How do you evaluate the integral #int xe^(-x^2)#?

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Answer 1 · 2017-02-27T10:14:25

The answer is #=-1/2e^(-x^2)+C#

We perform this integral by substitution

Let #u=-x^2#

#du=-2xdx#

#xdx=-(du)/2#

Therefore,

#intxe^(-x^2)dx=int-1/2e^(u)du#

#=-1/2e^(u)#

#=-1/2e^(-x^2)+C#