## Calculus (3rd Edition)

$\frac{1}{6}(e^{12}-1)$
Let $u= 3y^2$, then $du=6ydy$ and $u$ takes the values from $0$ to $12$; hence we have $$\int_0^2 ye^{ 3y^2} dy=\frac{1}{6}\int_0^{12} e^{ u} du\\ =\frac{1}{6}\left[e^{u}\right]_0^{12}=\frac{1}{6}(e^{12}-1) .$$