Calculus 2 Quiz
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INTEGRATION FOUNDATIONS
Question 1
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Let
g
(
x
)
=
∫
0
x
1
t
2
+
1
d
t
g(x) = \int_0^x \frac{1}{t^2 + 1} dt
g
(
x
)
=
∫
0
x
t
2
+
1
1
d
t
. If
g
(
a
)
=
π
6
g(a) = \frac{\pi}{6}
g
(
a
)
=
6
π
for some positive value
a
a
a
, then
g
′
(
a
)
g'(a)
g
′
(
a
)
equals:
A
1
2
\frac{1}{2}
2
1
B
3
2
\frac{\sqrt{3}}{2}
2
3
C
2
3
\frac{2}{3}
3
2
D
3
4
\frac{3}{4}
4
3
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