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Math Help - Find the average value

  1. #1
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    Question Find the average value

    Find the average value of the function on the given interval:

    h(u) = (3-2u)^-1 ; [-1,1]

    Let t = 3-2u => du = -dt/2

    => -1/4 integral of 1/t dt = -1/4 ln|t| = -1.4 * ln(1/(3-2u)) from -1 to 1

    then I got result = 0.402, correct?
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  2. #2
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    You missed that: \overline{f(x)} = \int_a^b f(x)x dx
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  3. #3
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    no, f(ave) = 1/(b-a) * integral of fx from b to a
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  4. #4
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    Quote Originally Posted by huybinhs View Post
    Let t = 3-2u => du = -dt/2

    -1/4 integral of 1/t dt = -1/4 ln|t|
    I think it should be \displaystyle \frac{-1}{2} \int \frac{1}{t}dt = \frac{-1}{2} \ln |t|
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  5. #5
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    no b/c we also have du = -dt/2 so => 1/2 * (-1/2) = -1.4 infront, right?
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  6. #6
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    Sorry, you are right, I was in statistics. Your solution is indeed: \frac{\ln 5}{4}
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  7. #7
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    Thanks for double check my work
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  8. #8
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    So what did you get for your new integral from -1 to 1?
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  9. #9
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    I did not change the limit because I replace 1 + x^2 = u after that
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