James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 5.2 The Definite Integral 383
Figure 7 illustrates the calculation by showing the positive and negative terms in
the right Riemann sum R n for n − 40. The values in the table show the Riemann sums
approaching the exact value of the integral, 26.75, as n l `.
y
n
R n
FIGURE 7
R 40 < 26.3998
5 y=˛-6x
0
3 x
40 26.3998
100 26.6130
500 26.7229
1000 26.7365
5000 26.7473
n
A much simpler method for evaluating the integral in Example 2 will be given in
Section 5.4.
Because f sxd − e x is positive, the
integral in Example 3 represents
the area shown in Figure 8.
y
Example 3
(a) Set up an expression for y 3 1 e x dx as a limit of sums.
(b) Use a computer algebra system to evaluate the expression.
SOLUTION
(a) Here we have f sxd − e x , a − 1, b − 3, and
y=´
Dx − b 2 a
n
− 2 n
10
0 1 3
FIGURE 8
x
So x 0 − 1, x 1 − 1 1 2yn, x 2 − 1 1 4yn, x 3 − 1 1 6yn, and
From Theorem 4, we get
x i − 1 1 2i
n
y 3
e x dx − lim
1
n l ` on f sx i d Dx
i−1
− lim
n l ` on fS1 1 2i
i−1
2
− lim
n l ` n on e 112iyn
i−1
nD 2 n
A computer algebra system is able to
find an explicit expression for this sum
because it is a geometric series. The
limit could be found using l’Hospital’s
Rule.
(b) If we ask a computer algebra system to evaluate the sum and simplify, we obtain
o n
e 112iyn − e s3n12dyn 2 e sn12dyn
i−1
e 2yn 2 1
Now we ask the computer algebra system to evaluate the limit:
y 3
2
e x dx − lim
1
n l ` n ? e s3n12dyn 2 e sn12dyn
e 2yn 2 1
− e 3 2 e
We will learn a much easier method for the evaluation of integrals in the next section.
n
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