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Mathematics
Mathematics
Problem
Let
g
g
g
be defined for
x
≥
−
1
x\ge -1
x
≥
−
1
by
g
(
x
)
=
∫
−
1
x
f
(
t
)
d
t
g(x)=\displaystyle\int_{-1}^{x} f(t)\,dt
g
(
x
)
=
∫
−
1
x
f
(
t
)
d
t
, where
f
f
f
is continuous. Which of the following equals
g
′
(
2
)
g'(2)
g
′
(
2
)
? a)
∫
−
1
2
f
′
(
t
)
d
t
\displaystyle\int_{-1}^{2} f'(t)\,dt
∫
−
1
2
f
′
(
t
)
d
t
b)
f
′
(
2
)
f'(2)
f
′
(
2
)
c)
f
(
2
)
f(2)
f
(
2
)
d)
g
(
2
)
g(2)
g
(
2
)
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