00:01
Okay, so we'd like to solve this integral over the region 0 to 2 in x and 1 to 2 and y.
00:05
So this gives us an option whether or not we'd like to integrate respect to y or x first.
00:12
And since we have y terms in both, we're going to go ahead and integrate respect to y just first, just to make this easier.
00:20
So that means that our outside will be our x bounds, and the inside will be our y bounds, and then we'll just have the same integrand inside with a da equal to dy dx.
00:35
So in order to evaluate this, we have to go ahead and start with the integral.
00:42
So then we'll keep the outer the same and evaluate this as y squared over 2.
00:49
And the integral of y to the minus 2 is just negative 1 over y.
00:55
So then this is just the same thing as negative x, y, to the minus 1...