Answer
$$\frac{52}{9}$$
Work Step by Step
Given $$\int_0^2y^2\sqrt{1+y^3}dy$$
Let $u=1+y^3\ \rightarrow \ \ du=3y^2dy$ at $y=0\to u=1$ at $y=2\to u=9$
\begin{align*}
\int_0^2y^2\sqrt{1+y^3}dy&=\frac{1}{3}\int_1^9u^{1/2}du\\
&=\frac{2}{9}u^{3/2}\bigg|_{1}^{9}\\
&=\frac{52}{9}
\end{align*}
