That's not the correct order of integration. The double integral after reversing integration order is
Now computations are quite straightforward.
I am suppose to reverse the order of integration and then evaluate the following integral.
integral from 0 to 1 of the integral from y^(1/2) to 1 of (2 + x^3)^(1/2) dx dy
When I do this I end up with the square root of x to 1 for the bounds and have no value to plug in for the square root of x!!!!! I know I am doing something wrong.
I am getting integral from x^(1/2) to 1 of integral from 0 to 1 of (2 + x^3)^(1/2) dy dx
which becomes
integral from x^(1/2) to 1 of (2 + x^3 )^(1/2) -- 2^(1/2) dx
but then I have 2/3(2 + x^5) -- 2^(1/2)x evaluated from x^(1/2) to one. But alas, I have no value to plug in for the square root of x!!!! Can someone tell me where I have gone wrong??? Thanks for looking at this. Frostking
Well we have to ways: make a sketch or set up the inequalities where the double integral is taken and play.
I'll take the second one: we have thus by taking the square root in the first inequality yields and by comparing it with the second inequality we get so our new bounds are