00:01
Hi, from the question given that we need to evaluate the following double integral over the region r.
00:10
So given double integral is double integral over the region r, e to the power of 2x plus 4y da, where r is equal to set of all x comma y such that 0 is less than are equal to x is less than r equal to ln of 5, that is natural algorithm of 5.
00:33
1 is less than or equal to y is less than or equal to ln of 3 that is natural algorithm of 3.
00:40
Now this implies double integral e to the power of 2x plus 4y for da, dx and dy.
00:55
Here the order of integration is dx, dy.
00:58
So here x varies from 0 to ln of 5 and y varies from 1 to 1 to 8.
01:06
L n of 3 so this implies integral 1 to ln of 3 integral 0 to ln of 5 this can be written as e to the power of 2x times e to the power of 4 y d x d y since we know that e to the power of m plus n is equal to e to the power of m into the power of n so this can be written as integral 1 to 1 to l .n of 3, e to the power of 4y, d .y, integral 0 to ln of 5, e to the power of 2x, dx.
01:48
So this implies when we're integrating this, we obtain e to the power of 4 y divided by 4 with limit 1 to ln of 3 and e to the power of 2x divided by 2 with limit 0 to ln of 5.
02:03
Now apply the limit, so we obtained e to the power of 4 ln of 3 divided by 4 minus e to the power of 4 times of 1 divided by 4.
02:17
Multiplied with e to the power of 2 ln of 5 divided by 2 minus e to the power of 2 times of 0 is 0.
02:27
So, e to the power of 0 divided by 2...