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

418 Chapter 5 Integrals
y
_a
0
a x
a
(a) ƒ even, j ƒ dx=2 j ƒ dx
_a
y
0
a
In the first integral on the far right side we make the substitution u − 2x. Then
du − 2dx and when x − 2a, u − a. Therefore
2y 2a
0
and so Equation 8 becomes
9
f sxd dx − 2y a
y a 2a
0
f s2ud s2dud − y a
f s2ud du
f sxd dx − y a
f s2ud du 1 y a
f sxd dx
(a) If f is even, then f s2ud − f sud so Equation 9 gives
y a 2a
0
f sxd dx − y a
f sud du 1 y a
f sxd dx − 2 y a
f sxd dx
(b) If f is odd, then f s2ud − 2f sud and so Equation 9 gives
0
0
0
0
0
_a
FIGURE 3
0
a
(b) ƒ odd, j ƒ dx=0
_a
a
x
y a 2a
f sxd dx − 2y a
0
f sud du 1 y a
f sxd dx − 0
Theorem 7 is illustrated by Figure 3. For the case where f is positive and even, part
(a) says that the area under y − f sxd from 2a to a is twice the area from 0 to a because
of sym metry. Recall that an integral y b f sxd dx can be expressed as the area above the
a
x-axis and below y − f sxd minus the area below the axis and above the curve. Thus
part (b) says the integral is 0 because the areas cancel.
0
n
Example 10 Since f sxd − x 6 1 1 satisfies f s2xd − f sxd, it is even and so
y 2 22
sx 6 1 1d dx − 2 y 2
sx 6 1 1d dx
0
− 2f 1 7 x 7 1 xg 0
2
− 2( 128
7 1 2) − 284
7 n
Example 11 Since f sxd − stan xdys1 1 x 2 1 x 4 d satisfies f s2xd − 2f sxd, it is odd
and so
y 1 tan x
21 1 1 x 2 1 x dx − 0 4 n
1–6 Evaluate the integral by making the given substitution.
1. y cos 2x dx, u − 2x
2. y xe 2x 2 dx, u − 2x 2
3. y x 2 sx 3 1 1 dx, u − x 3 1 1
4. y sin 2 cos d, u − sin
5. y
x 3
x 4 2 5 dx, u − x4 2 5
6. y s2t 1 1 dt, u − 2t 1 1
7–48 Evaluate the indefinite integral.
7. y xs1 2 x 2 dx 8. y x 2 e x3 dx
9. y s1 2 2xd 9 dx 10. y sint s1 1 cost dt
11. y cossty2d dt 12. y sec 2 2 d
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