# Reversal of nested integrals

• February 25th 2009, 04:01 PM
Frostking
Reversal of nested integrals
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
• February 25th 2009, 05:08 PM
Krizalid
That's not the correct order of integration. The double integral after reversing integration order is $\int_{0}^{1}{\int_{\sqrt{y}}^{1}{\sqrt{2+x^{3}}\,d x}\,dy}=\int_{0}^{1}{\int_{0}^{x^{2}}{\sqrt{2+x^{3 }}\,dy}\,dx}.$

Now computations are quite straightforward.
• February 25th 2009, 08:15 PM
Frostking
Nested integral bounds question
Thanks so much for your assistance. I was able to evaluate the integral just fine with your help but I am afraid I do not understand how you obtained 0 and x^2 as bounds for the inner integral. Frostking
• February 25th 2009, 08:21 PM
Krizalid
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 $0\le y\le1,\,\sqrt y\le x\le1$ thus by taking the square root in the first inequality yields $0\le\sqrt y\le1$ and by comparing it with the second inequality we get $0\le\sqrt y\le x\le1$ so our new bounds are $0\le x\le1,\,0\le y\le x^2.$