00:01
In this question, we are asked to find the given double interval over the region d, where d the region bounded by 0 and 3 from the x axis and x to and x and 3x plus 5 over the y axis.
00:14
This, we can write this interval as the double integral of x cube y, and we are going to replace da by dydx.
00:24
The outer interval is over x, so the outer limits are going to be from 0 to 3.
00:29
And since the inner interval is over y, the inner limits of iteration are going to be from x to 3x plus 5.
00:38
Then, since the inner interval doesn't depend on x, we are going to factor out x to get x cube times the integral of y -di -y from x to 3x plus 5, and the d x integral from 0 to 3.
00:56
And now we are going to iterate the inner interval.
01:02
The interval of y -d -y is y squared over 2, in substitution from x to 3x plus.
01:11
You are going to factor out one half and plug in the limits of iteration for y to get x cube multiplied by 3x plus 5 squared minus x squared dx this equals to one half times the interval of x cube and the expression in parentheses could be expanded as 9x squared plus 30 x plus 25 minus x squared d x in substitution from 0 to 3.
01:51
Now that's a regular single integral.
01:54
We can write it as 1 .5 times the integral of 9.
01:58
Sorry, 9x squared minus x squared becomes 8x squared...