AP Calculus AB Quiz
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APPLICATIONS OF INTEGRATION
Question 1
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What is the correct volume setup when the region under
y
=
x
y=\sqrt{x}
y
=
x
from
x
=
1
x=1
x
=
1
to
x
=
5
x=5
x
=
5
is revolved about
y
=
−
2
y=-2
y
=
−
2
?
A
V
=
π
∫
1
5
(
x
)
2
d
x
V=\pi\int_{1}^{5}(\sqrt{x})^2\,dx
V
=
π
∫
1
5
(
x
)
2
d
x
B
V
=
π
∫
1
5
(
−
2
−
x
)
2
d
x
V=\pi\int_{1}^{5}(-2-\sqrt{x})^2\,dx
V
=
π
∫
1
5
(
−
2
−
x
)
2
d
x
C
V
=
π
∫
1
5
(
x
+
2
)
2
d
x
V=\pi\int_{1}^{5}(\sqrt{x}+2)^2\,dx
V
=
π
∫
1
5
(
x
+
2
)
2
d
x
D
V
=
π
∫
1
5
(
2
−
x
)
2
d
x
V=\pi\int_{1}^{5}(2-\sqrt{x})^2\,dx
V
=
π
∫
1
5
(
2
−
x
)
2
d
x
E
V
=
π
∫
−
2
5
(
y
+
2
)
2
d
y
V=\pi\int_{-2}^{\sqrt{5}}(y+2)^2\,dy
V
=
π
∫
−
2
5
(
y
+
2
)
2
d
y
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