Question

How do you find the integral of `int 2x e^ (-x^2)dx` from negative infinity to infinity?

Answer 1

See the explanation section, below.

Explanation:

First notice that the integrand is continuous on `(-oo,oo)` so we only nee to evaluate two integrals.

Choose a number in `(-oo,oo)` to separate the integral. `0` is usually easy to work with. So

`int_-oo^oo 2x e^ (-x^2)dx = int_-oo^0 2x e^ (-x^2)dx +int_0^oo 2x e^ (-x^2)dx`

(Provided that both integrals exist.)

`int_-oo^0 2x e^ (-x^2)dx = lim_(ararr-oo) int_a^0 2x e^ (-x^2)dx`

`= lim_(ararr-oo) [- e^ (-x^2)]_a^0`

`= lim_(ararr-oo) [-1- (- e^ (-a^2))] = -1`

And

`int_0^oo 2x e^ (-x^2)dx = lim_(brarroo) int_0^b 2x e^ (-x^2)dx`

`= lim_(brarroo) [- e^ (-x^2)]_0^b`

`= lim_(brarroo) [(- e^ (-b^2))-(-1)] = 1`

Both limits DO exists, so,

`int_-oo^oo 2x e^ (-x^2)dx = int_-oo^0 2x e^ (-x^2)dx +int_0^oo 2x e^ (-x^2)dx`

`= -1+1=0`