How Can You Derive the Average Velocity in a Maxwell-Boltzmann Distribution?

[AStaunton]

maxwell Boltzmann (very sorry!)

I attempted to post this already but the latex code didn't work out and also the edit command didn't seem to work...so I am reposting, very sorry for the duplicate but do not know how to remove original!:

trying to show v_average=sqrt(8kT/(pi)m) using max boltz. distrib.:

$$f(v)=4\pi(\frac{m}{2\pi mkT})^{\frac{3}{2}}v{}^{2}e^{-\frac{mv^{2}}{2kT}}$$

to start with the formula is:

$$\langle v\rangle=\int_{0}^{\infty}vf(v)dv$$

to solve this need substitution x=v^2 and this gives dx=2vdv which gives 1/2dx=vdv and so leads to:

$$4\pi(\frac{m}{2\pi kT})^{\frac{3}{2}}\frac{1}{2}\int_{0}^{\infty}xe^{-mx/kT}dx$$

can now integrate the above by parts:

u=x dv=e^-mx/kT
du=dx v=-(kT/m)e^-mx/kT


the uv expression evaluates to 0 and so the final part of interest I am left with is:

$$\frac{kT}{m}\int_{0}^{\infty}e^{-mx/kT}dx=-\frac{k^{2}T^{2}}{m^{2}}e^{-mx/kT}|_{0}^{\infty}=-\frac{k^{2}T^{2}}{m^{2}}$$

not forgetting the constant term that was brought outside integral at very start:

$$(2\pi(\frac{m}{2\pi kT})^{\frac{3}{2}})(-\frac{k^{2}T^{2}}{m^{2}})=-\sqrt{\frac{kT}{2m\pi}}$$

this is my final answer and is in disagreement with the answer I should have gotten for two reasons:
the sign is clearly wrong and also instead of sqrt2 on the bottom I should have sqrt8 on the top:

$$\sqrt{\frac{8kT}{m\pi}}$$

Can anyone suggest where problem is, as I have rechecked a few times and I keep coming out with the same answer.

 

[AStaunton]

OK, just figured saw why I got the sqrt2 on the bottom instead of sqrt8 on the top:
the expression has e^-mv^2/2kT I thought it was simply e^-mv^2/kT.

Still can't see where the sign went wrong though