300 STEP 4. Review the Knowledge You Need to Score High
∫
sin
(
x
2
)
dx=
∫
sinu(2du)
= 2
∫
sinudu=−2 cosu+c
=−2 cos
(
x
2
)
+c
A=
∫π
π/ 2
sin
(
x
2
)
dx=
[
−2 cos
(
x
2
)]π
π/ 2
=− 2
[
cos
(
π
2
)
−cos
(
π/ 2
2
)]
=− 2
(
cos
(
π
2
)
−cos
(
π
4
))
=− 2
(
0 −
√
2
2
)
=
√
2
- (See Figure 12.9-8.)
Using the Disc Method:
V=π
∫ 3
0
(
x^2 + 4
) 2
dx
=π
∫ 3
0
(
x^4 + 8 x^2 + 16
)
dx
=π
[
x^5
5
+
8 x^3
3
+ 16 x
] 3
0
=π
[
35
5
+
8(3)^3
3
+16(3)
]
− 0 =
843
5
π
0 3
4
y y = x
(^2) + 4
Not to Scale
x
Figure 12.9-8
Area
∫k
1
1
x
dx=lnx]k 1 =lnk−ln 1=lnk.
Set lnk=1. Thuselnk=e^1 ork=e.
- (See Figure 12.9-9.)
1
0
–1
y = 1
y = –1
x
x = y^2 + 1
y
Figure 12.9-9
Using the Disc Method:
V=π
∫ 1
− 1
(
y^2 + 1
) 2
dy
=π
∫ 1
− 1
(
y^4 + 2 y^2 + 1
)
dy
=π
[
y^5
5
+
2 y^3
3
+y
] 1
− 1
=π
[(
15
5
+
2(1)^3
3
+ 1
)
−
(
(−1)^5
5
+
2(−1)^3
3
+(−1)
)]
=π
(
28
15
+
28
15
)
=
56 π
15
Note: You can use the symmetry of the
region and find the volume by
2 π
∫ 1
0
(
y^2 + 1
) 2
dy.