View Single Post
 P: 1,031 Okay so: $$= \int^{\infty}_0 x^2 P(x)$$ P(x) = B^2 x e^{-2\beta x} $$= \int^{\infty}_0 x^2B^2 x e^{-2\beta x}$$ $$= \int^{\infty}_0 x^3B^2 e^{-2\beta x}$$ $$= B^2 \int^{\infty}_0 x^3e^{-2\beta x}$$ I have a feeling that this integral will have to be done three times... Okay so first time: $$\left[\int^{\infty}_0 x^3e^{-2\beta x}\right]$$ $$f(x) = x^3, f'(x) = 3x^2$$ $$g'(x) = e^{-2\beta x}, g(x) = -\frac{1}{2\beta}e^{-2\beta x}$$ Thus: $$\left[ -x^3\frac{1}{2\beta}e^{-2\beta x} - \int -3x^2\frac{1}{2\beta}e^{-2\beta x} \right]$$ $$\left[ -x^3\frac{1}{2\beta}e^{-2\beta x} + \frac{3}{2\beta}\int x^2e^{-2\beta x} \right]$$ Okay so far?