# determine whether the integral converges or diverges. if it converges, find the limit from(-invinity to +invinity) xe^-x^2 dx

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`int_-oo^oo x e^(-x^2) dx`

Let's ignore the endpoints for a moment and just find an antiderivative for: `x e^(-x^2)`

This is a u-substitution problem:

`u=-x^2`

`du=-2xdx`

`-(1)/(2) du = xdx`

And the problem becomes finding an antiderivative of: `-(1)/(2)e^u`

An antiderivative is also: `-(1)/(2)e^u`

Substituting back in for x gives: `-(1)/(2)e^(-x^2)`

Now, to evaluate an integral with both endpoints "improper", set up a double limit:

`int_-oo^oo x e^(-x^2)dx `

`= lim_(a->-oo,b->oo) int_a^b x e^(-x^2)dx`

(I don't know how to make that `a->-oo, b->oo` in one nicely-stacked column, but pretend that's how it looks)

`=lim_(a->-oo, b->oo) ` `-(1)/(2)e^(-x^2)|_a^b` ``

`=lim_(a->-oo, b->oo)` `-(1)/(2)e^(-b^2)+(1)/(2)e^(-a^2)`

Now take the limits:

As `a->-oo` , `-a^2 -> -oo` , so `e^(-a^2)->0`

Similarly for b:

As `b->oo` , `-b^2->-oo` , so `e^(-b^2)->0`

Thus we have `-(1)/(2)e^(-b^2)+(1)/(2)e^(-a^2)->0`

So

`int_-oo^oo xe^(-x^2)dx=0`

(If this seems unintuitive, notice that you are integrating an odd function. So for every "piece" of positive area that you have on one side, you have an equal negative "piece" on the other side. So as long as you have convergence, your integral is zero)