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  1. #1
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    laws for changing integrating orger

    \int_{0}^{2}\int_{y^{2}}^{\sqrt[3]{y^{2}}}ye^{x^{2}}dxdy

    we solve it by changing the order of ontegration
    at this integral
    y goes from 0 till 2
    x goes from x=y^{2} till x=\sqrt[3]{y^{2}}
    now we what first integrate by y then by x
    so when we integrate by y: our intervals are y=\sqrt{x} till y=\sqrt[2]{y^{3}}

    what are the intervals when we integrate by x

    ?
    Last edited by transgalactic; June 15th 2011 at 01:40 PM. Reason: typo
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  2. #2
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    Re: laws for changing integrating orger

    Quote Originally Posted by transgalactic View Post
    \int_{0}^{2}\int_{y^{2}}^{\sqrt[3]{y^{2}}}ye^{x^{2}}dxdy

    we solve it by changing the order of ontegration
    at this integral
    y goes from 0 till 2
    x goes from x=y^{2} till x=\sqrt[3]{y^{2}}
    now we what first integrate by y then by x
    so when we integrate by y: our intervals are y=\sqrt{x} till y=\sqrt[2]{y^{3}}

    what are the intervals when we integrate by x

    ?
    Did you draw the region defined by the terminals of the given double integral? It should then be obvious.
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  3. #3
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    Re: laws for changing integrating orger

    if i draw it than it will be obvios

    but not every formula i could draw and i dont want to remmember every type

    how to solve it mathematickly without using graph
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  4. #4
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    Re: laws for changing integrating orger

    Quote Originally Posted by transgalactic View Post
    \int_{0}^{2}\int_{y^{2}}^{\sqrt[3]{y^{2}}}ye^{x^{2}}dxdy

    we solve it by changing the order of ontegration
    at this integral
    y goes from 0 till 2
    x goes from x=y^{2} till x=\sqrt[3]{y^{2}}
    now we what first integrate by y then by x
    so when we integrate by y: our intervals are y=\sqrt{x} till y=\sqrt[2]{y^{3}}

    what are the intervals when we integrate by x

    ?
    Your terminals tell you \displaystyle y^2 \leq x \leq \sqrt[3]{y^2} and \displaystyle 0 \leq y \leq 2.

    From the second inequality

    \displaystyle 0 \leq y \leq 2 \implies 0 \leq y^2 \implies 0 \leq x

    and

    \displaystyle 0 \leq y \leq 2 \implies \sqrt[3]{y^2} \leq \sqrt[3]{4} \implies x \leq \sqrt[3]{4}.

    So you have new \displaystyle x terminals \displaystyle 0 \leq x \leq \sqrt[3]{4}.


    Now from the first inequality

    \displaystyle y^2 \leq x \leq \sqrt[3]{y^2} \implies y \leq \sqrt{x}

    and

    \displaystyle y^2 \leq x \leq \sqrt[3]{y^2} \implies \sqrt{x^3} \leq y.

    So you now have \displaystyle y terminals \displaystyle \sqrt{x^3} \leq y \leq \sqrt{x}.


    Therefore

    \displaystyle \int_0^2{\int_{y^2}^{\sqrt[3]{y^2}}{y\,e^{x^2}\,dx}\,dy} = \int_0^{\sqrt[3]{4}}{\int_{\sqrt{x^3}}^{\sqrt{x}}{y\,e^{x^2}\,dy}\  ,dx}
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    Re: laws for changing integrating orger

    cool proof
    thank
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