Definite integral, wrong answer

**sinewave85** · June 27th 2009, 10:30 AM

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!

**Shyam** · June 27th 2009, 10:38 AM

Originally Posted by sinewave85

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.

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

$= \frac{1} {2}\left[ {e^x + e^{ - x} } \right]_{ - \ln 2}^{\ln 2} \hfill \\$

$= \frac{1} {2}\left[ {e^{\ln 2} + e^{ - \ln 2} - e^{\ln 2} - e^{ - \ln 2} } \right] \hfill \\$

$= \frac{1} {2}\left[ {2 + 2^{ - 1} - 2 - 2^{ - 1} } \right] \hfill \\$

$= 0 \hfill \\$

**sinewave85** · June 27th 2009, 10:41 AM

Originally Posted by Shyam

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?

**TWiX** · June 27th 2009, 10:45 AM

in the second line

All signs between terms should be negative sgins :>

**Shyam** · June 27th 2009, 10:49 AM

Originally Posted by TWiX

in the second line

All signs between terms should be negative sgins :>

Please see my previous post.#2

**TWiX** · June 27th 2009, 10:52 AM

xD
Actually, i mean the 2nd integral lol
because its integration is -e^(-x) :>

**sinewave85** · June 27th 2009, 10:53 AM

Originally Posted by Shyam

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

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

$= \frac{1} {2}\left[ {e^{\ln 2} + e^{ - \ln 2} - e^{\ln 2} - e^{ - \ln 2} } \right] \hfill \\$

$= \frac{1} {2}\left[ {2 + 2^{ - 1} - 2 - 2^{ - 1} } \right] \hfill \\$

$= 0 \hfill \\$

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?

**CaptainBlack** · June 27th 2009, 11:00 AM

Originally Posted by sinewave85

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

**Shyam** · June 27th 2009, 11:03 AM

Originally Posted by sinewave85

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.

$\int\limits_{ - \ln 2}^{\ln 2} {\frac{1} {2}\left[ {e^x - e^{ - x} } \right]} dx \hfill \\$

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

$= \frac{1} {2}\left\{ {\left[ {e^x } \right]_{ - \ln 2}^{\ln 2} + \left[ { - e^{ - x} } \right]_{\ln 2}^{ - \ln 2} } \right\} \hfill \\$

$= \frac{1} {2}\left[ {e^{\ln 2} - e^{ - \ln 2} - e^{\ln 2} + e^{ - \ln 2} } \right] \hfill \\$

$= \frac{1} {2}\left[ {2 - 2^{ - 1} - 2 + 2^{ - 1} } \right] \hfill \\$

$= 0 \hfill \\$

Did you figure out your mistake now? sinewave85?

**Amer** · June 27th 2009, 11:05 AM

Originally Posted by sinewave85

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 $e^{-x} = -e^{-x}$

**sinewave85** · June 27th 2009, 11:18 AM

Originally Posted by Shyam

you have a sign mistake in your second integral, while integrating.

$\int\limits_{ - \ln 2}^{\ln 2} {\frac{1} {2}\left[ {e^x - e^{ - x} } \right]} dx \hfill \\$

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

$= \frac{1} {2}\left\{ {\left[ {e^x } \right]_{ - \ln 2}^{\ln 2} + \left[ { - e^{ - x} } \right]_{\ln 2}^{ - \ln 2} } \right\} \hfill \\$

$= \frac{1} {2}\left[ {e^{\ln 2} - e^{ - \ln 2} - e^{\ln 2} + e^{ - \ln 2} } \right] \hfill \\$

$= \frac{1} {2}\left[ {2 - 2^{ - 1} - 2 + 2^{ - 1} } \right] \hfill \\$

$= 0 \hfill \\$

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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