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Math Help - integrate

  1. #1
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    integrate

    Show that the integral of e^(-kx)dx from 0 to 1 is 1/k(1-e^-k).
    When I integrate from and calculate from 0 to 1, I get k(1-e^-k) instead of 1/k.

    My solution is...

    [-ke^(-kx)] upper limit 1, lower limit 0
    = -ke^(-k) +k
    = k(-e^(-k) +1)

    Where did I go wrong?
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  2. #2
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     \int e^{-kx}~dx = \frac{-1}{k}e^{-kx} \neq -ke^{-kx}
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  3. #3
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    I don't follow...
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  4. #4
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    Quote Originally Posted by bhuang View Post
    I don't follow...
    let u = -kx \implies du = -k dx

     \therefore dx = \frac{du}{-k}

    and your integral becomes:

    \int \frac{e^u \times du}{-k} = \frac{-1}{k} e^u

    substitute back u = -kx, you have:

     \frac{-1}{k} e^{-kx}. Plug in the upper and lower limits in this expression.
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  5. #5
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    Thanks! I can do it through substitution, but I'm just wondering why the way I did it did not work. I just followed the rule that the derivative of a function in the form: f(x) = e^(g(x)) is e^(g(x)) x g'(x).
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  6. #6
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    Quote Originally Posted by bhuang View Post
    Thanks! I can do it through substitution, but I'm just wondering why the way I did it did not work. I just followed the rule that the derivative of a function in the form: f(x) = e^(g(x)) is e^(g(x)) x g'(x).
    Because that is the rule for a derivative, you are integrating.
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  7. #7
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    woooooooooow :| thanks!
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