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 7268 views around the world You can reuse this answer Creative Commons License