430 STEP 5. Build Your Test-Taking Confidence
- The correct answer is shown below.
[−3,3] by [−1,5]
(A) Setx^2 = 4 ⇒x=± 2.
Area ofR=
∫ 2
− 2
(4−x^2 )dx= 4 x−
x^3
3
] 2
− 2
=
(
4(2)−
23
3
)
−
(
4(−2)−
(−2)^3
3
)
=
16
3
−
(
−
16
3
)
=
32
3
.
(B) Sincey=x^2 is an even function,x= 0
dividesRinto two regions of equal area.
Thus,a=0.
(C) AreaR 1 =AreaR 2 =
16
3
.
y
x
y = b
R 1
R 2
y = 4
y = x^2
- √b 0 √b
AreaR 2 =
∫√b
−√b
(b−x^2 )dx
= 2
∫√b
0
(b−x^2 )dx
= 2
[
bx−
x^3
3
]√b
0
= 2
⎡
⎣b
(√
b
)
−
(√
b^3
)
3
⎤
⎦
= 2
(
b^3 /^2 −
b^3 /^2
3
)
= 2
(
2 b^3 /^2
3
)
=
4 b^3 /^2
3
.
Set
4 b^3 /^2
3
=
16
3
⇒b^3 /^2 =4orb= 42 /^3.
(D) Washer Method
V=π
∫ 2
− 2
(4^2 −(x^2 )^2 )dx
=π
∫ 2
− 2
(16−x^4 )dx
=π
[
16 x−
x^5
5
] 2
− 2
=
256 π
5