00:01
Okay, so we're going to evaluate the following iterated integrals.
00:05
So the key here is that when we're integrating with respect to one variable, we treat the other variable as a constant.
00:11
So in this first one, we're going to be integrating with respect to x first.
00:17
So we're treating any y as constant.
00:20
This is going to be integral from 0 to 1.
00:25
Okay, integrating with respect to x, the antiderivative of 1 is x.
00:28
The antiderivative of x cubed is negative x to the 4th over 4.
00:33
And then here, y is just a coefficient on x is treated as a constant.
00:38
So the antiderivative of the x is going to become x squared over 2.
00:43
We need to evaluate that from the x bounds at 0 and 1.
00:50
Okay, these are x bounds, so we're plugging them in for x.
00:52
So here is 1.
00:55
X is 1.
00:56
We get 1 minus a quarter plus y over 2.
01:01
And then when x is 0, this will become 0.
01:03
So it would be minus 0, but i need to write that.
01:06
That's the integral from 0 to 1 of 3 quarters plus y over 2.
01:14
So now we integrate with respect to y, so it will be 3 quarters of y plus y squared over 4 evaluated from y is 0 to y is 1.
01:26
So at y is 1 we're going to get 3 quarters plus 1 quarter and at 0 we get 0 so we minus 0, so that's just equal to 1.
01:40
So the iterated integral evaluates to 1.
01:42
In the first example.
01:45
Right.
01:45
And the second example, we're integrating with respect to x first again.
01:51
So it's going to be 0 to pi over 2.
01:57
Then sine y is just going to be a coefficient because y is treated the constant.
02:01
So it's going to be sine y.
02:03
And then the antiderivative of cosine x becomes sine x.
02:07
We want to evaluate that from x as negative pi over 2 to x as pi over 2.
02:18
Okay, so plugging that in for the sign value, we're going to get 0 to pi over 2, sine y.
02:26
Okay, sine of pi over 2 is 1, so we just get a sine y, then minus sine y.
02:34
Evaluated negative pi over 2, sine of negative pi over 2 is negative 1, b .y.
02:42
So it becomes the integral of 0 to pi over 2 of 2 of 2 copies of sine y, b.
02:49
Y.
02:50
The antiderivative of sign is going to be negative cosine.
02:55
We want to evaluate this from 1.
02:56
Y value 0 to y value pi over 2.
03:00
At pi over 2 the cosine value is 0.
03:04
So it's going to be 0 minus negative 2.
03:08
And at 0, the cosine value is 1.
03:11
So this comes out to 2.
03:13
Okay, so that iterated integral is value of 2.
03:19
I'm going to copy this down to do c and d...