INTEGRALS

5

5.3

The 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 1 FTC

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

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( ) ( )

x

g 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 1 FTC

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 1 FTC

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 1 FTC

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

So, we start subtracting areas, as

follows.

Example 1 FTC

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 1 FTC

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, 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

h

f 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

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

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

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

Proof FTC1

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].

Proof FTC1

By Property 8 of integrals, we have:

That is,

( )

x h

xmh f t dt Mh

Proof FTC1

( ) ( ) ( )

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 3 FTC1

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

Proof FTC1

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

If x = a or b, then Equation 4 can be

interpreted as a one-sided limit.

Then, Theorem 4 in Section 2.8 (modified

for one-sided limits) shows that g is continuous

on [a, b].

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

Proof—Equation 5 FTC1

Find the derivative of the function

As is continuous, the FTC1 gives:

Example 2

2

0( ) 1

x

g 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)

x

S 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( ) ( )

x

S x f t dt

Example 3 FRESNEL 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 3 FRESNEL FUNCTION

The other figure shows a larger part

of the graph of S.

Example 3 FRESNEL 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 3 FRESNEL 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 4 FTC1

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 dx

d dusec t dt

du dx

duu

dx

x x

Example 4 FTC1

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

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

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’s 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:

Other common notations are and .

( )] ( ) ( )b

aF x F b F a

( ) ( )] where 'b

b

aa

f x dx F x F f

( ) |baF x [ ( )]b

aF 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 3

12

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

The given integral is an abbreviation for .

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

dx

x6

3

1dx

x

FTC2

Therefore, 6

6

33

1ln ]

ln 6 ln 3

6ln

3

ln 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 sin0 sin

bbA xdx 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?

31

3

211

1 1 41

1 3 3

x

dxx

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 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

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’s 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.