Chapter 7 - Exponential Functions - 7.1 Derivative of f(x)=bx and the Number e - Exercises - Page 328: 78

$\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 $$ \int_0^2 ye^{ 3y^2} dy=\frac{1}{6}\int_0^{12} e^{ u} du\\ =\frac{1}{6}\left[e^{u}\right]_0^{12}=\frac{1}{6}(e^{12}-1) . $$