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Math Help - Reversal of nested integrals

  1. #1
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    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
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  2. #2
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    Krizalid's Avatar
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    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.
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  3. #3
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    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
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  4. #4
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    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.
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