how to integrate this? xe^-2x from 0 to infinity i used ibp with dv = e^-2x and u = x then v = -0.5e^-2x and du = 1 I'm not getting the right answer though...
Follow Math Help Forum on Facebook and Google+
Your u and v subs are fine. What did you do next?
= uv - int vdu int vdu = 1/4e^-2x so i get -0.5xe^-2x - 0.25e^-2x | 0 to infinity not 100% sure how to evaluate the first term at infinity though, i believe it goes to 0 but i might be wrong...
Originally Posted by Kuma = uv - int vdu int vdu = 1/4e^-2x so i get -0.5xe^-2x - 0.25e^-2x | 0 to infinity not 100% sure how to evaluate the first term at infinity though, i believe it goes to 0 but i might be wrong... That's right (and it does go to 0. For the first term e^(-2x) will shrink faster than x will increase) edit: wolfram agrees - http://www.wolframalpha.com/input/?i...0+and+infinity
View Tag Cloud