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Math Help - Double integral - absolute values

  1. #1
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    Double integral - absolute values

    \int_{0}^{1} \int_{0}^{1} |x - y| dydx

    - x + y if y < x
    x - y if y >= x

    so i used this:
    \int_{0}^{1}(\int_{x}^{1} (x - y) + \int_{0}^{x}(-x +y) dy)dx

    \int_{0}^{1} [xy - \frac{y^2}{2}]_{x}^{1} + [-xy + \frac{y^2}{2}]_{0}^{x}dx
    \int_{0}^{1} x - \frac{1}{2} - x^2 + \frac{x^2}{2} + (-x^2 + \frac{x^2}{2} + 0 + 0 )dx

    \int_{0}^{1} x - \frac{1}{2} - x^2dx

    in solving this:
    i got a result of - 1/3


    but the answer at back of book is 1/3
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  2. #2
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    Remember, Adam, absolute values are always positive.
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  3. #3
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    x - y for y in [0,x]
    x is the maximum value
    x is greater than or equal to y
    x - y >= 0 so |x-y| = x-y

    x - y for y in [x,1]
    x is the minimum value
    x is less than or equal to y
    x - y <= 0 so |x-y| = -x+y

    It looks to me like you just have it backwards. Give it another go.
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  4. #4
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    back of book is wrong:

    My prof told me to integrate inside 1st:

    \int_{0}^{1} \int_{0}^{1} |x - y| dy dx

    \int_{0}^{1} |xy - \frac{y^2}{2}| dx

    \int_{0}^{1} |x - \frac{1}{2} |dx
    ....
    ....
    ....

    then the answer is 1/4 solving this

    so is this method correct?
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