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).

FTC

( ) ( )x

ag x f t dt

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

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

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

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

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

For t > 3, f(t) is negative.

So, we start subtracting areas, as follows.

Example 1FTC

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

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

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 tdt

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

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

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

For small h, we 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

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

Fundamental Theorem of Calculus Version 1

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

In words, the FTC1 says that the derivative of a

definite integral with respect to its upper limit is

the integrand evaluated at the upper limit.

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

FTC1

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

A formula of the form

may seem like a strange way of defining a

function.

However, books on physics, chemistry, and statistics are full of such functions.

( ) ( )x

ag x f t dt

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:

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

Example 3

2( ) sin( / 2)S x t

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

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

The other figure shows a larger part of the graph

of S.

Example 3FRESNEL FUNCTION

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

FRESNEL FUNCTION

So, this gives a visual confirmation of the FTC1.

Example 3

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

duu

dx

x x

Example 4FTC1

In Section 5.2, we computed integrals from the

definition as a limit of Riemann sums and saw that

this procedure is sometimes long and difficult.

The second part of the FTC (FTC2), which follows easily from the first part, provides us with a much simpler method for the evaluation of integrals.

FTC1

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

Fundamental Theorem of Calculus Version 2

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

The FTC2 states that, if we know an antiderivative

F of f, then we can evaluate

simply by subtracting the values of F at the

endpoints of the interval [a, b].

( )b

af x dx

FTC2

It is very surprising that , which

was defined by a complicated procedure

involving all the values of f(x) for a ≤ x ≤ b,

can be found by knowing the values of F(x) at

only two points, a and b.

( )b

af x dx

FTC2

At first glance, the theorem may be surprising.

However, it becomes plausible if we interpret it in physical terms.

FTC2

If v(t) is the velocity of an object and s(t) is its

position at time t, then v(t) = s’(t).

So, s is an antiderivative of v.

FTC2

In Section 5.1, we considered an object that

always moves in the positive direction.

Then, we guessed that the area under the velocity

curve equals the distance traveled.

In symbols,

That is exactly what the FTC2 says in this context.

( ) ( ) ( )b

av t dt s b s a

FTC2

Evaluate the integral

The function f(x) = ex is continuous everywhere and we know that an antiderivative is F(x) = ex.

So, the FTC2 gives:

Example 5

3

1

xe dx

3

1

3

(3) (1)xe dx F F

e e

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) = ex, instead of ex + 7 or ex + C.

Example 5

FTC2

We often use the notation

So, the equation of the FTC2 can be written as:

( ) | ( ) ( )baF x F b F a

( ) ( ) | where 'b

baa

f x dx F x F f

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 3 312

00

1 0 1

3 3 3 3

xA x dx

FTC2

If you compare the calculation in Example 6 with

the one in Example 2 in Section 5.1, you will see

the FTC gives a much shorter method.

FTC2

Evaluate

An antiderivative of f(x) = 1/x is F(x) = ln |x|.

As 3 ≤ x ≤ 6, we can write F(x) = ln x.

Example 7

6

3

1dx

x

FTC2

Therefore, 6

633

1ln ]

ln 6 ln 3

6ln

3ln 2

dx xx

Example 7

FTC2

Find the area under the cosine curve from 0 to b,

where 0 ≤ b ≤ π/2.

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

Example 8

00cos sin ] sin sin 0 sin

bbA 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 8

FTC2

When the French mathematician Gilles de

Roberval first found the area under the sine and

cosine curves in 1635, this was a very challenging

problem that required a great deal of ingenuity.

FTC2

If we didn’t have the benefit of the FTC, we would

have to compute a difficult limit of sums using

either:

Obscure trigonometric identities

A computer algebra system (CAS), as in Section 5.1

FTC2

It was even more difficult for Roberval.

The apparatus of limits had not been invented in 1635.

FTC2

However, in the 1660s and 1670s, when the FTC

was discovered by Barrow and exploited by

Newton and Leibniz, such problems became very

easy.

You can see this from Example 8.

FTC2

What is wrong with this calculation?

313

211

1 1 41

1 3 3

xdx

x

Example 9

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 cannot 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

INVERSE PROCESSES

We end this section by bringing together the two

parts of the FTC.

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 is in the form F(b) - F(a).

'( ) ( ) ( )b

aF x dx F b F a

INVERSE PROCESSES

Taken together, the two parts of the FTC say that

differentiation and integration are inverse

processes.

Each undoes what the other does.

SUMMARY

The FTC is unquestionably the most important

theorem in calculus.

Indeed, it ranks as one of the great accomplishments of the human mind.

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.