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Math Help - Simple integration

  1. #1
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    Simple integration

    \int_{0}^{1}\int_{0}^{1} min(x,y) dxdy
    = \int_{0}^{1}\int_{y}^{1} y dxdy
    = \int_{0}^{1} y(1-y)dy
    = 1/6
    What's wrong with it?
    Thank you.
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  2. #2
    Junior Member Youkla's Avatar
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    Re: Simple integration

    Can you write this in LaTex? I can't seem to follow what you have there.
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  3. #3
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    Re: Simple integration

    Quote Originally Posted by xiongs8421 View Post
    \int_{0}^{1}\int_{0}^{1} min(x,y) dxdy
    = \int_{0}^{1}\int_{y}^{1} y dxdy
    = \int_{0}^{1} y(1-y)dy
    = 1/6
    What's wrong with it?
    Thank you.
    On the square you can write the function

    \text{min}(x,y)=\begin{cases} x, \qaud \text{ if } y \ge x \\ y, \qaud \text{ if } y \le x \end{cases}

    So you can break this into two different integrals over each triangle. Or you can do one of the integrals and multiply it by 2. Why does this work?
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