Lesson 26: The Fundamental Theorem of Calculus (Section 4 version)

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Section5.4TheFundamentalTheoremofCalculus

V63.0121, CalculusI

April23, 2009

Announcements

I Quiz6nextweekon§§5.1–5.2

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Outline

Recall: TheEvaluationTheorema/k/a2FTC

TheFirstFundamentalTheoremofCalculusTheAreaFunctionStatementandproofof1FTCBiographies

Differentiationoffunctionsdefinedbyintegrals“Contrived”examplesErfOtherapplications

Worksheet

Summary

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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

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

af(x)dx = F(b) − F(a).

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TheIntegralasTotalChange

Anotherwaytostatethistheoremis:∫ b

aF′(x)dx = F(b) − F(a),

or theintegralofaderivativealonganintervalisthetotalchangebetweenthesidesofthatinterval. Thishasmanyramifications:

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aF′(x)dx = F(b) − F(a),


TheoremIf v(t) representsthevelocityofaparticlemovingrectilinearly,then ∫ t1

t0v(t)dt = s(t1) − s(t0).

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aF′(x)dx = F(b) − F(a),


TheoremIf MC(x) representsthemarginalcostofmaking x unitsofaproduct, then

C(x) = C(0) +

∫ x

0MC(q)dq.

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aF′(x)dx = F(b) − F(a),


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

m(x) =

∫ x

0ρ(s)ds.

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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

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Outline




Worksheet

Summary

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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 → ∞.

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Anareafunction, continued

So

g(x) =x4

4.

Thismeansthatg′(x) = x3.

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Theareafunction

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

g(x) =

∫ x

af(t)dt.

I Whenis g increasing?

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

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Theareafunction


g(x) =

∫ x

af(t)dt.

I Whenis g increasing?I Whenis g decreasing?

I Overasmallinterval, what’stheaveragerateofchangeof g?

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Theareafunction


g(x) =

∫ x

af(t)dt.

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

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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).

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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.

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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.


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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.


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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.


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MeettheMathematician: JamesGregory

I Scottish, 1638-1675I AstronomerandGeometer

I Conceivedtranscendentalnumbersandfoundevidencethatπ wastranscendental

I Provedageometricversionof1FTC asalemmabutdidn’ttakeitfurther

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MeettheMathematician: IsaacBarrow

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

I Hadafamousstudent

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MeettheMathematician: IsaacNewton

I English, 1643–1727I ProfessoratCambridge(England)

I PhilosophiaeNaturalisPrincipiaMathematicapublished1687

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MeettheMathematician: GottfriedLeibniz

I German, 1646–1716I Eminentphilosopheraswellasmathematician

I Contemporarilydisgracedbythecalculusprioritydispute

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DifferentiationandIntegrationasreverseprocesses

Puttingtogether1FTC and2FTC,wegetabeautifulrelationshipbetweenthetwofundamentalconceptsincalculus.

Iddx

∫ x

af(t)dt = f(x)

I ∫ b

aF′(x)dx = F(b) − F(a).

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Outline




Worksheet

Summary

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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.

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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

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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 .

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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 .

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Otherfunctionsdefinedbyintegrals

I Thefuturevalueofanasset:

FV(t) =

∫ ∞

tπ(τ)e−rτ dτ

where π(τ) istheprofitabilityattime τ and r isthediscountrate.

I Theconsumersurplusofagood:

CS(q∗) =

∫ q∗

0(f(q) − p∗)dq

where f(q) isthedemandfunctionand p∗ and q∗ theequilibriumpriceandquantity.

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Outline




Worksheet

Summary

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Worksheet

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.Image: ErickCifuentes

http://www.flickr.com/photos/homiesinheaven/1409350817/

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Summary

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Summary

I FTC linksintegrationanddifferentiationI Whendifferentiatingintegralfunctions, donotforgetthechainrule

I Factsabouttheintegralfunctioncanbegleanedfromtheintegrand

Lesson 26: The Fundamental Theorem of Calculus (Section 4 version)

Education

Transcript of Lesson 26: The Fundamental Theorem of Calculus (Section 4 version)