How do you evaluate the integral #int xe^-x dx# from 0 to #oo#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Eddie Sep 24, 2016 1 Explanation: #I (alpha) = int_0^oo e^(-alpha x) dx = [-1/alpha e^(- alpha x)]_0^oo= 1/alpha# #(dI)/(d alpha) = int_0^oo -x e^(-alpha x) dx = -1/alpha^2# #implies int_0^oo x e^(-alpha x) dx = 1/alpha^2 # #implies int_0^oo x e^(-x) dx = 1# Answer link Related questions What is the difference between definite and indefinite integrals? What is the integral of #ln(7x)#? Is f(x)=x^3 the only possible antiderivative of f(x)=3x^2? If not, why not? How do you find the integral of #x^2-6x+5# from the interval [0,3]? What is a double integral? What is an iterated integral? How do you evaluate the integral #1/(sqrt(49-x^2))# from 0 to #7sqrt(3/2)#? How do you integrate #f(x)=intsin(e^t)dt# between 4 to #x^2#? How do you determine the indefinite integrals? How do you integrate #x^2sqrt(x^(4)+5)#? See all questions in Definite and indefinite integrals Impact of this question 2573 views around the world You can reuse this answer Creative Commons License