1.
3.
R4
[(5x − x2 ) − x]dx =
0
h
2
(4x
−
x
)dx
=
2x2 −
0
R4
h
y
2
(e
−
y
+
2)dy
=
ey −
y=−1
R1
y3
3
11. y 2 − 1 = 1 − y 2 iff y = ±1.
i1
h
2y 3
= 83 .
2y − 3
+ 2y
i1
x3
3
i4
=
x=0
= e − e−1 +
10
.
3
y=−1
R1
y=−1
32
.
3
R1
[1 − y 2 − (y 2 − 1)]dy =
y=−1
(2 − 2y 2 )dy =
y=−1
R3
13. 12 − x2 = x2 − 6 iff x = ±3.
h
i3
3
18x − 2x3
= 72.
R3
[12 − x2 − (x2 − 6)]dx =
x=−3
x=−3
(18 − 2x2 )dx =
x=−3
16.
R 2π
0
(2 − 2 cos x)dx = [2x − 2 sin x]2π
x=0 = 4π.
R
R π3
= − ln |u| + C = − ln | cos x| + C. We have −π
| tan x −
tan xdx = −du
u
3
R0
2 sin x|dx = −π 2(tan x − 2 sin x)dx = [−2 ln | cos x| + 4 cos x]0x= −π = 2 − 2 ln 2.
21. Note
R
3
3
29.
R1
0
(2x −
x
)dx
3
+
R3
1
(−0.5x + 2.5 −
x
)dx
3
=
h
5x
6
i1
2
+
h
x=0
√
−1± 9
4
−5x2
12
+ 2.5x
i3
= 2.5.
x=1
Rπ
= −1 or 0.5 iff x = π6 . 02 | sin x−
31. sin x−cos 2x = sin x−1+2 sin2 x = 0 iff sin x =
π
Rπ
Rπ
6
+
cos 2x|dx = 06 (− sin x + cos 2x)dx + π2 (sin x − cos 2x)dx = [cos x + 0.5 sin 2x]x=0
π
2
[− cos x − 0.5 sin 2x]x=
π =
6
2
1. A(x) = π(2 − 0.5x) .
R2
1
√
3 3
2
6
− 1.
h
2
A(x)dx = π 4x − x +
x3
12
i2
19π
.
12
=
x=1
R9
√
9
5. A(y) = π(2 y)2 = 4πy. 0 A(y)dy = 2π [y 2 ]y=0 = 162π.
R1
7. x3 = x iff x = 0, ±1. A(x) = π(x2 − x6 ).
0
R2
9. y 2 = 2y iff y = 0, 2. A(y) = π(4y 2 − y 4 ).
0
A(x)dx = π
A(y)dy = π
11. y 2 = (x2 )2 = x iff x = 0, 1. A(x) = π((1 − x2 )2 − (1 −
h 5
i1
R1
x
2x3
x2
4x3/2
A(x)dx
=
π
−
−
+
= 11π
.
5
3
2
3
30
0
h
h
x3
3
−
4y 3
3
−
x7
7
i1
y5
5
=
4π
.
21
=
64π
.
15
x=0
i2
y=0
√ 2
√
x) ) = π(x4 − 2x2 − x + 2 x).
x=0
−2x2
R1
−2x2
R1
2
e−2x dx ≈ 3.75825.
R1
R1
2
2
2
2
2
2
(b) A(x) = π((e−x +1)2 −12 ) = πe−x (2+e−x ). −1 πe−x (2+e−x )dx = 2π 0 (2e−x +
2
e−2x )dx ≈ 13.14312.
p
√
33. y = ±0.5 4 − x2 . x = ± 4 − 4y 2 .
31. (a) A(x) = πe
.
−1
πe
dx = 2π
0
1
√
√
√
√
R2
(a) A(x) = π((2+0.5 4 − x2 )2 −(2−0.5 4 − x2 )2 ) = 4π 4 − x2 . −2 4π 4 − x2 dx =
R2√
8π 0 4 − x2 dx ≈ 78.95684.
Rπ
Rπ
Rπ
2
2
2
Remark 1. Observe that θ=0
sin2 θdθ = θ=0
cos2 θdθ. This gives us θ=0
(sin2 θ +
√
π
π
R
R
R
2
2
2
cos2 θ = π2 ⇒ θ=0
cos2 θ = π4 . Hence x=0 4 − x2 dx =
cos2 θ)dθ = 2 θ=0
Rπ
2
2 cos θ · 2 cos θdθ = π, i.e., the volume = 8π 2 .
θ=0
p
p
p
p
R1
(b) A(y) = π((2+ 4 − 4y 2 )2 −(2− 4 − 4y 2 )2 ) = 8π 4 − 4y 2 . −1 8π 4 − 4y 2 dy =
R1p
32π 0 1 − y 2 dy ≈ 78.95684.
R1 p
Rπ
2
Remark 2. y=0 1 − y 2 dy = θ=0
cos θ · cos θdθ = π4 , i.e., the volume = 8π 2 .
41. Rotate the region above the x-axis bounded by x = y 2 , x = y 4 about the y-axis.
2