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  1. #1
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    Exclamation double integration

    ∫01 ∫y√y (2x2 + y) dy dx
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  2. #2
    Eater of Worlds
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    I am going to take aleap here and try to decipher that.

    \int_{0}^{1}\int_{y}^{\sqrt{y}}(2x^{2}+y)dydx?.
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  3. #3
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    That's right! I have confused myself on how to work it out.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by tulip View Post
    ∫01 ∫y√y (2x2 + y) dy dx
    \int_{0}^{1}\int_{y}^{\sqrt{y}}(2x^2+y)\,dy\,dx

    Its not going to work out...could it be this?

    \int_{0}^{1}\int_{y}^{\sqrt{y}}(2x^2+y)\,dx\,dy
    \implies \int_{0}^{1}\left.\left[\frac{2}{3}x^3+yx\right]\right|_{y}^{\sqrt{y}}\,dy
    \implies \int_{0}^{1}\left[\frac{5}{3}y^{\frac{3}{2}}-\frac{2}{3}y^3-y^2\right]\,dy
    \implies \left.\left[\frac{2}{3}y^{\frac{5}{2}}-\frac{1}{6}y^4-\frac{1}{3}y^3\right]\right|_{0}^{1}
    \implies \frac{2}{3}-\frac{1}{6}-\frac{1}{3}=\color{red}\boxed{\frac{1}{6}}.

    Hope this helped you out!
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  5. #5
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    Quote Originally Posted by tulip View Post
    ∫01 ∫y√y (2x2 + y) dy dx
    Are we meant to guess that you mean \int_{0}^{1} \int_{y}^{\sqrt{y}} (2x^2 + y) \, dy \, dx.

    If this is the question, where exactly are you stuck? Please show your working.
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