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

380 Chapter 5 Integrals
If the subinterval widths are Dx 1 , Dx 2 , . . . , Dx n , we have to ensure that all these
widths approach 0 in the limiting process. This happens if the largest width, max Dx i ,
approaches 0. So in this case the definition of a definite integral becomes
y b
f sxd dx −
a
lim
max Dxi l 0
on f sx*d i Dx i
i−1
Note 5 We have defined the definite integral for an integrable function, but not all
functions are integrable (see Exercises 71–72). The following theorem shows that the
most commonly occurring functions are in fact integrable. The theorem is proved in
more advanced courses.
3 Theorem If f is continuous on fa, bg, or if f has only a finite number of jump
discontinuities, then f is integrable on fa, bg; that is, the definite integral y b a
f sxd dx
exists.
If f is integrable on fa, bg, then the limit in Definition 2 exists and gives the same
value no matter how we choose the sample points x i *. To simplify the calculation of the
integral we often take the sample points to be right endpoints. Then x i * − x i and the definition
of an integral simplifies as follows.
4 Theorem If f is integrable on fa, bg, then
y b
f sxd dx − lim
a
nl` o n
f sx i d Dx
i−1
where
Dx − b 2 a
n
and
x i − a 1 i Dx
Example 1 Express
as an integral on the interval f0, g.
lim
n l ` on sx 3 i 1 x i sin x i d Dx
i−1
SOLUTION Comparing the given limit with the limit in Theorem 4, we see that they
will be identical if we choose f sxd − x 3 1 x sin x. We are given that a − 0 and b − .
Therefore, by Theorem 4, we have
lim
n l ` on
i−1
sx 3 i 1 x i sin x i d Dx − y
sx 3 1 x sin xd dx
0
n
Later, when we apply the definite integral to physical situations, it will be important to
recognize limits of sums as integrals, as we did in Example 1. When Leibniz chose the
notation for an integral, he chose the ingredients as reminders of the limiting process. In
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