00:01
All right, so for this question, we are integrating over a triangular region.
00:08
The bounds of integration have been provided, which is nice, and we just want to evaluate this double integral.
00:16
So let's go ahead and do that.
00:20
So i'm integrating an expression squared.
00:24
So you could write it as a u -substitution, but we can do the direct integral here.
00:30
So we're going to have the expression to the third over three.
00:34
So we're going to get a factor of one -third that pops out.
00:40
And normally, if i was taking an x partial derivative, i would have to multiply by the six on the inside for chain rule.
00:48
Since we're integrating, we actually will get to divide out by six.
00:54
So we get an additional factor of one over six that pops out.
00:58
We have the integral from zero to one.
01:02
And then left inside, we should only have the expression cubed, evaluated from our y minus one to one minus y.
01:23
Okay, so let's see what this will simplify to.
01:25
Our constants on the outside become one over 18, integral zero to one.
01:34
And we are substituting only into the value for x.
01:44
Okay, so i have six times the one minus y plus nine y quantity cubed minus six y minus one plus nine y also cubed, and this will be dy.
02:12
So i think we'll have a little bit of a nicer time if we simplify this.
02:22
So let's see.
02:23
For the y, we get nine y minus six y.
02:30
Nine y minus six y will be three y.
02:36
And we also get a plus six minus.
02:42
Now here we get nine y plus six y.
02:49
So we get 15 y and then minus six cubed...