I am trying to show $\int_0^{\infty} x^2 e^{-2 x} dx = 1/4 $
Integration by parts gets the indefinite integral $$\int x^2 e^{-2 x} dx = \frac{-1}{4} e^{-2 x} (2 x^2+2 x+1)+constant$$
In order to take the limit at $\infty$ I split it up;
$$ = \frac{-1}{2}\left[x^2 e^{-2 x}\right]_0^\infty + \frac{-1}{2}\left[x e^{-2 x}\right]_0^\infty + \frac{-1}{4}\left[ e^{-2 x}\right]_0^\infty$$
However, as the title suggests, I can't work out how to take $x^2 e^{-2 x}$ and $x e^{-2 x}$ to infinity.
I have tried L'Hopital's rule, without success. I have also tried Taylor expanding.Possible one of these is correct, I just haven't applied it correctly.
L'Hôpital works perfectly:
$$\lim_{x\to\infty}x^2e^{-2x} = \lim_{x\to\infty}\frac{x^2}{e^{2x}} = \lim_{x\to\infty}\frac{2x }{2e^{2x}}=\lim_{x\to\infty}\frac{2 }{4e^{2x}}=0.$$