How do you integrate int xe^(-x^2/2) from [0,sqrt2]?

1 Answer
Feb 9, 2017

int_0^sqrt2 xe^(-x^2/2)dx = (e-1)/e

Explanation:

Evaluate:

int_0^sqrt2 xe^(-x^2/2)dx = int_0^sqrt2 e^(-x^2/2)d(x^2/2)

int_0^sqrt2 xe^(-x^2/2)dx =[-e^(-x^2/2)]_0^sqrt2

int_0^sqrt2 xe^(-x^2/2)dx =-e^(-1)+e^0

int_0^sqrt2 xe^(-x^2/2)dx = 1 -1/e = (e-1)/e