Evaluate the integral $int \ xe^(-ax^2) \ dx$ ?

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Answer 1 · 2017-11-25T19:30:46

$int \ xe^(-ax^2) \ dx = -1/(2a) \ e^(-ax^2) +C$

I assume we seek:

$I = int \ xe^(-ax^2) \ dx$

We can perform a substitution:

$u = -ax^2 => (du)/(dx) = -2ax$

Then we can rewrite the integral and perform a substitution

$I = -1/(2a) \ int \ (-2ax)e^(-ax^2) \ dx$
$\ \ = -1/(2a) \ int \ e^u \ du$
$\ \ = -1/(2a) \ e^u +C$

And restoring the substitution we have:

$I = -1/(2a) \ e^(-ax^2) +C$

Evaluate the integral int \ xe^(-ax^2) \ dx int \ xe^(-ax^2) \ dx ?