How do you evaluate the integral #int x^2e^xdx# from #-oo# to 0?

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Answer 1 · 2016-09-28T12:29:00

#=2#

you can IBP it

Indefinite Integral
#int x^2 e^x dx = int x^2 d/dx(e^x) dx #

#= int x^2 d/dx(e^x) dx #

#= x^2 e^x - int d/dx( x^2) e^x dx #

#= x^2 e^x - int 2 x e^x dx #

#= x^2 e^x - int 2 x d/dx( e^x) dx #

#= x^2 e^x - ( 2 x e^x - int d/dx(2 x) e^x dx )#

#= x^2 e^x - ( 2 x e^x - 2int e^x dx )#

#= x^2 e^x - 2 x e^x + 2 e^x + C#

#= e^x (x^2 - 2 x + 2 ) + C#

Definite Integral

#= [ e^x (x^2 - 2 x + 2 ) ]_(-oo)^0#

#= 2#