Integrating $\int^2_0 xe^{x^2}dx$ Well what I was thinking was to integrate the indefinite integral first. 
$u=x^2$, $x=\sqrt u$
$du=2xdx = 2\sqrt {u} dx$ 
$dx= \frac{1}{2\sqrt{u}}du$
$\int xe^{x^2} dx = \int \sqrt{u}\frac{1}{2\sqrt{u}} du =\frac{1}{2}\int e^u du = \frac{1}{2}e^u =\frac{1}{2}e^{x^2} +C$
Now I can evaluate $\frac{1}{2}e^{x^2}\Big|_0^2= \frac{1}{2} e^{4} -\frac{1}{2} e^0 =\frac{1}{2}e^4-1$
so my answer should be $$\frac{1}{2}e^4-1$$
Is this correct? It's been a while since I've done stuff like this.

You mean $\frac12 e^4-\frac12$.
