Section 5.4The Fundamental Theorem of Calculus
Math 1a
December 12, 2007
Announcements
I my next office hours: Today 1–3 (SC 323)
I MT II is graded. Come to OH to talk about it
I Final seview sessions: Wed 1/9 and Thu 1/10 in Hall D, Sun1/13 in Hall C, all 7–8:30pm
I Final tentatively scheduled for January 17, 9:15am
Outline
The Area Function
FTC1StatementProofBiographies
Differentiation of functions defined by integrals“Contrived” examplesErfOther applications
FTC2
Facts about g from fA problem
An area function
Let f (t) = t2 and define g(x) =
∫ x
0t3 dt. Can we evaluate the
integral in g(x)?
0 x
Dividing the interval [0, x ] into n pieces
gives ∆x =x
nand xi = 0 + i∆x =
ix
n.
So
Rn =x
n· x3
n3+
x
n· (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→∞.
An area function
Let f (t) = t2 and define g(x) =
∫ x
0t3 dt. Can we evaluate the
integral in g(x)?
0 x
Dividing the interval [0, x ] into n pieces
gives ∆x =x
nand xi = 0 + i∆x =
ix
n.
So
Rn =x
n· x3
n3+
x
n· (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→∞.
An area function, continued
So
g(x) =x4
4.
This means thatg ′(x) = x3.
An area function, continued
So
g(x) =x4
4.
This means thatg ′(x) = x3.
The area function
Let f be a function which is integrable (i.e., continuous or withfinitely many jump discontinuities) on [a, b]. Define
g(x) =
∫ t
af (t) dt.
I When is g increasing?
I When is g decreasing?
I Over a small interval, what’s the average rate of change of g?
The area function
Let f be a function which is integrable (i.e., continuous or withfinitely many jump discontinuities) on [a, b]. Define
g(x) =
∫ t
af (t) dt.
I When is g increasing?
I When is g decreasing?
I Over a small interval, what’s the average rate of change of g?
The area function
Let f be a function which is integrable (i.e., continuous or withfinitely many jump discontinuities) on [a, b]. Define
g(x) =
∫ t
af (t) dt.
I When is g increasing?
I When is g decreasing?
I Over a small interval, what’s the average rate of change of g?
Outline
The Area Function
FTC1StatementProofBiographies
Differentiation of functions defined by integrals“Contrived” examplesErfOther applications
FTC2
Facts about g from fA problem
Theorem (The First Fundamental Theorem of Calculus)
Let f be an integrable function on [a, b] and define
g(x) =
∫ x
af (t) dt.
If f is continuous at x in (a, b), then g is differentiable at x and
g ′(x) = f (x).
Proof.Let h > 0 be given so that x + h < b. We have
g(x + h)− g(x)
h=
1
h
∫ x+h
xf (t) dt.
Let Mh be the maximum value of f on [x , x + h], and mh theminimum value of f on [x , x + h]. From §5.2 we have
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 tend to f (x). Zappa-dappa.
Proof.Let h > 0 be given so that x + h < b. We have
g(x + h)− g(x)
h=
1
h
∫ x+h
xf (t) dt.
Let Mh be the maximum value of f on [x , x + h], and mh theminimum value of f on [x , x + h]. From §5.2 we have
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 tend to f (x). Zappa-dappa.
Proof.Let h > 0 be given so that x + h < b. We have
g(x + h)− g(x)
h=
1
h
∫ x+h
xf (t) dt.
Let Mh be the maximum value of f on [x , x + h], and mh theminimum value of f on [x , x + h]. From §5.2 we have
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 tend to f (x). Zappa-dappa.
Proof.Let h > 0 be given so that x + h < b. We have
g(x + h)− g(x)
h=
1
h
∫ x+h
xf (t) dt.
Let Mh be the maximum value of f on [x , x + h], and mh theminimum value of f on [x , x + h]. From §5.2 we have
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 tend to f (x). Zappa-dappa.
Proof.Let h > 0 be given so that x + h < b. We have
g(x + h)− g(x)
h=
1
h
∫ x+h
xf (t) dt.
Let Mh be the maximum value of f on [x , x + h], and mh theminimum value of f on [x , x + h]. From §5.2 we have
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 tend to f (x). Zappa-dappa.
Proof.Let h > 0 be given so that x + h < b. We have
g(x + h)− g(x)
h=
1
h
∫ x+h
xf (t) dt.
Let Mh be the maximum value of f on [x , x + h], and mh theminimum value of f on [x , x + h]. From §5.2 we have
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 tend to f (x). Zappa-dappa.
Proof.Let h > 0 be given so that x + h < b. We have
g(x + h)− g(x)
h=
1
h
∫ x+h
xf (t) dt.
Let Mh be the maximum value of f on [x , x + h], and mh theminimum value of f on [x , x + h]. From §5.2 we have
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 tend to f (x). Zappa-dappa.
Meet the Mathematician: Isaac Barrow
I English, 1630-1677
I Professor of Greek,theology, andmathematics atCambridge
I Had a famous student
Meet the Mathematician: Isaac Newton
I English, 1643–1727
I Professor at Cambridge(England)
I Philosophiae NaturalisPrincipia Mathematicapublished 1687
Meet the Mathematician: Gottfried Leibniz
I German, 1646–1716
I Eminent philosopher aswell as mathematician
I Contemporarily disgracedby the calculus prioritydispute
Outline
The Area Function
FTC1StatementProofBiographies
Differentiation of functions defined by integrals“Contrived” examplesErfOther applications
FTC2
Facts about g from fA problem
Differentiation of area functions
Example
Let g(x) =
∫ x
0t3 dt. We know g ′(x) = x3. What if instead we
had
h(x) =
∫ 3x
0t3 dt.
What is h′(x)?
SolutionWe can think of h as the composition 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.
Differentiation of area functions
Example
Let g(x) =
∫ x
0t3 dt. We know g ′(x) = x3. What if instead we
had
h(x) =
∫ 3x
0t3 dt.
What is h′(x)?
SolutionWe can think of h as the composition 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. What is h′(x)?
SolutionWe have
d
dx
∫ sin2 x
0(17t2 + 4t − 4) dt
=(17(sin2 x)2 + 4(sin2 x)− 4
)· d
dxsin2 x
=(17 sin4 x + 4 sin2 x − 4
)· 2 sin x cos x
Example
Let h(x) =
∫ sin2 x
0(17t2 + 4t − 4) dt. What is h′(x)?
SolutionWe have
d
dx
∫ sin2 x
0(17t2 + 4t − 4) dt
=(17(sin2 x)2 + 4(sin2 x)− 4
)· d
dxsin2 x
=(17 sin4 x + 4 sin2 x − 4
)· 2 sin x cos x
ErfHere’s a function with a funny name but an important role:
erf(x) =2√π
∫ x
0e−t2
dt.
It turns out erf is the shape of the bell curve. We can’t find erf(x),explicitly, but we do know its derivative.
erf ′(x) =2√π
e−x2.
Example
Findd
dxerf(x2).
SolutionBy the chain rule we have
d
dxerf(x2) = erf ′(x2)
d
dxx2 =
2√2π
e−(x2)22x =4√π
xe−x4.
ErfHere’s a function with a funny name but an important role:
erf(x) =2√π
∫ x
0e−t2
dt.
It turns out erf is the shape of the bell curve.
We can’t find erf(x),explicitly, but we do know its derivative.
erf ′(x) =2√π
e−x2.
Example
Findd
dxerf(x2).
SolutionBy the chain rule we have
d
dxerf(x2) = erf ′(x2)
d
dxx2 =
2√2π
e−(x2)22x =4√π
xe−x4.
ErfHere’s a function with a funny name but an important role:
erf(x) =2√π
∫ x
0e−t2
dt.
It turns out erf is the shape of the bell curve. We can’t find erf(x),explicitly, but we do know its derivative.
erf ′(x) =
2√π
e−x2.
Example
Findd
dxerf(x2).
SolutionBy the chain rule we have
d
dxerf(x2) = erf ′(x2)
d
dxx2 =
2√2π
e−(x2)22x =4√π
xe−x4.
ErfHere’s a function with a funny name but an important role:
erf(x) =2√π
∫ x
0e−t2
dt.
It turns out erf is the shape of the bell curve. We can’t find erf(x),explicitly, but we do know its derivative.
erf ′(x) =2√π
e−x2.
Example
Findd
dxerf(x2).
SolutionBy the chain rule we have
d
dxerf(x2) = erf ′(x2)
d
dxx2 =
2√2π
e−(x2)22x =4√π
xe−x4.
ErfHere’s a function with a funny name but an important role:
erf(x) =2√π
∫ x
0e−t2
dt.
It turns out erf is the shape of the bell curve. We can’t find erf(x),explicitly, but we do know its derivative.
erf ′(x) =2√π
e−x2.
Example
Findd
dxerf(x2).
SolutionBy the chain rule we have
d
dxerf(x2) = erf ′(x2)
d
dxx2 =
2√2π
e−(x2)22x =4√π
xe−x4.
ErfHere’s a function with a funny name but an important role:
erf(x) =2√π
∫ x
0e−t2
dt.
It turns out erf is the shape of the bell curve. We can’t find erf(x),explicitly, but we do know its derivative.
erf ′(x) =2√π
e−x2.
Example
Findd
dxerf(x2).
SolutionBy the chain rule we have
d
dxerf(x2) = erf ′(x2)
d
dxx2 =
2√2π
e−(x2)22x =4√π
xe−x4.
Other functions defined by integrals
I The future value of an asset:
FV (t) =
∫ ∞t
π(τ)e−rτ dτ
where π(τ) is the profitability at time τ and r is the discountrate.
I The consumer surplus of a good:
CS(p∗) =
∫ p∗
0f (p) dp
where f (p) is the demand function and p∗ is the equilibriumprice (depends on supply)
Outline
The Area Function
FTC1StatementProofBiographies
Differentiation of functions defined by integrals“Contrived” examplesErfOther applications
FTC2
Facts about g from fA problem
Theorem (The Second Fundamental Theorem of Calculus,Weak Form)
If f is continuous on [a, b] and f = F ′ for another function F , then∫ b
af (t) dt = F (b)− F (a).
Proof.Let g be the area function. Since f is continuous on [a, b], g isdifferentiable on (a, b), and g ′ = f = F ′ on (a, b). Henceg(x) = F (x) + C for all x in [a, b] (remember this requires theMean Value Theorem!). Since g(a) = 0, we have C = −F (a).Therefore
g(b) = F (b)− F (a).
Theorem (The Second Fundamental Theorem of Calculus,Weak Form)
If f is continuous on [a, b] and f = F ′ for another function F , then∫ b
af (t) dt = F (b)− F (a).
Proof.Let g be the area function. Since f is continuous on [a, b], g isdifferentiable on (a, b), and g ′ = f = F ′ on (a, b). Henceg(x) = F (x) + C for all x in [a, b] (remember this requires theMean Value Theorem!). Since g(a) = 0, we have C = −F (a).Therefore
g(b) = F (b)− F (a).
Outline
The Area Function
FTC1StatementProofBiographies
Differentiation of functions defined by integrals“Contrived” examplesErfOther applications
FTC2
Facts about g from fA problem
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
What is the particle’s velocityat time t = 5?
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
What is the particle’s velocityat time t = 5?
SolutionRecall that by the FTC wehave
s ′(t) = f (t).
So s ′(5) = f (5) = 2.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
Is the acceleration of the par-ticle at time t = 5 positive ornegative?
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
Is the acceleration of the par-ticle at time t = 5 positive ornegative?
SolutionWe have s ′′(5) = f ′(5), whichlooks negative from thegraph.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
What is the particle’s positionat time t = 3?
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
What is the particle’s positionat time t = 3?
SolutionSince on [0, 3], f (x) = x, wehave
s(3) =
∫ 3
0x dx =
9
2.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
At what time during the first 9seconds does s have its largestvalue?
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
At what time during the first 9seconds does s have its largestvalue?
Solution
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
At what time during the first 9seconds does s have its largestvalue?
SolutionThe critical points of s arethe zeros of s ′ = f .
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
At what time during the first 9seconds does s have its largestvalue?
SolutionBy looking at the graph, wesee that f is positive fromt = 0 to t = 6, then negativefrom t = 6 to t = 9.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
At what time during the first 9seconds does s have its largestvalue?
SolutionTherefore s is increasing on[0, 6], then decreasing on[6, 9]. So its largest value isat t = 6.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
Approximately when is the ac-celeration zero?
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
Approximately when is the ac-celeration zero?
Solutions ′′ = 0 when f ′ = 0, whichhappens at t = 4 and t = 7.5(approximately)
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
When is the particle movingtoward the origin? Away fromthe origin?
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
When is the particle movingtoward the origin? Away fromthe origin?
SolutionThe particle is moving awayfrom the origin when s > 0and s ′ > 0.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
When is the particle movingtoward the origin? Away fromthe origin?
SolutionSince s(0) = 0 and s ′ > 0 on(0, 6), we know the particle ismoving away from the originthen.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
When is the particle movingtoward the origin? Away fromthe origin?
SolutionAfter t = 6, s ′ < 0, so theparticle is moving toward theorigin.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
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 or neg-ative) of the origin does theparticle lie at time t = 9?
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
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 or neg-ative) of the origin does theparticle lie at time t = 9?
SolutionWe have s(9) =∫ 6
0f (x) dx +
∫ 9
6f (x) dx,
where the left integral ispositive and the right integralis negative.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
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 or neg-ative) of the origin does theparticle lie at time t = 9?
SolutionIn order to decide whethers(9) is positive or negative,we need to decide if the firstarea is more positive than thesecond area is negative.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
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 or neg-ative) of the origin does theparticle lie at time t = 9?
SolutionThis appears to be the case,so s(9) is positive.
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