What is #intx^2*e^-x#? Please explain?

1 Answer
George C.
Nov 14, 2015

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

Explanation:

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

#d/(dx) (u*v) = uv' + u'v#

So:

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

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

So:

#d/(dx) (-(x^2+2x+2) e^(-x) + C)#

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

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

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