Answer
$$\int_{x^{3}}^{\sqrt{x}}\left(x^{2}+3 y^{2}\right) d y=x^{5 / 2}+x^{3 / 2}-x^{5}-x^{9}$$
Work Step by Step
Given $$\int_{x^{3}}^{\sqrt{x}}\left(x^{2}+3 y^{2}\right) d y$$
So,
\begin{align}
&\int_{x^{3}}^{\sqrt{x}}\left(x^{2}+3 y^{2}\right) d y\\
&=\left[x^{2} y+y^{3}\right]_{x^{3}}^{\sqrt{x}}\\
&=\left(x^{2} \sqrt{x}+(\sqrt{x})^{3}\right)-\left(x^{2} x^{3}+\left(x^{3}\right)^{3}\right)\\
&=x^{5 / 2}+x^{3 / 2}-x^{5}-x^{9}
\end{align}
