If
f
(
x
)
=
∫
0
x
(
t
3
+
4
t
2
+
7
)
d
t
\displaystyle {f{{\left({x}\right)}}}={\int_{{0}}^{{x}}}{\left({t}^{{3}}+{4}{t}^{{2}}+{7}\right)}{\left.{d}{t}\right.}
f
(
x
)
=
∫
0
x
(
t
3
+
4
t
2
+
7
)
d
t
then
f
′
′
(
x
)
=
\displaystyle {f}{''}{\left({x}\right)}=
f
′′
(
x
)
=
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