, 6), weknowtheparticleismovingawayfromtheoriginthen.

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)


SolutionAfter t = 6, s′ < 0, sotheparticleismovingtowardtheorigin.

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)

On which side (positive ornegative) of the origin doestheparticlelieattime t = 9?

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)


SolutionWehave s(9) =∫ 6

0f(x)dx +

∫ 9

6f(x)dx,

wheretheleftintegralispositiveandtherightintegralisnegative.

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)


SolutionInordertodecidewhethers(9) ispositiveornegative,weneedtodecideifthefirstareaismorepositivethanthesecondareaisnegative.

. . . . . .

Factsabout g from f



∫ t



. ..1

..2

..3

..4

..5

..6

..7

..8

..9

.1

.2

.3

.4

.• .(1,1)

.• .(2,2)

.• .(3,3).• .(5,2)


SolutionThisappearstobethecase,so s(9) ispositive.

Lesson 28: The Fundamental Theorem of Calculus

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