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Mathematics
Problem
When rewritten in terms of
u
=
3
x
2
+
1
u=3x^{2}+1
u
=
3
x
2
+
1
, which of the following gives the correct new limits of integration for
∫
0
2
sin
(
3
x
2
+
1
)
6
x
d
x
\displaystyle \int_{0}^{2} \sin(3x^{2}+1)\,6x\,dx
∫
0
2
sin
(
3
x
2
+
1
)
6
x
d
x
expressed as
∫
c
d
sin
(
u
)
d
u
\displaystyle\int_{c}^{d}\sin(u)\,du
∫
c
d
sin
(
u
)
d
u
? a)
c
=
−
1
,
d
=
13
c=-1,\;d=13
c
=
−
1
,
d
=
13
b)
c
=
0
,
d
=
12
c=0,\;d=12
c
=
0
,
d
=
12
c)
c
=
1
,
d
=
13
c=1,\;d=13
c
=
1
,
d
=
13
d)
c
=
1
,
d
=
7
c=1,\;d=7
c
=
1
,
d
=
7
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