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Math Help - average value problem

  1. #1
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    average value problem

    FInd an interval like [0,a] so that 1/x^1/3 is 1

    I have that 1/a time int (0 to a) is 1/x^1/3 so then

    1/a * 1.5* a^2/3 =1 and I get a=9/4, but this does not work when I plug it back in....what am I doing wrong here?

    Then it tells you to find interval like [0,a] for 1/nth root of x.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by zhupolongjoe View Post
    FInd an interval like [0,a] so that 1/x^1/3 is 1

    I have that 1/a time int (0 to a) is 1/x^1/3 so then

    1/a * 1.5* a^2/3 =1 and I get a=9/4, but this does not work when I plug it back in....what am I doing wrong here?

    Then it tells you to find interval like [0,a] for 1/nth root of x.
    Could you state the question as it comes out of the book? What is the integrand supposed to be originally?
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  3. #3
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    The questions is "find an interval of form [0,a] s.th. the average value of 1/cubed root of x = 1
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by zhupolongjoe View Post
    The questions is "find an interval of form [0,a] s.th. the average value of 1/cubed root of x = 1
    \Rightarrow\frac{1}{a}\int_0^ax^{-1/3}=1

    \Rightarrow\frac{1}{a}\left(\frac{3}{2}x^{2/3}\right)\left|\right.^a_0=1

    I think this is right...which is what you have...

    =\frac{1}{a}\left(\frac{3}{2}a^{2/3}-\frac{3}{2}(0)^{2/3}\right)=\frac{3}{2\sqrt[3]{a}}=1\Rightarrow{a}=\frac{27}{16}
    Last edited by VonNemo19; September 11th 2009 at 03:17 PM.
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