. . . . . .
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