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Thread: Integration of 1 over e

  1. #1
    M.R
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    Integration of 1 over e

    Hi,

    I am trying to integrate $\displaystyle \int \frac{1}{e^x} dx$

    I have tried using substitution of $\displaystyle u = x$ or $\displaystyle u=e^x$ both does not help.

    On a side note, is it correct that $\displaystyle \frac{1}{e^x} = e^{-x}$?
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Integration of 1 over e

    Hi M.R!

    Quote Originally Posted by M.R View Post
    Hi,

    I am trying to integrate $\displaystyle \int \frac{1}{e^x} dx$

    I have tried using substitution of $\displaystyle u = x$ or $\displaystyle u=e^x$ both does not help.

    On a side note, is it correct that $\displaystyle \frac{1}{e^x} = e^{-x}$?
    Yes, it is correct that $\displaystyle \frac{1}{e^x} = e^{-x}$.

    Do you know what the derivative of $\displaystyle e^{x}$ is?
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  3. #3
    M.R
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    Re: Integration of 1 over e

    Yes. Derivative of $\displaystyle e^x = e^x$ and derivative of $\displaystyle e^{-x} = -e^{-x}$

    So it follows that $\displaystyle \int e^{-x} dx = -e^{-x}$ ?
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  4. #4
    Super Member ILikeSerena's Avatar
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    Re: Integration of 1 over e

    Yes, although you should add an integration constant to an indeterminate integral.

    That is:

    $\displaystyle \int e^{-x} dx = -e^{-x} + C$
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  5. #5
    Master Of Puppets
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    Re: Integration of 1 over e

    Correct, but don't forget the $\displaystyle +C$ , you can check it by finding $\displaystyle \frac{d}{dx}\left(-e^{-x}\right)$
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  6. #6
    M.R
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    Re: Integration of 1 over e

    Thank you
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