00:01
Okay, we're asked to evaluate a double interval.
00:03
I'm going to start with the inside interval.
00:05
I'm calling it a, and that is the integral from negative 1 to 1 of x minus y, d .y.
00:31
Okay, and because x is held constant, i just consider a constant.
00:35
I get x, y, minus y squared over 2 when i perform the integration.
00:43
Now we're going to evaluate that expression.
00:50
The limit as y approaches negative 1, subtracted from the limit as y approaches 1.
01:02
When y is 1, i have x minus 1 squared as 1, 1 1 1 1 1 1 1 1 1.
01:13
And from that i'm going to subtract.
01:18
Y equals negative 1 1 1⁄2 gives me minus x.
01:25
Y is negative 1 squared is 1 .5...