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

368 Chapter 5 Integrals
From the values in the table in Example 1, it looks as if R n is approaching 1 3 as n
increases. We confirm this in the next example.
Example 2 For the region S in Example 1, show that the sum of the areas of the
upper approximating rectangles approaches 1 3 , that is,
lim
n l ` Rn − 1 3
y
y=≈
(1, 1)
SOLUTION R n is the sum of the areas of the n rectangles in Figure 7. Each rectangle
has width 1yn and the heights are the values of the function f sxd − x 2 at the points
1yn, 2yn, 3yn, . . . , nyn; that is, the heights are s1ynd 2 , s2ynd 2 , s3ynd 2 , . . . , snynd 2 . Thus
R n − 1 n
S 1 nD2
1 1 n
S 2 nD2
1 1 n
S 3 nD2
1 ∙ ∙ ∙ 1 1 n
S n nD2
0
1
n
FIGURE 7
1
x
− 1 n ∙ 1 n 2 s12 1 2 2 1 3 2 1 ∙ ∙ ∙ 1 n 2 d
− 1 n 3 s12 1 2 2 1 3 2 1 ∙ ∙ ∙ 1 n 2 d
Here we need the formula for the sum of the squares of the first n positive integers:
1
1 2 1 2 2 1 3 2 1 ∙ ∙ ∙ 1 n 2 −
nsn 1 1ds2n 1 1d
6
Perhaps you have seen this formula before. It is proved in Example 5 in Appendix E.
Putting Formula 1 into our expression for R n , we get
Here we are computing the limit of
the sequence hR n j. Sequences and their
limits were discussed in A Preview of
Calculus and will be studied in detail in
Section 11.1. The idea is very similar to
a limit at infinity (Section 2.6) except
that in writing lim n l ` we restrict n to
be a positive integer. In particular, we
know that
1
lim
nl ` n − 0
When we write lim n l ` R n − 1 3 we
mean that we can make R n as close to 1 3
as we like by taking n sufficiently large.
Thus we have
R n − 1 nsn 1 1ds2n 1 1d sn 1 1ds2n 1 1d
3
∙ −
n 6
6n 2
sn 1 1ds2n 1 1d
lim
n l ` Rn − lim
n l ` 6n 2
D
S1 1 1 1
6 nDS2 nD
1
1
− lim
n
S DS n 1 1 2n 1 1
l ` 6 n n
1
− lim
n l `
− 1 6 ? 1 ? 2 − 1 3
It can be shown that the lower approximating sums also approach 1 3 , that is,
lim
n l ` Ln − 1 3
n
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