User:R3m0t/Sandbox

$$\int_0^\infty x \exp{\left (-\tfrac{3}{2} x^2 \right ) } \,dx = \int_0^{-\infty} -\frac{1}{3} {dv\over dx} \exp{v} \,dx

= -\frac{1}{3} \int_0^{-\infty} \exp{v} \,dv = -\frac{1}{3} \int_0^{-\infty}  \exp{v} \,dv

= -\frac{1}{3} \left [ \exp{v} \right ]_0^{-\infty} = \frac{1}{3}$$

$$\int_0^\infty x^3 \exp{\left ( -\tfrac{3}{2} x^2 \right ) } \,dx = $$


 * Take $$u=x^2, \frac{dv}{dx} = x \exp{ \left ( -\tfrac{3}{2} x^2 \right ) }$$ and find:


 * $$\frac{du}{dx} = 2x, v=-\tfrac{1}{3}\exp{\left ( - \tfrac{3}{2} x^2 \right ) }$$ then sub in:

$$ = \left [ - \frac{1}{3} x^2 \exp { \left ( - \tfrac{3}{2} x^2 \right ) } \right ]_0^\infty - \int_0^\infty -\frac{2}{3} x \exp{\left (-\tfrac{3}{2} x^2 \right ) } \,dx

= 0 + \frac{2}{3} \cdot \frac{1}{3}$$