. . . . . .

Section5.4TheFundamentalTheoremofCalculus

V63.0121.027, CalculusI

December8, 2009

Announcements

I FinalExam: Friday12/18, 2:00-3:50pm, TischUC50

. . . . . .

Redemptionpolicies

I Currentdistributionofgrade: 40%final, 25%midterm, 15%quizzes, 10%writtenHW,10%WebAssign

I Rememberwedropthelowestquiz, lowestwrittenHW,and5 lowestWebAssign-ments

I [new!] Ifyourfinalexamscorebeatsyourmidtermscore,wewillre-weightitby50%andmakethemidterm15%

. . . . . .

Outline

Recall: TheEvaluationTheorema/k/a2FTC

TheFirstFundamentalTheoremofCalculusTheAreaFunctionStatementandproofof1FTCBiographies

Differentiationoffunctionsdefinedbyintegrals“Contrived”examplesErfOtherapplications

. . . . . .

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:

. . . . . .

TheIntegralasTotalChange

Anotherwaytostatethistheoremis:∫ b

aF′(x)dx = F(b)− F(a),

or theintegralofaderivativealonganintervalisthetotalchangebetweenthesidesofthatinterval. Thishasmanyramifications:

TheoremIf v(t) representsthevelocityofaparticlemovingrectilinearly,then ∫ t1

t0v(t)dt = s(t1)− s(t0).

. . . . . .

TheIntegralasTotalChange

Anotherwaytostatethistheoremis:∫ b

aF′(x)dx = F(b)− F(a),

or theintegralofaderivativealonganintervalisthetotalchangebetweenthesidesofthatinterval. Thishasmanyramifications:

TheoremIf MC(x) representsthemarginalcostofmaking x unitsofaproduct, then

C(x) = C(0) +∫ x

0MC(q)dq.

. . . . . .

TheIntegralasTotalChange

Anotherwaytostatethistheoremis:∫ b

aF′(x)dx = F(b)− F(a),

or theintegralofaderivativealonganintervalisthetotalchangebetweenthesidesofthatinterval. Thishasmanyramifications:

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

Recall: TheEvaluationTheorema/k/a2FTC

TheFirstFundamentalTheoremofCalculusTheAreaFunctionStatementandproofof1FTCBiographies

Differentiationoffunctionsdefinedbyintegrals“Contrived”examplesErfOtherapplications

. . . . . .

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 ∆t =xnand ti = 0+ i∆t =

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 ∆t =xnand ti = 0+ i∆t =

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.

I Thevariableis x; t isa“dummy”variablethat’sintegratedover.

I Picturechanging x andtakingmoreoflessoftheregionunderthecurve.

I Question: Whatdoes f tellyouabout g?

. . . . . .

Envisioningtheareafunction

ExampleSuppose f(t) isthefunctiongraphedbelow

..t

.v

..t0

..t1

..t2

..t3

.

.

..c

I Let g(x) =∫ x

t0f(t)dt. Whatcanyousayabout g?

. . . . . .

featuresof g from f

Interval sign monotonicity monotonicity concavity

of f of g of f of g

[t0, t1] + ↗ ↗ ⌣

[t1, c] + ↗ ↘ ⌢

[c, t2] − ↘ ↘ ⌢

[t2, t3] − ↘ ↗ ⌣

[t3,∞) − ↘ → none

Weseethat g isbehavingalotlikeanantiderivativeof f.

. . . . . .

featuresof g from f

Interval sign monotonicity monotonicity concavity

of f of g of f of g

[t0, t1] + ↗ ↗ ⌣

[t1, c] + ↗ ↘ ⌢

[c, t2] − ↘ ↘ ⌢

[t2, t3] − ↘ ↗ ⌣

[t3,∞) − ↘ → none

Weseethat g isbehavingalotlikeanantiderivativeof f.

. . . . . .

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).

. . . . . .

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).

. . . . . .

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).

. . . . . .

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).

. . . . . .

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).

. . . . . .

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).

. . . . . .

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).

. . . . . .

MeettheMathematician: JamesGregory

I Scottish, 1638-1675I AstronomerandGeometer

I Conceivedtranscendentalnumbersandfoundevidencethatπ wastranscendental

I Provedageometricversionof1FTC asalemmabutdidn’ttakeitfurther

. . . . . .

MeettheMathematician: IsaacBarrow

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

I Hadafamousstudent

. . . . . .

MeettheMathematician: IsaacNewton

I English, 1643–1727I ProfessoratCambridge(England)

I PhilosophiaeNaturalisPrincipiaMathematicapublished1687

. . . . . .

MeettheMathematician: GottfriedLeibniz

I German, 1646–1716I Eminentphilosopheraswellasmathematician

I Contemporarilydisgracedbythecalculusprioritydispute

. . . . . .

DifferentiationandIntegrationasreverseprocesses

Puttingtogether1FTC and2FTC,wegetabeautifulrelationshipbetweenthetwofundamentalconceptsincalculus.

Iddx

∫ x

af(t)dt = f(x)

I ∫ b

aF′(x)dx = F(b)− F(a).

. . . . . .

DifferentiationandIntegrationasreverseprocesses

Puttingtogether1FTC and2FTC,wegetabeautifulrelationshipbetweenthetwofundamentalconceptsincalculus.

Iddx

∫ x

af(t)dt = f(x)

I ∫ b

aF′(x)dx = F(b)− F(a).

. . . . . .

Outline

Recall: TheEvaluationTheorema/k/a2FTC

TheFirstFundamentalTheoremofCalculusTheAreaFunctionStatementandproofof1FTCBiographies

Differentiationoffunctionsdefinedbyintegrals“Contrived”examplesErfOtherapplications

. . . . . .

Differentiationofareafunctions

Example

Let h(x) =∫ 3x

0t3 dt. Whatis h′(x)?

Solution(Using2FTC)

h(x) =t4

4

∣∣∣∣3x0

=14(3x)4 = 1

4 · 81x4, so h′(x) = 81x3.

Solution(Using1FTC)

Wecanthinkof h asthecomposition g ◦ k, where g(u) =∫ u

0t3 dt

and k(x) = 3x. Then

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

. . . . . .

Differentiationofareafunctions

Example

Let h(x) =∫ 3x

0t3 dt. Whatis h′(x)?

Solution(Using2FTC)

h(x) =t4

4

∣∣∣∣3x0

=14(3x)4 = 1

4 · 81x4, so h′(x) = 81x3.

Solution(Using1FTC)

Wecanthinkof h asthecomposition g ◦ k, where g(u) =∫ u

0t3 dt

and k(x) = 3x. Then

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

. . . . . .

Differentiationofareafunctions

Example

Let h(x) =∫ 3x

0t3 dt. Whatis h′(x)?

Solution(Using2FTC)

h(x) =t4

4

∣∣∣∣3x0

=14(3x)4 = 1

4 · 81x4, so h′(x) = 81x3.

Solution(Using1FTC)

Wecanthinkof h asthecomposition g ◦ k, where g(u) =∫ u

0t3 dt

and k(x) = 3x.

Then

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

. . . . . .

Differentiationofareafunctions

Example

Let h(x) =∫ 3x

0t3 dt. Whatis h′(x)?

Solution(Using2FTC)

h(x) =t4

4

∣∣∣∣3x0

=14(3x)4 = 1

4 · 81x4, so h′(x) = 81x3.

Solution(Using1FTC)

Wecanthinkof h asthecomposition g ◦ k, where g(u) =∫ u

0t3 dt

and k(x) = 3x. Then

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

. . . . . .

Differentiationofareafunctions, ingeneral

I by1FTCddx

∫ k(x)

af(t)dt = f(k(x))k′(x)

I byreversingtheorderofintegration:

ddx

∫ b

h(x)f(t)dt = − d

dx

∫ h(x)

bf(t)dt = −f(h(x))h′(x)

I bycombiningthetwoabove:

ddx

∫ k(x)

h(x)f(t)dt =

ddx

(∫ k(x)

0f(t)dt+

∫ 0

h(x)f(t)dt

)= f(k(x))k′(x)− f(h(x))h′(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

. . . . . .

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

Findthederivativeof F(x) =∫ ex

x3sin4 t dt.

Solution

ddx

∫ ex

x3sin4 t dt = sin4(ex) · ex − sin4(x3) · 3x2

Noticehereit’smucheasierthanfindinganantiderivativeforsin4.

. . . . . .

Example

Findthederivativeof F(x) =∫ ex

x3sin4 t dt.

Solution

ddx

∫ ex

x3sin4 t dt = sin4(ex) · ex − sin4(x3) · 3x2

Noticehereit’smucheasierthanfindinganantiderivativeforsin4.

. . . . . .

Example

Findthederivativeof F(x) =∫ ex

x3sin4 t dt.

Solution

ddx

∫ ex

x3sin4 t dt = sin4(ex) · ex − sin4(x3) · 3x2

Noticehereit’smucheasierthanfindinganantiderivativeforsin4.

. . . . . .

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 .

. . . . . .

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 .

. . . . . .

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 .

. . . . . .

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 .

. . . . . .

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 .

. . . . . .

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 .

. . . . . .

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.

. . . . . .

Surplusbypicture

..quantity(q)

.price(p)

.

.demand f(q)

.supply

.equilibrium

..q∗

..p∗

.consumersurplus