Maxwellian distribution — UPSC Physics Optional CSE 1988 Paper I

A gas possesses a Maxwellian velocity distribution function. Show that the fraction of molecules in a given volume that possess a velocity (+v_x) in one direction only and whose magnitude is greater than some selected value v_o is \int_{v_o}^{\infty} f(V_x) dv_x = \frac{1}{2} \left[ 1 - \frac{1}{\sqrt{\pi}} \text{erf} \left( \frac{1/4 m v_o^2}{KT} \right)^{1/2} \right] Symbols have their usual meanings.