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The First Fundamental Theorem of Calculus looks at the area function and its derivative. It so happens that the derivative of the area function is the original integrand.
Citation preview
. . . . . .
Section5.4TheFundamentalTheoremofCalculus
V63.0121, CalculusI
April23, 2009
Announcements
I Quiz6nextweekon§§5.1–5.2
. . . . . .
Outline
Recall: TheEvaluationTheorema/k/a2FTC
TheFirstFundamentalTheoremofCalculusTheAreaFunctionStatementandproofof1FTCBiographies
Differentiationoffunctionsdefinedbyintegrals“Contrived”examplesErfOtherapplications
Worksheet
Summary
. . . . . .
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
Worksheet
Summary
. . . . . .
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.
I Whenis g increasing?
I Whenis g decreasing?I Overasmallinterval, what’stheaveragerateofchangeof g?
. . . . . .
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?
. . . . . .
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?
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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
Worksheet
Summary
. . . . . .
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 .
. . . . . .
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.
. . . . . .
Outline
Recall: TheEvaluationTheorema/k/a2FTC
TheFirstFundamentalTheoremofCalculusTheAreaFunctionStatementandproofof1FTCBiographies
Differentiationoffunctionsdefinedbyintegrals“Contrived”examplesErfOtherapplications
Worksheet
Summary
. . . . . .
Worksheet
.
.Image: ErickCifuentes
. . . . . .
Outline
Recall: TheEvaluationTheorema/k/a2FTC
TheFirstFundamentalTheoremofCalculusTheAreaFunctionStatementandproofof1FTCBiographies
Differentiationoffunctionsdefinedbyintegrals“Contrived”examplesErfOtherapplications
Worksheet
Summary
. . . . . .
Summary
I FTC linksintegrationanddifferentiationI Whendifferentiatingintegralfunctions, donotforgetthechainrule
I Factsabouttheintegralfunctioncanbegleanedfromtheintegrand