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

1 Answer
Andrea S.
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#

Related questions
See all questions in Integrals of Exponential Functions
Impact of this question
7403 views around the world
You can reuse this answer
Creative Commons License