INTEGRALSINTEGRALS

5

5.3The Fundamental

Theorem of Calculus

INTEGRALS

In this section, we will learn about:

The Fundamental Theorem of Calculus

and its significance.

The Fundamental Theorem of Calculus

(FTC) is appropriately named.

It establishes a connection between the two branches of calculus—differential calculus and integral calculus.

FUNDAMENTAL THEOREM OF CALCULUS

FTC

Differential calculus arose from the tangent

problem.

Integral calculus arose from a seemingly

unrelated problem—the area problem.

Newton’s mentor at Cambridge, Isaac Barrow

(1630–1677), discovered that these two

problems are actually closely related.

In fact, he realized that differentiation and integration are inverse processes.

FTC

The FTC gives the precise inverse

relationship between the derivative

and the integral.

FTC

It was Newton and Leibniz who exploited this

relationship and used it to develop calculus

into a systematic mathematical method.

In particular, they saw that the FTC enabled them to compute areas and integrals very easily without having to compute them as limits of sums—as we did in Sections 5.1 and 5.2

FTC

The first part of the FTC deals with functions

defined by an equation of the form

where f is a continuous function on [a, b]

and x varies between a and b.

( ) ( )x

ag x f t dt

Equation 1FTC

Observe that g depends only on x, which appears as the variable upper limit in the integral.

If x is a fixed number, then the integral is a definite number.

If we then let x vary, the number also varies and defines a function of x denoted by g(x).

( ) ( )x

ag x f t dt

( )x

af t dt

( )x

af t dt

FTC

If f happens to be a positive function, then g(x)

can be interpreted as the area under the

graph of f from a to x, where x can vary from a

to b.

Think of g as the ‘area so far’ function, as seen here.

FTC

Figure 5.3.1, p. 314

If f is the function

whose graph is shown

and ,

find the values of:

g(0), g(1), g(2), g(3),

g(4), and g(5).

Then, sketch a rough graph of g.

Example 1

0( ) ( )

xg x f t dt

FTC

Figure 5.3.2, p. 314

First, we notice that:

0

0(0) ( ) 0g f t dt

FTC Example 1

From the figure, we see that g(1) is

the area of a triangle:

1

0

12

(1) ( )

(1 2)

1

g f t dt

Example 1FTC

Figure 5.3.3a, p. 314

To find g(2), we add to g(1) the area of

a rectangle:2

0

1 2

0 1

(2) ( )

( ) ( )

1 (1 2)

3

g f t dt

f t dt f t dt

Example 1FTC

Figure 5.3.3b, p. 314

We estimate that the area under f from 2 to 3

is about 1.3.

So, 3

2(3) (2) ( )

3 1.3

4.3

g g f t dt

Example 1FTC

Figure 5.3.3c, p. 314

Thus,4

3(4) (3) ( ) 4.3 ( 1.3) 3.0g g f t dt

FTC Example 1

5

4(5) (4) ( ) 3 ( 1.3) 1.7g g f t dt

Figure 5.3.3d, p. 314 Figure 5.3.3e, p. 314

We use these values to sketch the graph

of g. Notice that, because f(t)

is positive for t < 3, we keep adding area for t < 3.

So, g is increasing up to x = 3, where it attains a maximum value.

For x > 3, g decreases because f(t) is negative.

Example 1FTC

Figure 5.3.4, p. 314

If we take f(t) = t and a = 0, then,

using Exercise 27 in Section 5.2,

we have:2

0( )

2

x xg x t dt

FTC

Notice that g’(x) = x, that is, g’ = f.

In other words, if g is defined as the integral of f by Equation 1, g turns out to be an antiderivative of f—at least in this case.

FTC

If we sketch the derivative

of the function g, as in the

first figure, by estimating

slopes of tangents, we get

a graph like that of f in the

second figure.

So, we suspect that g’ = f in Example 1 too.

FTC

Figure 5.3.4, p. 314

Figure 5.3.2, p. 314

To see why this might be generally true, we

consider a continuous function f with f(x) ≥ 0.

Then, can be interpreted as the area under the graph of f from a to x.

( ) ( )x

ag x f t dt

FTC

Figure 5.3.1, p. 314

To compute g’(x) from the definition of

derivative, we first observe that, for h > 0,

g(x + h) – g(x) is obtained by subtracting

areas.

It is the area under the graph of f from x to x + h (the gold area).

FTC

Figure 5.3.5, p. 315

For small h, you can see that this area is

approximately equal to the area of the

rectangle with height f(x) and width h:

So,

FTC

( ) ( ) ( )g x h g x hf x

( ) ( )

( )

g x h g x

hf x

Figure 5.3.5, p. 315

Intuitively, we therefore expect that:

The fact that this is true, even when f is not necessarily positive, is the first part of the FTC (FTC1).

0

( ) ( )'( ) lim ( )

h

g x h g xg x f x

h

FTC

FTC1

If f is continuous on [a, b], then the function g

defined by

is continuous on [a, b] and differentiable on

(a, b), and g’(x) = f(x).

( ) ( )x

ag x f t dt a x b

If x and x + h are in (a, b), then

( ) ( )

( ) ( )

( ) ( ) ( ) (Property5)

( )

x h x

a a

x x h x

a x a

x h

x

g x h g x

f t dt f t dt

f t dt f t dt f t dt

f t dt

ProofFTC1

So, for h ≠ 0,

Proof—Equation 2

( ) ( ) 1( )

x h

x

g x h g xf t dt

h h

FTC1

For now, let us assume that h > 0.

Since f is continuous on [x, x + h], the Extreme Value Theorem says that there are numbers u and v in [x, x + h] such that f(u) = m and f(v) = M.

m and M are the absolute minimum and maximum values of f on [x, x + h].

ProofFTC1

Figure 5.3.6, p. 316

By Property 8 of integrals, we have:

That is,

( )

x h

xmh f t dt Mh

ProofFTC1

( ) ( ) ( )

x h

xf u h f t dt f v h

Since h > 0, we can divide this inequality

by h:

1( ) ( ) ( )

x h

xf u f t dt f v

h

FTC1 Proof

Now, we use Equation 2 to replace the middle

part of this inequality:

Inequality 3 can be proved in a similar manner for the case h < 0.

( ) ( )( ) ( )

g x h g xf u f v

h

Proof—Equation 3FTC1

Now, we let h → 0.

Then, u → x and v → x, since u and v lie

between x and x + h.

Therefore,

and

because f is continuous at x.

0lim ( ) lim ( ) ( )h u x

f u f u f x

ProofFTC1

0lim ( ) lim ( ) ( )h v x

f v f v f x

From Equation 3 and the Squeeze

Theorem, we conclude that:

Proof—Equation 4

0

( ) ( )'( ) lim ( )

h

g x h g xg x f x

h

FTC1

Using Leibniz notation for derivatives, we can

write the FTC1 as

when f is continuous.

Roughly speaking, Equation 5 says that, if we first integrate f and then differentiate the result, we get back to the original function f.

( ) ( )x

a

df t dt f x

dx

Equation 5FTC1

Find the derivative of the function

As is continuous, the FTC1 gives:

Example 2

2

0( ) 1

xg x t dt

2( ) 1f t t 2'( ) 1g x x

FTC1

FRESNEL FUNCTION

For instance, consider the Fresnel function

It is named after the French physicist Augustin Fresnel (1788–1827), famous for his works in optics.

It first appeared in Fresnel’s theory of the diffraction of light waves.

More recently, it has been applied to the design of highways.

2

0( ) sin( / 2)

xS x t dt

Example 3

FRESNEL FUNCTION

The FTC1 tells us how to differentiate

the Fresnel function:

S’(x) = sin(πx2/2)

This means that we can apply all the methods of differential calculus to analyze S.

Example 3

The figure shows the graphs of

f(x) = sin(πx2/2) and the Fresnel function

A computer was used to graph S by computing the value of this integral for many values of x.

0( ) ( )

xS x f t dt

Example 3FRESNEL FUNCTION

Figure 5.3.7, p. 317

It does indeed look as if S(x) is the area

under the graph of f from 0 to x (until x ≈ 1.4,

when S(x) becomes a difference of areas).

Example 3FRESNEL FUNCTION

Figure 5.3.7, p. 317

The other figure shows a larger part

of the graph of S.

Example 3FRESNEL FUNCTION

Figure 5.3.7, p. 317Figure 5.3.8, p. 317

If we now start with the graph of S here and

think about what its derivative should look like,

it seems reasonable that S’(x) = f(x).

For instance, S is increasing when f(x) > 0 and decreasing when f(x) < 0.

Example 3FRESNEL FUNCTION

Figure 5.3.7, p. 317

Find

Here, we have to be careful to use the Chain Rule in conjunction with the FTC1.

4

1sec

xdt dt

dx

Example 4FTC1

Let u = x4.

Then,

4

1 1

1

4 3

sec sec

(Chain Rule)

sec (FTC1)

sec( ) 4

x u

u

d dt dt t dt

dx dxd du

sec t dtdu dx

duudx

x x

Example 4FTC1

FTC2

If f is continuous on [a, b], then

where F is any antiderivative of f,

that is, a function such that F’ = f.

( ) ( ) ( )b

af x dx F b F a

FTC2

Let

We know from the FTC1 that g’(x) = f(x),

that is, g is an antiderivative of f.

( ) ( )x

ag x f t dt

Proof

FTC2

If F is any other antiderivative of f on [a, b],

then we know from Corollary 7 in Section 4.2

that F and g differ by a constant:

F(x) = g(x) + C

for a < x < b.

Proof—Equation 6

FTC2

However, both F and g are continuous on

[a, b].

Thus, by taking limits of both sides of

Equation 6 (as x → a+ and x → b- ),

we see it also holds when x = a and x = b.

Proof

FTC2

If we put x = a in the formula for g(x),

we get:

Proof

( ) ( ) 0a

ag a f t dt

FTC2

So, using Equation 6 with x = b and x = a,

we have:

( ) ( ) [ ( ) ] [ ( ) ]

( ) ( )

( )

( )b

a

F b F a g b C g a C

g b g a

g b

f t dt

Proof

FTC2

Evaluate the integral

The function f(x) = x3 is continuous on [-2, 1] and we know from Section 4.9 that an antiderivative is F(x) = ¼x4.

So, the FTC2 gives:

Example 51 3

2 x dx

1 3

2

4 41 14 4

154

(1) ( 2)

1 2

x dx F F

FTC2

Notice that the FTC2 says that we can use any antiderivative F of f.

So, we may as well use the simplest one, namely F(x) = ¼x4, instead of ¼x4 + 7 or ¼x4 + C.

Example 5

FTC2

We often use the notation

So, the equation of the FTC2 can be written

as:

Other common notations are and .

( )] ( ) ( )baF x F b F a

( ) ( )] where 'b b

aaf x dx F x F f

( ) |baF x [ ( )]baF x

FTC2

Find the area under the parabola y = x2

from 0 to 1.

An antiderivative of f(x) = x2 is F(x) = (1/3)x3. The required area is found using the FTC2:

Example 6

13 3 31 2

00

1 0 1

3 3 3 3

xA x dx

FTC2

Find the area under the cosine curve

from 0 to b, where 0 ≤ b ≤ π/2.

Since an antiderivative of f(x) = cos x is F(x) = sin x, we have:

Example 7

00cos sin

sin sin 0

sin

b b

A x dx x

b

b

FTC2

In particular, taking b = π/2, we have

proved that the area under the cosine curve

from 0 to π/2 is sin(π/2) =1.

Example 7

Figure 5.3.9, p. 319

FTC2

What is wrong with this calculation?

313

211

1 1 41

1 3 3

x

dxx

Example 8

FTC2

To start, we notice that the calculation must

be wrong because the answer is negative

but f(x) = 1/x2 ≥ 0 and Property 6 of integrals

says that when f ≥ 0.( ) 0b

af x dx

Example 9

FTC2

The FTC applies to continuous functions.

It can’t be applied here because f(x) = 1/x2

is not continuous on [-1, 3].

In fact, f has an infinite discontinuity at x = 0.

So, does not exist.3

21

1dx

x

Example 9

FTC

Suppose f is continuous on [a, b].

1.If , then g’(x) = f(x).

2. , where F is

any antiderivative of f, that is, F’ = f.

( ) ( )x

ag x f t dt

( ) ( ) ( )b

af x dx F b F a

INVERSE PROCESSES

We noted that the FTC1 can be rewritten

as:

This says that, if f is integrated and then the result is differentiated, we arrive back at the original function f.

( ) ( )x

a

df t dt f x

dx

INVERSE PROCESSES

As F’(x) = f(x), the FTC2 can be rewritten

as:

This version says that, if we take a function F, first differentiate it, and then integrate the result, we arrive back at the original function F.

However, it’s in the form F(b) - F(a).

'( ) ( ) ( )b

aF x dx F b F a

SUMMARY

Before it was discovered—from the time

of Eudoxus and Archimedes to that of Galileo

and Fermat—problems of finding areas,

volumes, and lengths of curves were so

difficult that only a genius could meet

the challenge.

SUMMARY

Now, armed with the systematic method

that Newton and Leibniz fashioned out of

the theorem, we will see in the chapters to

come that these challenging problems are

accessible to all of us.