Question

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

Answer 1

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

Explanation:

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`