Single Variable Calculus, Chapter 5, Review Exercises, Section Review Exercises, Problem 16

Find the intergral $\displaystyle \int^2_0 y^2\sqrt{1+y^3}dy$, if it exists.
If we let $u = 1 + y^3$, then $du = 3y^2 dy$, so $\displaystyle y^2 dy = \frac{du}{3}$, when $y = 0$, $u = 1$ and when $y = 2$, $u = 9$. Therefore,
$\displaystyle = \int^2_0 y^2 \sqrt{1+y^3} dy$
Make sure tha the upper and lower limits are also in terms of $u$, so...

$\begin{equation} \begin{aligned} &= \int^{1+2^3}_{1+0^3} \sqrt{u} \left( \frac{du}{3} \right)\\ \\ &= \frac{1}{3} \int^9_1 \sqrt{u} du\\ \\ &= \frac{1}{3} \left[ \frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1} \right]^9_1\\ \\ &= \frac{1}{3} \left[ \frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]^9_1\\ \\ &= \frac{1}{3} \left[ u^{\frac{3}{2}} \right]^9_1\\ \\ &= \frac{2}{9} \left[ 9^{\frac{3}{2}} - 1^{\frac{3}{2}} \right]\\ \\ &= \frac{2}{9} [ 27 - 1 ]\\ \\ &= \frac{52}{9} \end{aligned} \end{equation}$