00:01
This improper integral, we're going to split it up into two parts.
00:02
So we'll take the limit as a goes to negative infinity of a to zero of x, e to the negative x squared dx, and we'll add it to the limit as b goes to infinity from zero to b of xe to the negative x squared dx.
00:20
So we're just choosing zero as an arbitrary point to split it up into two integrals.
00:26
This antiderivative is done by integration by substitution.
00:30
So we'll let u equal to x squared, which means du by dx is equal to 2x, which tells us dx is equal to du over 2x.
00:42
So that tells us the original integrals can be written in this form.
00:46
Limit as a goes to negative infinity.
00:49
The integral is no longer from a to zero because those are x values.
00:53
Now they're going to be u values, which will be a squared.
00:55
Squared to 0 of 1 over 2, e to the negative u, d .u...