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Thread: Average Value of an abs. value function

  1. #1
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    Average Value of an abs. value function

    Q: Find the average value of $\displaystyle f(x) = | 7 - x | $ $\displaystyle [4,10] $

    My solution: I used $\displaystyle \frac {1}{b-a} \int_{a}^{b} f(x)dx $

    So I have $\displaystyle \frac {1}{10-4} \int_{4}^{10} (7 - x) dx $, but the result is 0...

    KK
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  2. #2
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    You need to find, first
    $\displaystyle \int_4^{10}|7-x|dx$
    The function is continous therefore you can subdivide this in any way you wish,
    Do this,
    $\displaystyle \int_4^7 |7-x|dx+\int_7^{10}|7-x|dx$
    Now on the interval,
    $\displaystyle f(x)=|7-x|=7-x$ on $\displaystyle [4,7]$
    And,
    $\displaystyle f(x)=|7-x|=x-7$ on $\displaystyle [7,10]$
    Thus,
    $\displaystyle \int_4^7 7-x dx+\int_7^{10} x-7dx$
    --------
    the result is 0...
    Impossible, since the function is continous and,
    $\displaystyle |7-x|>0$
    Then the dominance rule says,
    $\displaystyle \int_4^{10}|7-x|dx>0$
    And when you divide a non-zero number by the length of this interval you do not end with a negative number
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