00:01
In this problem, we're given the improper integral x times each the negative x squared dx from negative infinity to infinity.
00:09
And we're asked to find whether this is convergent or divergent.
00:13
So, since we have two bounds that go to infinity, i'm going to rewrite this as two integrals breaking up that bound.
00:21
So 1 is going to have the bounds negative infinity to 0 of our function, plus the bounds 0 to infinity of our function.
00:39
So because we have now two integrals with the same terms, i'm going to begin by just solving the indefinite integral x times e to the negative x squared dx, so that we can use it for both of our integrals.
00:55
So in order to solve this, we're going to use u substitution where u equals negative x squared and du equals negative 2x d x.
01:14
So if we plug this in to our integral, we get a negative 1 half on the side and then we have the integral e to the u, d u.
01:26
And this will just equal 1 half, e to the u plus c.
01:32
If we substitute back in our value for u, that gives us negative 1 half, e to the negative x squared plus c...