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Math Help - double integration

  1. #1
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    double integration

    can someone help me solve it?
    (integral from 0 to 4) dx (integral form x/2 to x^1/2) dy (x^2 - y^1/2)=

    I ended up with:

    (76/56) - (64*2^1/2)/21 + 64/(15*2^1/2)


    help!!!
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  2. #2
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    Hey, maybe it's your integral?
    <br />
 \displaystyle\int_{0}^{4} dx \displaystyle\int_{x/2}^{x^{1/2}} x^2 - y^{1/2}.dy
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  3. #3
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    Quote Originally Posted by Dogod11 View Post
    Hey, maybe it's your integral?
    <br />
\displaystyle\int_{0}^{4} dx \displaystyle\int_{x/2}^{x^{1/2}} x^2 - y^{1/2}.dy
    yes, this is exactly my integral, can you help me with this?
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  4. #4
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    Hello, well, not so difficult, is integrated first with respect to y, we can separate into several integrals as follows:

    \displaystyle\int_{x/2}^{x^{1/2}} x^2.dy - \displaystyle\int_{x/2}^{x^{1/2}}y^{1/2}.dy =

    x^2y\bigr]_{x/2}^{x^{1/2}} -  \displaystyle\frac{2}{3} .y^{3/2}\bigr]_{x/2}^{x^{1/2}}.

    Evaluating......

    x^2( \sqrt[ ]{x} -   \displaystyle\frac{x}{2} ) - (- \displaystyle\frac{1}{6}  \sqrt[ ]{2} x^{3/2} +  \displaystyle\frac{2}{3} x^{3/4}).<br />
    Simplify the latter expression, and the integrity of zero to four.

    Greetings
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