User:R3m0t/Sandbox

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\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}