Question

Show that the integral from 0 to infinity of (x^2)e^-x^2 = (1/2) the integral for 0 to infinity of e^-x^2, using appropriate methods for improper integrals.

Answer 1

@reemii @ganeshie8

Answer 2

@Australopithecus @aaronq @Hero @jim_thompson5910 @rebeccaxhawaii @JFraser @Daniellelovee @Jadeishere @jabez177 @Jack_Prism @j

Answer 3

Is there a way to do this without the error function?

Answer 4

Integration by parts.

Answer 5

\(\int_{0}^{+\infty} x^2 e^{-x^2}dx = \int_{0}^{+\infty} x\cdot \bigl( xe^{-x^2}\bigr) \,dx\).

with \(u(x) = x\), \(v'(x) = xe^{-x^2}\).