Lesson 32: The Fundamental Theorem Of Calculus

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


g(x) =

∫ t

af (t) dt.




The area function


g(x) =

∫ t

af (t) dt.




Outline

The Area Function



FTC2


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.


g(x + h)− g(x)

h=

1

h

∫ x+h

xf (t) dt.


mh · h ≤

∫ x+h

xf (t) dt

≤ Mh · h

So

mh ≤g(x + h)− g(x)

h≤ Mh.



g(x + h)− g(x)

h=

1

h

∫ x+h

xf (t) dt.


mh · h ≤

∫ x+h

xf (t) dt

≤ Mh · h

So

mh ≤g(x + h)− g(x)

h≤ Mh.



g(x + h)− g(x)

h=

1

h

∫ x+h

xf (t) dt.


mh · h ≤

∫ x+h

xf (t) dt ≤ Mh · h

So

mh ≤g(x + h)− g(x)

h≤ Mh.



g(x + h)− g(x)

h=

1

h

∫ x+h

xf (t) dt.


mh · h ≤∫ x+h


So

mh ≤g(x + h)− g(x)

h≤ Mh.



g(x + h)− g(x)

h=

1

h

∫ x+h

xf (t) dt.


mh · h ≤∫ x+h


So

mh ≤g(x + h)− g(x)

h≤ Mh.



g(x + h)− g(x)

h=

1

h

∫ x+h

xf (t) dt.


mh · h ≤∫ x+h


So

mh ≤g(x + h)− g(x)

h≤ Mh.


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



FTC2


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.


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).


d


d

dxx2 =

2√2π

e−(x2)22x =4√π

xe−x4.


erf(x) =2√π

∫ x

0e−t2

dt.


erf ′(x) =

2√π

e−x2.

Example

Findd

dxerf(x2).


d


d

dxx2 =

2√2π

e−(x2)22x =4√π

xe−x4.


erf(x) =2√π

∫ x

0e−t2

dt.


erf ′(x) =2√π

e−x2.

Example

Findd

dxerf(x2).


d


d

dxx2 =

2√2π

e−(x2)22x =4√π

xe−x4.


erf(x) =2√π

∫ x

0e−t2

dt.


erf ′(x) =2√π

e−x2.

Example

Findd

dxerf(x2).


d


d

dxx2 =

2√2π

e−(x2)22x =4√π

xe−x4.


erf(x) =2√π

∫ x

0e−t2

dt.


erf ′(x) =2√π

e−x2.

Example

Findd

dxerf(x2).


d


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



FTC2


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



FTC2


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)




∫ t



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?




∫ t



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.




∫ t



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?




∫ t



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.




∫ t



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?




∫ t



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.




∫ t



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?




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


Solution




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


SolutionThe critical points of s arethe zeros of s ′ = f .




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


SolutionBy looking at the graph, wesee that f is positive fromt = 0 to t = 6, then negativefrom t = 6 to t = 9.




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


SolutionTherefore s is increasing on[0, 6], then decreasing on[6, 9]. So its largest value isat t = 6.




∫ t



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?




∫ t



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)




∫ t



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?




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


SolutionThe particle is moving awayfrom the origin when s > 0and s ′ > 0.




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


SolutionSince s(0) = 0 and s ′ > 0 on(0, 6), we know the particle ismoving away from the originthen.




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


SolutionAfter t = 6, s ′ < 0, so theparticle is moving toward theorigin.




∫ t



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?




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


SolutionWe have s(9) =∫ 6

0f (x) dx +

∫ 9

6f (x) dx,

where the left integral ispositive and the right integralis negative.




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


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.




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


SolutionThis appears to be the case,so s(9) is positive.

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Lesson 32: The Fundamental Theorem Of Calculus