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Math Help - find the average value of the function - with an e

  1. #1
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    find the average value of the function - with an e

    Hi

    The goal is to find the average value of the function h (t) = 1/(e^2t) over the interval (0,1)

    So havg (1-0) = S a=0; b=1 e^-2t dt

    So I am assuming I should do some substitution

    let u = e^-2t then du =-2e^-2t dt so du/-2 = e^-2t dt

    Then havg = S a=0;b=1 du/-2 = -1/2 S du = -1/2 (u) a=0;b=1

    So then -1/2 (e^-2t) a=0;b=1

    = -1/2 (e^-2 - e^0)

    = -1/(2e) + 1

    Did I do this correctly???

    Thanks

    Calculus beginner - who no longer likes the letter e
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  2. #2
    Super Member redsoxfan325's Avatar
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    Swampscott, MA
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    Quote Originally Posted by calcbeg View Post
    Hi

    The goal is to find the average value of the function h (t) = 1/(e^2t) over the interval (0,1)

    So havg (1-0) = S a=0; b=1 e^-2t dt

    So I am assuming I should do some substitution

    let u = e^-2t then du =-2e^-2t dt so du/-2 = e^-2t dt

    Then havg = S a=0;b=1 du/-2 = -1/2 S du = -1/2 (u) a=0;b=1

    So then -1/2 (e^-2t) a=0;b=1

    = -1/2 (e^-2 - e^0) So far, so good

    = -1/(2e) + 1 you simplified incorrectly

    Did I do this correctly???

    Thanks

    Calculus beginner - who no longer likes the letter e
    -\frac{1}{2}(e^{-2}-e^0)=-\frac{1}{2}\left(\frac{1}{e^2}-1\right)=\frac{1}{2}-\frac{1}{2e^2}

    And e is great: \frac{d}{dx}[e^x]=\int e^x\,dx=\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x

    Learn to love it, because it never goes away.
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