The average speed of molecules in an ideal gas is
$$
\bar{v}=\frac{4}{\sqrt{\pi}}\left(\frac{M}{2 R T}\right)^{3 / 2} \int_{0}^{\infty} v^{3} e^{-A b^{2} /(2 R T)} d v
$$
where $M$ is the molecular weight of the gas, $R$ is the gas constant, $T$ is the gas temperature, and $v$ is the molecular speed. Show that
$$
\bar{v}=\sqrt{\frac{8 R T}{\pi M}}
$$