Section 5.4 The 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, Sun 1/13 in Hall C, all 7–8:30pm I Final tentatively scheduled for January 17, 9:15am
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
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
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.