$\int_{-3}^0 x \sqrt{x+4} d x$
let $x+4=t$
$$
\begin{aligned}
& \Rightarrow x=t-4 \\
& \Rightarrow \mathrm{d} x=\mathrm{d} t
\end{aligned}
$$
when $x=-3, t=1$
when $x=0, t=4$
$\begin{aligned} & \int_1^4(t-4) \sqrt{t} \mathrm{~d} t=\int_1^4\left(t^{3 / 2}-4 t^{1 / 2}\right) \mathrm{d} t=\left[\frac{2}{5} t^{5 / 2}-\frac{8}{3} t^{3 / 2}\right]_1^4 \\ & =\frac{2}{5}(32-1)-\frac{8}{3}(8-1)=\frac{186-280}{15}=-\frac{94}{15}\end{aligned}$