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

382 Chapter 5 Integrals
x 6 − 3.0. So the Riemann sum is
R 6 − o 6
f sx i d Dx
i−1
y
5 y=˛-6x
0
FIGURE 5
3 x
− f s0.5d Dx 1 f s1.0d Dx 1 f s1.5d Dx 1 f s2.0d Dx 1 f s2.5d Dx 1 f s3.0d Dx
− 1 2 s22.875 2 5 2 5.625 2 4 1 0.625 1 9d
− 23.9375
Notice that f is not a positive function and so the Riemann sum does not represent
a sum of areas of rectangles. But it does represent the sum of the areas of the blue
rectangles (above the x-axis) minus the sum of the areas of the gold rectangles
(below the x-axis) in Figure 5.
(b) With n subintervals we have
Dx − b 2 a
n
− 3 n
So x 0 − 0, x 1 − 3yn, x 2 − 6yn, x 3 − 9yn, and, in general, x i − 3iyn. Since we are
using right endpoints, we can use Theorem 4:
In the sum, n is a constant (unlike i),
so we can move 3yn in front of the
o sign.
y 3
sx 3 2 6xd dx − lim
0
n l ` o n
f sx i d Dx − lim
i−1
n l ` o n
i−1
− lim
n l `
− lim
n l `
3
n i−1FS 3i 3
nD on 2 6S 3i
3
n i−1F 27
on
n 3 i 3 2 18 n i G
fS nD 3i 3 n
nDG
(Equation 9 with c − 3yn)
y
5 y=˛-6x
0
A
A¡
3 x
− lim
n l `F 81
o n
i 3 2 54
o iG n
n 4 i−1 n 2 (Equations 11 and 9)
i−1
− lim
n l `H F 81
n 4
nsn 1 1dG
2
− lim
n l `F 81 S1 1 1 2
4 nD
− 81 4
2 27 − 2
27
4 − 26.75
2
2 54 nsn 1 1d
2 J (Equations 7 and 5)
n 2
2 27S1 1
nDG
1
FIGURE 6
y 3
sx 3 2 6xd dx − A 1 2 A 2 − 26.75
0
This integral can’t be interpreted as an area because f takes on both positive and negative
values. But it can be interpreted as the difference of areas A 1 2 A 2 , where A 1 and
A 2 are shown in Figure 6.
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