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Math Help - Definite integral, wrong answer

  1. #1
    Member sinewave85's Avatar
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    Definite integral, wrong answer

    I know my answer is wrong -- I just don't know why. What did I miss?

    \int_{-\ln2}^{\ln2}\frac{1}{2}(e^{x}-e^{-x})=\frac{1}{2}\left(\int_{-\ln2}^{\ln2}e^{x} + \int_{\ln2}^{-\ln2}e^{-x}\right)

    \int_{-\ln2}^{\ln2}\frac{1}{2}(e^{x}-e^{-x})=\frac{1}{2}\left(e^{\ln{2}}-e^{-\ln{2}}+ e^{-(-\ln{2})}-e^{-\ln{2}}\right)

    \int_{-\ln2}^{\ln2}\frac{1}{2}(e^{x}-e^{-x})=\frac{1}{2}\left(2-2^{-1}+2-2^{-1}\right)=\frac{3}{2}

    The answer is supposed to be zero. Any help is appreciated!
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  2. #2
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    Quote Originally Posted by sinewave85 View Post
    I know my answer is wrong -- I just don't know why. What did I miss?

    \int_{-\ln2}^{\ln2}\frac{1}{2}(e^{x}-e^{-x})=\frac{1}{2}\left(\int_{-\ln2}^{\ln2}e^{x} + \int_{\ln2}^{-\ln2}e^{-x}\right)

    \int_{-\ln2}^{\ln2}\frac{1}{2}(e^{x}-e^{-x})=\frac{1}{2}\left(e^{\ln{2}}-e^{-\ln{2}}+ e^{-(-\ln{2})}-e^{-\ln{2}}\right)

    \int_{-\ln2}^{\ln2}\frac{1}{2}(e^{x}-e^{-x})=\frac{1}{2}\left(2-2^{-1}+2-2^{-1}\right)=\frac{3}{2}

    The answer is supposed to be zero. Any help is appreciated!
    you have a sign mistake while integrating second integral.

    =\int_{-\ln2}^{\ln2}\frac{1}{2}(e^{x}-e^{-x})

     = \frac{1}<br />
{2}\left[ {e^x  + e^{ - x} } \right]_{ - \ln 2}^{\ln 2}  \hfill \\

    = \frac{1}<br />
{2}\left[ {e^{\ln 2}  + e^{ - \ln 2}  - e^{\ln 2}  - e^{ - \ln 2} } \right] \hfill \\

      = \frac{1}<br />
{2}\left[ {2 + 2^{ - 1}  - 2 - 2^{ - 1} } \right] \hfill \\

      = 0 \hfill \\
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  3. #3
    Member sinewave85's Avatar
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    Quote Originally Posted by Shyam View Post
    you have a sign mistake while integrating second integral.
    I figured as much -- but I just don't see where I made the mistake. Could someone point it out to me?
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    in the second line

    All signs between terms should be negative sgins :>
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    Quote Originally Posted by TWiX View Post
    in the second line

    All signs between terms should be negative sgins :>
    Please see my previous post.#2
    Last edited by Shyam; June 27th 2009 at 11:02 AM.
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    xD
    Actually, i mean the 2nd integral lol
    because its integration is -e^(-x) :>
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  7. #7
    Member sinewave85's Avatar
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    Quote Originally Posted by Shyam View Post
    you have a sign mistake while integrating second integral.

    =\int_{-\ln2}^{\ln2}\frac{1}{2}(e^{x}-e^{-x}) How do you get from here

     = \frac{1}<br />
{2}\left[ {e^x  + e^{ - x} } \right]_{ - \ln 2}^{\ln 2}  \hfill \\ to here?

    = \frac{1}<br />
{2}\left[ {e^{\ln 2}  + e^{ - \ln 2}  - e^{\ln 2}  - e^{ - \ln 2} } \right] \hfill \\

      = \frac{1}<br />
{2}\left[ {2 + 2^{ - 1}  - 2 - 2^{ - 1} } \right] \hfill \\

      = 0 \hfill \\
    By what property did you get from the first to the second line? How do you drop the negative on the second e without reversing the parameters for the integral?
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  8. #8
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    Quote Originally Posted by sinewave85 View Post
    I know my answer is wrong -- I just don't know why. What did I miss?

    \int_{-\ln2}^{\ln2}\frac{1}{2}(e^{x}-e^{-x})=\frac{1}{2}\left(\int_{-\ln2}^{\ln2}e^{x} + \int_{\ln2}^{-\ln2}e^{-x}\right)

    \int_{-\ln2}^{\ln2}\frac{1}{2}(e^{x}-e^{-x})=\frac{1}{2}\left(e^{\ln{2}}-e^{-\ln{2}}+ e^{-(-\ln{2})}-e^{-\ln{2}}\right)

    \int_{-\ln2}^{\ln2}\frac{1}{2}(e^{x}-e^{-x})=\frac{1}{2}\left(2-2^{-1}+2-2^{-1}\right)=\frac{3}{2}

    The answer is supposed to be zero. Any help is appreciated!
    let:

    f(x)=e^x-e^{-x}

    then:

    f(-x)=e^{-x}-e^x=-f(x)

    so f is odd and so your integral must be zero as it is the integrand is continuous and is integrated over a symmetric interval about 0.

    CB
    Last edited by CaptainBlack; June 27th 2009 at 11:59 PM.
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    Quote Originally Posted by sinewave85 View Post
    I figured as much -- but I just don't see where I made the mistake. Could someone point it out to me?
    you have a sign mistake in your second integral, while integrating.

     \int\limits_{ - \ln 2}^{\ln 2} {\frac{1}<br />
{2}\left[ {e^x  - e^{ - x} } \right]} dx \hfill \\

     = \frac{1}<br />
{2}\left[ {\int\limits_{ - \ln 2}^{\ln 2} {e^x dx + \int\limits_{\ln 2}^{ - \ln 2} {e^{ - x} dx} } } \right] \hfill \\

     = \frac{1}<br />
{2}\left\{ {\left[ {e^x } \right]_{ - \ln 2}^{\ln 2}  + \left[ { - e^{ - x} } \right]_{\ln 2}^{ - \ln 2} } \right\} \hfill \\

     = \frac{1}<br />
{2}\left[ {e^{\ln 2}  - e^{ - \ln 2}  - e^{\ln 2}  + e^{ - \ln 2} } \right] \hfill \\

     = \frac{1}<br />
{2}\left[ {2 - 2^{ - 1}  - 2 + 2^{ - 1} } \right] \hfill \\

     = 0 \hfill \\

    Did you figure out your mistake now? sinewave85?
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  10. #10
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by sinewave85 View Post
    I know my answer is wrong -- I just don't know why. What did I miss?

    \int_{-\ln2}^{\ln2}\frac{1}{2}(e^{x}-e^{-x})=\frac{1}{2}\left(\int_{-\ln2}^{\ln2}e^{x} + \int_{\ln2}^{-\ln2}e^{-x}\right)

    \int_{-\ln2}^{\ln2}\frac{1}{2}(e^{x}-e^{-x})=\frac{1}{2}\left(e^{\ln{2}}-e^{-\ln{2}}{\color{red}-} e^{-(-\ln{2})}-e^{-\ln{2}}\right)

    \int_{-\ln2}^{\ln2}\frac{1}{2}(e^{x}-e^{-x})=\frac{1}{2}\left(2-2^{-1}{\color{red}-}(2-2^{-1})\right)=\frac{3}{2}

    The answer is supposed to be zero. Any help is appreciated!
    that is your mistake the integrate of e^{-x} = -e^{-x}
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  11. #11
    Member sinewave85's Avatar
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    Quote Originally Posted by Shyam View Post
    you have a sign mistake in your second integral, while integrating.

     \int\limits_{ - \ln 2}^{\ln 2} {\frac{1}<br />
{2}\left[ {e^x  - e^{ - x} } \right]} dx \hfill \\

     = \frac{1}<br />
{2}\left[ {\int\limits_{ - \ln 2}^{\ln 2} {e^x dx + \int\limits_{\ln 2}^{ - \ln 2} {e^{ - x} dx} } } \right] \hfill \\

     = \frac{1}<br />
{2}\left\{ {\left[ {e^x } \right]_{ - \ln 2}^{\ln 2}  + \left[ { - e^{ - x} } \right]_{\ln 2}^{ - \ln 2} } \right\} \hfill \\

     = \frac{1}<br />
{2}\left[ {e^{\ln 2}  - e^{ - \ln 2}  - e^{\ln 2}  + e^{ - \ln 2} } \right] \hfill \\

     = \frac{1}<br />
{2}\left[ {2 - 2^{ - 1}  - 2 + 2^{ - 1} } \right] \hfill \\

     = 0 \hfill \\

    Did you figure out your mistake now? sinewave85?
    Yes, I got it! Thanks for your patience, Shyam. And, thanks for the help, everybody!
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