We notice that the numerator of the fraction is nearly the derivative of the

denominator. Put in other words, if we were asked to integrate

2x

x2 + 3

102 CHAPTER 4 The Integral

then we would see that we are integrating an expression of the form

• (x)

•(x)

(which we in fact encountered among our differentiation rules in Section 2.5).

As we know, expressions like this arise from differentiating log •(x).

Returning to the original problem, we pose our initial guess as log[x 2 + 3].

Differentiation of this expression gives the answer 2x/[x 2 + 3]. This is close

to what we want, but we must adjust by a factor of 1/2. We write our ¬nal

answer as

1

x

dx = log[x 2 + 3] + C.

x2 + 3 2

You Try It: Calculate the inde¬nite integral

2 +5

xe3x dx.

EXAMPLE 4.3

Calculate the inde¬nite integral

(x 3 + x 2 + 1)50 · (6x 2 + 4x) dx.

SOLUTION

We observe that the expression 6x 2 + 4x is nearly the derivative of x 3 +

x 2 + 1. In fact if we set •(x) = x 3 + x 2 + 1 then the integrand (the quantity

that we are asked to integrate) is

[•(x)]50 · 2• (x).

It is natural to guess as our antiderivative [•(x)]51 . Checking our work, we

¬nd that

([•(x)]51 ) = 51[•(x)]50 · • (x).

We see that the answer obtained is quite close to the answer we seek; it is off by

a numerical factor of 2/51. With this knowledge, we write our ¬nal answer as

2

(x 3 + x 2 + 1)50 · (6x 2 + 4x) dx = · [x 3 + x 2 + 1]51 + C.

51

You Try It: Calculate the inde¬nite integral

x2

dx.

x3 + 5

103

CHAPTER 4 The Integral

4.2 Area

Consider the curve shown in Fig. 4.1. The curve is the graph of y = f (x). We set

for ourselves the task of calculating the area A that is (i) under the curve, (ii) above

the x-axis, and (iii) between x = a and x = b. Refer to Fig. 4.2 to see the geometric

region we are considering.

y = f (x)

Fig. 4.1

y = f (x)

a b

Fig. 4.2

We take it for granted that the area of a rectangle of length and width w is

— w. Now our strategy is to divide the base interval [a, b] into equal subintervals.

Fix an integer k > 0. We designate the points

P = {x0 , x1 , x2 , . . . , xk },

with x0 = a and xk = b. We require that |xj ’ xj ’1 | = |b ’ a|/k ≡ x for

j = 1, . . . , k. In other words, the points x0 , x1 , . . . , xk are equally spaced. We call

104 CHAPTER 4 The Integral

the set P a partition. Sometimes, to be more speci¬c, we call it a uniform partition

(to indicate that all the subintervals have the same length). Refer to Fig. 4.3.

b_a

k

x0 = a xj xk = b

Fig. 4.3

The idea is to build an approximation to the area A by erecting rectangles over

the segments determined by the partition. The ¬rst rectangle R1 will have as base

the interval [x0 , x1 ] and height chosen so that the rectangle touches the curve at its

upper right hand corner; this means that the height of the rectangle is f (x1 ). The

second rectangle R2 has as base the interval [x1 , x2 ] and height f (x2 ). Refer to

Fig. 4.4.

y = f (x)

x0 = a x1 x2 xk = b

Fig. 4.4

Continuing in this manner, we construct precisely k rectangles, R1 , R2 , . . . , Rk ,

as shown in Fig. 4.5. Now the sum of the areas of these rectangles is not exactly

equal to the area A that we seek. But it is close. The error is the sum of the little

semi-triangular pieces that are shaded in Fig. 4.6. We can make that error as small

as we please by making the partition ¬ner. Figure 4.7 illustrates this idea.

Let us denote by R(f, P) the sum of the areas of the rectangles that we created

from the partition P. This is called a Riemann sum. Thus

k

R(f, P) = f (xj ) · x ≡ f (x1 ) · x + f (x2 ) · x + · · · + f (xk ) · x.

j =1

Here the symbol k =1 denotes the sum of the expression to its right for each of

j

the instances j = 1 to j = k.

105

CHAPTER 4 The Integral

y = f (x)

x0 = a x1 x2 xk = b

Fig. 4.5

y = f (x)

x0 = a x1 x2 xk = b

Fig. 4.6

y = f (x)

a b

Fig. 4.7

The reasoning just presented suggests that the true area A is given by

lim R(f, P).

k’∞

We call this limit the integral of f from x = a to x = b and we write it as

b

f (x) dx.

a

106 CHAPTER 4 The Integral

Thus we have learned that

b

area A = f (x) dx.

a

It is well to take a moment and comment on the integral notation. First, the

integral sign

is an elongated “S,” coming from “summation.” The dx is an historical artifact,

coming partly from traditional methods of developing the integral, and partly from

a need to know explicitly what the variable is. The numbers a and b are called the

limits of integration”the number a is the lower limit and b is the upper limit. The

function f is called the integrand.

Before we can present a detailed example, we need to record some important

information about sums:

N

I. We need to calculate the sum S = 1 + 2 + · · · + N = j =1 j . To achieve this

goal, we write

S =1+2 + · · · + (N ’ 1) + N

S = N + (N ’ 1) + · · · + 2 +1

Adding each column, we obtain

2S = (N + 1) + (N + 1) + · · · + (N + 1) + (N + 1) .

N times

Thus

2S = N · (N + 1)

or

N · (N + 1)

S= .

2

This is a famous formula that was discovered by Carl Friedrich Gauss (1777“1855)

when he was a child.

n

II. The sum S = 12 + 22 + · · · + N 2 = 2 is given by

j =1 j

2N 3 + 3N 2 + N

S= .

6

We shall not provide the details of the proof of this formula, but refer the interested

reader to [SCH2].

107

CHAPTER 4 The Integral

For our ¬rst example, we calculate the area under a parabola.

EXAMPLE 4.4

Calculate the area under the curve y = x 2 , above the x-axis, and between

x = 0 and x = 2.

SOLUTION

Refer to Fig. 4.8 as we reason along. Let f (x) = x 2 .

Fig. 4.8

Consider the partition P of the interval [0, 2] consisting of k + 1 points

x0 , x1 , . . . , xk . The corresponding Riemann sum is

k

R(f, P) = f (xj ) · x.

j =1

Of course

2’0 2

x= =

k k

and

2

xj = j · .

k

In addition,

2

4j 2

2

f (xj ) = j · = 2.

k k

108 CHAPTER 4 The Integral

As a result, the Riemann sum for the partition P is

k

4j 2 2

R(f, P) = ·

k2 k

j =1

k k

8j 2 8

= =3 j 2.

k3 k

j =1 j =1

Now formula II above enables us to calculate the last sum explicitly. The result

is that

8 2k 3 + 3k 2 + k

R(f, P) = 3 ·

6

k

84 4

= + + 2.

3 k 3k

In sum,

Y

2 84 4 8

x 2 dx = lim R(f, P) = lim ++ 2=.

FL

3 k 3k 3

k’∞ k’∞

0

We conclude that the desired area is 8/3.

AM

You Try It: Use the method presented in the last example to calculate the area

under the graph of y = 2x and above the x-axis, between x = 1 and x = 2. You

should obtain the answer 3, which of course can also be determined by elementary

TE

considerations”without taking limits.

The most important idea in all of calculus is that it is possible to calculate an

integral without calculating Riemann sums and passing to the limit. This is the

Fundamental Theorem of Calculus, due to Leibniz and Newton. We now state the

theorem, illustrate it with examples, and then brie¬‚y discuss why it is true.

Theorem 4.1 (Fundamental Theorem of Calculus)

Let f be a continuous function on the interval [a, b]. If F is any antiderivative of

f then

b

f (x) dx = F (b) ’ F (a).

a

EXAMPLE 4.5

Calculate

2

x 2 dx.

0

109

CHAPTER 4 The Integral

SOLUTION

We use the Fundamental Theorem. In this example, f (x) = x 2 . We need to

¬nd an antiderivative F . From our experience in Section 4.1, we can determine

that F (x) = x 3 /3 will do. Then, by the Fundamental Theorem of Calculus,

2 23 03 8

x dx = F (2) ’ F (0) = ’ =.

2

3 3 3

0

Notice that this is the same answer that we obtained using Riemann sums in

Example 4.4.

EXAMPLE 4.6

Calculate

π

sin x dx.

0

SOLUTION

In this example, f (x) = sin x. An antiderivative for f is F (x) = ’ cos x.

Then

π

sin x dx = F (π) ’ F (0) = (’ cos π ) ’ (’ cos 0) = 1 + 1 = 2.

0

EXAMPLE 4.7

Calculate

2

ex ’ cos 2x + x 3 ’ 4x dx.

1

SOLUTION

In this example, f (x) = ex ’ cos 2x + x 3 ’ 4x. An antiderivative for f is

F (x) = ex ’ (1/2) sin 2x + x 4 /4 ’ 2x 2 . Therefore

2

ex ’ cos 2x + x 3 ’ 4x dx = F (2) ’ F (1)

1

24

1

= e ’ sin(2 · 2) + ’ 2 · 22

2

2 4

14

1

’ e ’ sin(2 · 1) + ’ 2 · 12

1

2 4

1 9

= (e2 ’ e) ’ [sin 4 ’ sin 2] ’ .

2 4

110 CHAPTER 4 The Integral

You Try It: Calculate the integral

’1

x 3 ’ cos x + x dx.

’3

Math Note: Observe in this last example, in fact in all of our examples, you can

use any antiderivative of the integrand when you apply the Fundamental Theorem

of Calculus. In the last example, we could have taken F (x) = ex ’ (1/2) sin 2x +

x 4 /4 ’ 2x 2 + 5 and the same answer would have resulted. We invite you to provide

the details of this assertion.

Justi¬cation for the Fundamental Theorem Let f be a continuous function on

the interval [a, b]. De¬ne the area function F by

F (x) = area under f , above the x-axis, and between 0 and x.

Fig. 4.9

Let us use a pictorial method to calculate the derivative of F . Refer to Fig. 4.9

as you read on. Now

F (x + h) ’ F (x) [area between x and x + h, below f ]

=

h h

f (x) · h

≈

h

= f (x).

As h ’ 0, the approximation in the last display becomes nearer and nearer to

equality. So we ¬nd that

F (x + h) ’ F (x)

= f (x).

lim

h

h’0

But this just says that F (x) = f (x).

What is the practical signi¬cance of this calculation? Suppose that we wish to

calculate the area under the curve f , above the x-axis, and between x = a and

x = b. Obviously this area is F (b) ’ F (a). See Fig. 4.10. But we also know that

111