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Math Help - Weighted Average Intergral

  1. #1
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    Weighted Average Intergral

    Hi, i spent too much time on this one problem. Any help would be helpful.

    Let f(x) = x^2 for [0, 1]. The average value of f on [0, 1] is 1/3. Find a non negative weight function w such that the weighted average of f on [0, 1] is a) 1/2 b) 3/5 c) 2/3.

    This is the formula that should help out.

    weighted average = Integral( w(x)f(x) , x, a, b) / Integral( w(x), x, a, b)

    w(x) is the weighted function.

    Thanks for any help or advice you can give.
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  2. #2
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    Quote Originally Posted by hashshashin715 View Post
    Hi, i spent too much time on this one problem. Any help would be helpful.

    Let f(x) = x^2 for [0, 1]. The average value of f on [0, 1] is 1/3. Find a non negative weight function w such that the weighted average of f on [0, 1] is a) 1/2 b) 3/5 c) 2/3.

    This is the formula that should help out.

    weighted average = Integral( w(x)f(x) , x, a, b) / Integral( w(x), x, a, b)

    w(x) is the weighted function.

    Thanks for any help or advice you can give.
    Maybe you should try finding a simple weight function of the form w = x^n ....
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  3. #3
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    Oh i see. I know the answer is x, x^2, and x^3. So if I just assume it is a simple function then all i have to do is solve for n. OK, thanks.
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  4. #4
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    Quote Originally Posted by hashshashin715 View Post
    Oh i see. I know the answer is x, x^2, and x^3. So if I just assume it is a simple function then all i have to do is solve for n. OK, thanks.
    I haven't checked the others but I think you'll find that the
    non negative weight function w such that the weighted average of f on [0, 1] is a) 1/2
    is w = 1/x, not w = x ....
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