Integration Revision – Integration By Parts, Twice

So looking at the first one from a few days ago (Oct 13th):

$\int^{\infty}_{0}x^{2}\:\lambda\: e^{-\lambda x}\:dx=\frac{2}{\lambda^{2}}$

I was totally lost with this. But running through some old tricks made this seem a lot more approachable. First off, moving the constant out brings a bit more clarity to the integrand,

$\int^{\infty}_{0}x^{2}\lambda e^{-\lambda x}\:dx=\lambda\int^{\infty}_{0}x^{2} e^{-\lambda x}\:dx$

But given the $x^{2}$ and the exponential, we'll also probably need to use integration by parts:

$\int^{b}_{a} f(x)g^{\prime}(x)dx=[f(x)g(x)]^{b}_{a}-\int^{b}_{a} f^{\prime}(x)g(x)dx$

So, as above, letting:
$f(x)=x^{2}$
$g^{\prime}(x)=e^{-\lambda x}$ ,

we have:
$f^{\prime}(x)=2x$
$g(x)=-\frac{1}{\lambda}e^{-\lambda x}$

Then plugging $f$ , $f^{\prime}$ , $g$ and $g^{\prime}$ into our lovely integration by parts tool above gives:

$\lambda\int^{\infty}_{0}x^{2}\: e^{-\lambda x}\:dx$

$=\lambda\left(\left[-x^{2}\: \frac{1}{\lambda}\:e^{-\lambda x}\right]^{\infty}_{0}-\int^{\infty}_{0}-2x\:\frac{1}{\lambda}\: e^{-\lambda x}\:dx\right)$

$=\lambda\left(0-0+2\frac{1}{\lambda}\int^{\infty}_{0}x\: e^{-\lambda x}\:dx\right)$

$=2\int^{\infty}_{0}x\: e^{-\lambda x}\:dx$

Which looks very familiar with what we've started with, except $x^{2}$ is now just $x$ ! If we can reduce $x$ by another power, we'll end up with just $1$ which will surely give us a much easier integral to solve.