Answer
$F(x)=\displaystyle \frac{x^{2}}{4}(x^{2}+2)$
b. see "step by step"
Work Step by Step
a.
$F(x)= \displaystyle \int_{0}^{x}t(t^{2}+1)dt=\int_{0}^{x}(t^{3}+t)dt$
$=[\displaystyle \frac{1}{4}t^{4}+\frac{1}{2}t^{2}]_{0}^{\chi}$
$=\displaystyle \frac{1}{4}x^{4}+\frac{1}{2}x^{2}$
$F(x)=\displaystyle \frac{x^{2}}{4}(x^{2}+2)$
b.
$\displaystyle \frac{d}{dx}[\frac{1}{4}x^{4}+\frac{1}{2}x^{2}]=\frac{1}{4}(4x^{3})+\frac{1}{2}(2x)$
$=x^{3}+x$
$=x(x^{2}+1)$
$f(x)=x(x^{2}+1)\Rightarrow f(t)=t(t^{2}+1)$
