How do you find the integral of #int (x^2)(e)^(-x^3) dx# from negative infinity to infinity? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Sasha P. Sep 26, 2015 #+oo# Explanation: #-x^3=t => -3x^2dx=dt => x^2dx=-dt/3# #int x^2e^(-x^3)dx=-1/3inte^tdt=-1/3e^(-x^3)+C# #I=int_-oo^(+oo) x^2e^(-x^3)dx= -1/3e^(-x^3)|_-oo^(+oo)# #I=-1/3lim_(x->+oo) e^(-x^3) + 1/3lim_(x->-oo) e^(-x^3)# #I=0+oo=+oo# 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 3260 views around the world You can reuse this answer Creative Commons License