00:01
For this problem, we are asked to graph the region associated with the iterated integral shown for part a.
00:06
So we can read off from the innermost integral that we have the square root of y must be less than or equal to x, which in turn must be less than or equal to 1.
00:14
Then from the outer integral, we can see that we must have that zero is less than or equal to y is less than or equal to 1.
00:20
So plotting out that region, you can see that this is the result.
00:25
R.
00:25
And we have this curve here is x equals root y, then we have x equals 1.
00:31
So if we want to switch the order of integration, i'll note that we need to redefine x equals root y as y equals x squared.
00:40
So this will become the integral from 0 to 1, of the integral from 0 to x squared, of e to the power of y over x, d y, d x.
00:54
D x...
