00:01
Hi, from the question given that here we need to evaluate the given iterated integral.
00:10
So consider double integral that is integral 0 to ln of 5, that is natural logarithm of 5, integral 1 to ln of 2, e to the power of 3x plus 4y, d, d ,x.
00:28
Now, this implies integral 0 to ln of 2.
00:33
L n of 5 integral 1 to ln of 2 e to the power of 3x times e to the power of 4 y d y d x so this implies integral 0 to ln of 5 e to the power of 3x d x times integral 1 to ln of 2 e to the power of 4y, dy.
01:03
So this is equal to when we integrate this, we obtain e to the power of 3x divided by 3 with the limit 0 to ln of 5 multiplied with e to the power of 4y divided by 4 with limit 1 to ln of 2.
01:23
Now apply the limit, so we obtain e to the power of 3 ln of 5 divided by 3, minus e to the power of 3 times of 0 is 0.
01:37
So, e to the power of 0 is 1 divided by 3...