How do you integrate int xe^(-x^2/2) from [0,sqrt2]? Calculus Introduction to Integration Integrals of Exponential Functions 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 Answer link Related questions How do you evaluate the integral inte^(4x) dx? How do you evaluate the integral inte^(-x) dx? How do you evaluate the integral int3^(x) dx? How do you evaluate the integral int3e^(x)-5e^(2x) dx? How do you evaluate the integral int10^(-x) dx? What is the integral of e^(x^3)? What is the integral of e^(0.5x)? What is the integral of e^(2x)? What is the integral of e^(7x)? What is the integral of 2e^(2x)? See all questions in Integrals of Exponential Functions Impact of this question 7646 views around the world You can reuse this answer Creative Commons License