406 Chapter 6
Integration
Section 6-4
The Definite Integral
➤
➤
➤
Approximating Areas by Left and Right Sums
The Definite Integral as a Limit of Sums
Properties of the Definite Integral
The first three sections of this chapter focused on the indefinite integral. In this
section we introduce the definite integral. The definite integral is used to compute areas, probabilities, average values of functions, future values of continuous income streams, and many other quantities. Initially, the concept of the
definite integral may appear to be unrelated to the notion of the indefinite integral. There is, however, a close connection between the two integrals. The
fundamental theorem of calculus, discussed in Section 6-5, makes that connection precise.
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The Definite Integral
Section 6-4
407
➤ Approximating Areas by Left and Right Sums
How do we find the shaded area in Figure 1? That is, how do we find the area
bounded by the graph of f(x) = 0.25x2 + 1, the x axis, and the vertical lines
x = 1 and x = 5? [This cumbersome description is usually shortened to “the
area under the graph of f(x) = 0.25x2 + 1 from x = 1 to x = 5.”] Our standard geometric area formulas do not apply directly, but the formula for the
area of a rectangle can be used indirectly. To see how, we look at a method of
approximating the area under the graph by using rectangles. This method will
give us any accuracy desired, which is quite different from finding the area exactly. Our first area approximation is made by dividing the interval [1, 5] on
the x axis into four equal parts of length
f (x)
10
f (x) 0.25x2 1
5
0
1
5
¢x =
x
FIGURE 1 What is the shaded area?
5 -1
=1
4
(It is customary to denote the length of the subintervals by ¢x, which is read
“delta x,” since ¢ is the Greek capital letter delta.) We then place a rectangle
on each subinterval with a height determined by the function evaluated at the
left endpoint of the subinterval (see Fig. 2).
Summing the areas of the rectangles in Figure 2, we obtain a left sum of
four rectangles, denoted by L4, as follows:
L4 = f(1) 1 + f(2) 1 + f(3) 1 + f(4) 1
= 1.25 + 2.00 + 3.25 + 5 = 11.5
From Figure 3, since f(x) is increasing, it is clear that the left sum L4 underestimates the area, and we can write
11.5 = L4 6 Area
f (x)
f (x)
10
10
f (x) 0.25x2 1
5
5
0
1
2
3
4
5
FIGURE 2 Left rectangles
1
f (x) 0.25x2 1
x
0
1
2
3
4
5
x
FIGURE 3 Left and right
rectangles
If f(x) were decreasing over the interval [1, 5], would the left sum L4 overor underestimate the actual area under the curve? Explain.
Now suppose that we use the right endpoint of each subinterval to obtain
the height of the rectangle placed on top of it. Superimposing this result on
top of Figure 2, we obtain Figure 3.
Summing the areas of the higher rectangles in Figure 3, we obtain the right
sum of the four rectangles, denoted by R4, as follows (compare R4 with L4 and
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408 Chapter 6
Integration
note that R4 can be obtained from L4 by deleting one rectangular area and
adding one more):
R4 = f(2)1 + f(3)1 + f(4)1 + f(5)1
= 2.00 + 3.25 + 5.00 + 7.25 = 17.5
From Figure 3, since f(x) is increasing, it is clear that the right sum R4 overestimates the area, and we conclude that the actual area is between 11.5 and 17.5.
That is,
11.5 = L4 6 Area 6 R4 = 17.5
2
The first approximation of the area under the curve in (1) is fairly coarse,
but the method outlined can be continued with increasingly accurate results
by dividing the interval [1, 5] into more and more equal subintervals. Of
course, this is not a job for hand calculation but a job that computers are designed to do.* Figure 4 shows left and right rectangle approximations for 16
equal subdivisions.
For this case,
f (x)
10
If f(x) in Figure 3 were decreasing over the interval [1, 5], would the right
sum R4 over- or underestimate the actual area under the curve? Explain.
f (x) 0.25x2 1
¢x =
L16 =
=
R16 =
=
5
0
1
5
FIGURE 4
x
5 -1
= 0.25
16
f(1) ¢x + f(1.25) ¢x + p + f(4.75) ¢x
13.59
f(1.25) ¢x + f(1.50) ¢x + p + f(5) ¢x
15.09
Thus, we now know that the area under the curve is between 13.59 and 15.09.
That is,
13.59 = L16 6 Area 6 R16 = 15.09
For 100 equal subdivisions, computer calculations give us
14.214 = L100 6 Area 6 R100 = 14.454
The error in an approximation is the absolute value of the difference between the approximation and the actual value. In general, neither the actual
value nor the error in an approximation is known. However, it is often
possible to calculate an error bound, a positive number such that the error is
guaranteed to be less than or equal to that number.
The error in the approximation of the area under the graph of f from x = 1 to
x = 5 by the left sum L16 (or the right sum R16) is less than the sum of the areas
of the small rectangles in Figure 4. By stacking those rectangles (see Fig. 5), we
see that
Error = ƒ Area - L16 ƒ 6 ƒ f(5) - f(1) ƒ ¢x = 1.5
* The computer software that accompanies this book will perform these calculations (see the
Preface).
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The Definite Integral
Section 6-4
409
f(x)
10
f (x) 0.25x2 1
5
f (5) f (1)
x
0
1
5
x
FIGURE 5
Therefore, 1.5 is an error bound for the approximation of the area under f by
L16. We can apply the same stacking argument to any positive function that
is increasing on [a, b] or decreasing on [a, b], to obtain the error bound in
Theorem 1.
THEOREM 1
Error Bounds for Approximations of Area by Left or Right Sums
If f(x) 7 0 and is either increasing on [a, b] or decreasing on [a, b], then
f(b) - f(a) ƒ b -a
n
is an error bound for the approximation of the area between the graph of f
and the x axis, from x = a to x = b, by Ln or Rn.
Because the error bound of Theorem 1 approaches 0 as n S q, it can be
shown that left and right sums, for certain functions, approach the same limit
as n S q.
THEOREM 2
Limits of Left and Right Sums
If f(x) 7 0 and is either increasing on [a, b] or decreasing on [a, b], then its
left and right sums approach the same real number as n S q.
The number approached as n S q by the left and right sums in Theorem 2 is
the area between the graph of f and the x axis from x = a to x = b.
E XAMPLE 1 Approximating Areas Given the function f(x) = 9 - 0.25x2, we are interested in approximating the area under y = f(x) from x = 2 to x = 5.
(A) Graph the function over the interval [0, 6]; then draw left and right rectangles for the interval [2, 5] with n = 6.
(B) Calculate L6, R6, and error bounds for each.
(C) How large should n be chosen for the approximation of the area by Ln or
Rn to be within 0.05 of the true value?
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410 Chapter 6
Integration
Solution
(A) ¢x = 0.5:
f (x)
10
5
y f (x)
0
x
1
2
3
4
5
6
(B) L6 = f(2) ¢x + f(2.5) ¢x + f(3) ¢x + f(3.5) ¢x + f(4) ¢x
+ f(4.5) ¢x = 18.53
R6 = f(2.5) ¢x + f(3) ¢x + f(3.5) ¢x + f(4) ¢x
+ f(4.5) ¢x + f(5) ¢x = 15.91
Error bound for L6 and R6:
error ƒ f(5) - f(2) ƒ
5 -2
= ƒ 2.75 - 8 ƒ (0.5) = 2.625
6
(C) For Ln and Rn, find n such that error 0.05:
b -a
0.05
ƒ f(b) - f(a) ƒ
n
ƒ 2.75 - 8 ƒ
3
0.05
n
15.75 0.05n
n
✓1
Matched
Problem
15.75
= 315
0.05
■
Given the function f(x) = 8 - 0.5x2, we are interested in approximating the
area under y = f(x) from x = 1 to x = 3.
(A) Graph the function over the interval [0, 4]; then draw left and right rectangles for the interval [1, 3] with n = 4.
(B) Calculate L4, R4, and error bounds for each.
(C) How large should n be chosen for the approximation of the area by Ln or
Rn to be within 0.5 of the true value?
Insight
Note from Example 1(C) that a relatively large value of n (n = 315) is
required to approximate the area by Ln or Rn to within 0.05. In other words,
315 rectangles must be used, and 315 terms must be summed, to guarantee
that the error does not exceed 0.05. We can obtain a more efficient approximation of the area (in the sense that fewer terms must be summed to achieve
a given accuracy) by replacing rectangles by trapezoids. The resulting
trapezoidal rule, and other methods for approximating areas, are discussed in
Group Activity 1.
●
➤ The Definite Integral as a Limit of Sums
Left and right sums are special cases of more general sums, called Riemann
sums [named after the German mathematician Georg Riemann (1826–1866)],
that are used to approximate areas using rectangles.
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Section 6-4
The Definite Integral
411
Let f be a function defined on the interval [a, b]. We partition [a, b] into n
subintervals of equal length ¢x = (b - a)n with endpoints
a = x0 6 x1 6 x2 6 p 6 xn = b
Then, using summation notation (see Appendix B-1),
n
Left sum: Ln = f(x0)¢x + f(x1)¢x + p + f(xn -1)¢x = a f(xk -1)¢x
k =1
n
Right sum: Rn = f(x1)¢x + f(x2)¢x + p + f(xn)¢x = a f(xk)¢x
k =1
n
Riemann sum: Sn = f(c1)¢x + f(c2)¢x + p + f(cn)¢x = a f(ck)¢x
k =1
In a Riemann sum,* each ck is required to belong to the subinterval [xk -1, xk].
Left and right sums are the special cases of Riemann sums in which ck is the
left endpoint or right endpoint, respectively, of the subinterval. If f(x) 7 0,
then each term of a Riemann sum Sn represents the area of a rectangle having
height f(ck) and width ¢x (see Fig. 6). If f(x) has both positive and negative
values, then some terms of Sn represent areas of rectangles, and others represent the negatives of areas of rectangles, depending on the sign of f(ck) (see
Fig. 7).
f(x)
f(x)
0
0
a x0 c1 x1 c2
x2
c3 x3
FIGURE 6
c4 x4 b
a x0
x2 c3 x3
c1 x1c2
c4 x4
x
x
FIGURE 7
E XAMPLE 2 Riemann Sums Consider the function f(x) = 15 - x2 on [1, 5]. Partition
the interval [1, 5] into four subintervals of equal length. For each subinterval
[xk -1, xk], let ck be the midpoint. Calculate the corresponding Riemann sum
S4. (Riemann sums for which the ck are the midpoints of the subintervals are
called midpoint sums.)
Solution
5 -1
=1
4
S4 = f(c1) ¢x + f(c2) ¢x + f(c3) ¢x + f(c4) ¢x
= f(1.5)1 + f(2.5)1 + f(3.5)1 + f(4.5)1
= 12.75 + 8.75 + 2.75 - 5.25 = 19
¢x =
■
* The term Riemann sum is often applied to more general sums in which the subintervals [xk -1, xk]
are not required to have the same length. Such sums are not considered in this book.
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412 Chapter 6
Integration
✓2
Matched
Problem
Consider the function f(x) = x2 - 2x - 10 on [2, 8]. Partition the interval
[2, 8] into three subintervals of equal length. For each subinterval [xk -1, xk], let
ck be the midpoint. Calculate the corresponding Riemann sum S3.
By analyzing properties of a continuous function on a closed interval, it can
be shown that the conclusion of Theorem 2 is valid if f is continuous. Moreover,
not just left and right sums, but Riemann sums have the same limit as n S q.
THEOREM 3
Limit of Riemann Sums
If f is a continuous function on [a, b], then the Riemann sums for f on [a, b] approach a real number limit I as n S q.*
DEFINITION
Definite Integral
Let f be a continuous function on [a, b]. The limit I of Riemann sums for f on
[a, b], guaranteed by Theorem 2, is called the definite integral of f from a to b,
denoted
b
f(x) dx
3a
The integrand is f(x), the lower limit of integration is a, and the upper limit of
integration is b.
Because area is a positive quantity, the definite integral has the following
geometric interpretation:
b
3a
f(x) dx
represents the cumulative sum of the signed areas between the graph of f and
the x axis from x a to x b, where the areas above the x axis are counted
positively, and the areas below the x axis are counted negatively (see Fig. 8,
where A and B are the actual areas of the indicated regions).
f (x)
y f (x)
B
a
b
x
A
b
FIGURE 8
3a
f(x) dx = -A + B
* The precise meaning of this limit statement is as follows: For each e 7 0 there exists some d 7 0
such that ƒ Sn - I ƒ 6 e whenever Sn is a Riemann sum for f on [a, b] for which ¢x 6 d.
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The Definite Integral
Section 6-4
413
E XAMPLE 3 Definite Integrals Calculate the definite integrals by referring to Figure 9
with the indicated areas.
f (x)
b
(A)
y f (x)
f(x) dx
3a
(B)
3a
B
a
c
c
0
f(x) dx
b C
x
A
A 2.33
B 10.67
Area C 5.63
Area
Area
c
(C)
3b
f(x) dx
FIGURE 9
b
Solution
(A)
f(x) dx = -2.33 + 10.67 = 8.34
3a
c
(B)
3a
f(x) dx = -2.33 + 10.67 - 5.63 = 2.71
c
(C)
✓3
Matched
Problem
3b
■
f(x) dx = -5.63
Referring to the figure for Example 3, calculate the definite integrals.
0
(A)
3a
c
f(x) dx
(B)
30
b
f(x) dx
(C)
30
f(x) dx
➤ Properties of the Definite Integral
Because the definite integral is defined as the limit of Riemann sums, many
properties of sums are also properties of the definite integral. Note that Properties 3 and 4 parallel properties given in Section 6-1 for indefinite integrals.
PROPERTIES
Properties of Definite Integrals
a
1.
3a
f(x) dx = 0
b
2.
a
f(x) dx = - f(x) dx
3a
3b
b
3.
3a
b
kf(x) dx = k
3a
f(x) dx, k a constant
b
4.
3a
b
[f(x) ; g(x)] dx =
b
5.
3a
3a
b
f(x) dx ;
c
f(x) dx =
3a
f(x) dx +
3a
g(x) dx
b
3c
f(x) dx
Example 4 illustrates how properties of definite integrals can be used.
E XAMPLE 4 Using Properties of the Definite Integral If
2
30
2
x dx = 2
30
x2 dx =
8
3
3
32
x2 dx =
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19
3
414 Chapter 6
Integration
then
2
(A)
2
12x2 dx = 12
30
8
x2 dx = 12 a b = 32
3
30
2
(B)
2
(2x - 6x2) dx = 2
30
2
(C)
30
2
x dx - 6
3
x2 dx = -
33
32
x2 dx = -
8
x2 dx = 2(2) - 6 a b = -12
3
30
19
3
5
(D)
3x2 dx = 0
35
3
(E)
✓4
Matched
Problem
2
3x2 dx = 3
30
30
3
x2 dx + 3
Using the same integral values given in Example 4, find
3
(A)
2
6x2 dx
32
(B)
(C)
32
3x dx
3
3x dx
3-2
0
(9x2 - 4x) dx
30
-2
(D)
Answers to Matched Problems
8
19
x2 dx = 3 a b + 3 a b = 27
3
3
32
(E)
30
1. (A) ¢x = 0.5:
12x2 dx
2. S3 = 46
f (x)
3. (A) -2.33
10
4. (A) 38
(D) 0
(B) 5.04
(B) 16
(E) 108
5
y f (x)
0
1
2
3
4
x
(B) L4 = 12.625, R4 = 10.625; error for L4 and R4 = 2
(C) n 7 16 for Ln and Rn
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(C) 10.67
(C) -6
■