How do you integrate int xe^(-x^2/2)∫xe−x22 from [0,sqrt2][0,√2]?

1 Answer
Andrea S.
Feb 9, 2017

int_0^sqrt2 xe^(-x^2/2)dx = (e-1)/e∫√20xe−x22dx=e−1e

Explanation:

Evaluate:

int_0^sqrt2 xe^(-x^2/2)dx = int_0^sqrt2 e^(-x^2/2)d(x^2/2)∫√20xe−x22dx=∫√20e−x22d(x22)

int_0^sqrt2 xe^(-x^2/2)dx =[-e^(-x^2/2)]_0^sqrt2∫√20xe−x22dx=[−e−x22]√20

int_0^sqrt2 xe^(-x^2/2)dx =-e^(-1)+e^0∫√20xe−x22dx=−e−1+e0

int_0^sqrt2 xe^(-x^2/2)dx = 1 -1/e = (e-1)/e∫√20xe−x22dx=1−1e=e−1e

Related questions
See all questions in Integrals of Exponential Functions
Impact of this question
7997 views around the world
You can reuse this answer
Creative Commons License