00:01
Let's evaluate this iterated integral.
00:04
So here we have three integrals.
00:06
The outermost is 0 to 1.
00:08
The inner one is from y to 2y and the innermost integral is 0 to x plus y.
00:16
The expression inside the innermost integral is 6xy and we integrate first with respect to z and then with respect to x and then with respect to y.
00:27
So first let's focus on the innermost integral.
00:31
That is we have to integrate this expression with respect to z.
00:37
So when we integrate with respect to z, we treat x and y as constant.
00:42
Since we have to integrate this expression 6xy with respect to z, these two are treated as constant.
00:50
So we can write the two outermost integral, that is 0 to 1, and then y2, 2, 2, 2 .2.
00:57
To y.
00:58
Now let's write the result of the integration of 6xy.
01:02
When you integrate 6xy with respect to z, this will be 6xy times z.
01:08
Because 6xy is a constant and basically we are integrating 1 d z, which is 6xy z.
01:15
So this has to be evaluated within the limits 0 to x plus y.
01:21
Let's plug in the z within the limits x plus y.
01:27
The lower limit is 0.
01:30
So let's write this as 0 to 1 and then y 2 to y.
01:36
So this will be 6xy times if you replace z by the upper limit that is x plus y minus the lower limit minus 0.
01:48
So this basically is x plus y.
01:50
We can just remove this 0.
01:54
And now we have i just forgot to write down the dx and dye here.
02:00
This one is dx, and y.
02:03
Now we focus on this integral.
02:06
This is the inner integral and this has to be integrated with respect to x.
02:12
And so for that we consider this as the expression.
02:16
So first let's simplify this and then integrate with respect to x.
02:21
So therefore i write the outermost limits that is 0 to 1.
02:25
And then the inner integral is from y to 2y.
02:28
We are going to distribute this 6xy with the terms inside the bracket.
02:34
6xy times x is 6x squared y plus 6xxxxxxx.
02:40
Plus 6xy times y is 6xy squared.
02:44
So basically we have simplified this expression.
02:47
This has to be integrated with respect to x and then with respect to y.
02:53
So we are now focusing on this integral with respect to x.
02:59
And for this this expression we have to integrate with respect to x when we integrate with respect to x we have to treat y as constant so let's integrate this and so this equals the outer integral is from zero to one the integration of six x squared y with respect to x the six and y are constant so they will remain as it is 6 y times the integration of x squared using the power rule of integral this will be x to the power of 3 divided by 3 then integrate the next term so here this y squared and 6 are constants so they remain as it is times we have to integrate only x integration of x is x squared divided by 2 and this has to be evaluated within the limit from y to to y.
03:56
So let's simplify the expression inside and then substitute the limits.
04:02
Therefore here we have 0 to 1.
04:05
We can reduce this to simple terms.
04:08
This is 2y x cube or otherwise this is 2x cube y plus this one will be 3x squared y squared.
04:22
This has to be evaluated from y to 2y and also we will be...