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Thread: Can anyone help with this integration thanks

  1. #1
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    Can anyone help with this integration thanks


    $$\int_{20}^{10} -1/x dx$$
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  2. #2
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    Re: Can anyone help with this integration thanks

    $\displaystyle \int -\dfrac 1 x~dx = -\ln(x)$

    $\displaystyle \int_{20}^{10} -\dfrac 1 x~dx = -\ln(10)+\ln(20) = \ln\left(\dfrac{20}{10}\right)=\ln(2)$
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    Re: Can anyone help with this integration thanks

    thanks I was confused with the + and - signs with the ln cheers
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    Re: Can anyone help with this integration thanks

    woops actually my book says its -ln(2) is the book wrong?
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    Re: Can anyone help with this integration thanks

    Quote Originally Posted by bee77 View Post
    woops actually my book says its -ln(2) is the book wrong?
    the way you've written it $\ln(2)$ is correct.

    Check the limits. If you have them reversed the answer will be $-\ln(2)$
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    Re: Can anyone help with this integration thanks

    Quote Originally Posted by romsek View Post
    $\displaystyle \int -\dfrac 1 x~dx = -\ln(x)$
    $\displaystyle \int -\dfrac 1 x~dx = -\ln|x|$


    It was a fluke that the answer worked out.
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    Re: Can anyone help with this integration thanks

    Quote Originally Posted by bee77 View Post
    $\int_{20}^{10} -1/x dx$
    $ \begin{align*}\int_{20}^{10} -1/x dx&=\int_{10}^{20} 1/x dx \\&=\left. {\log(x)} \right|_{10}^{20}\\&=\log(2) \end{align*}$

    Quote Originally Posted by greg1313 View Post
    $\displaystyle \int -\dfrac 1 x~dx = -\ln|x|$
    It was a fluke that the answer worked out.
    It was by no means a fluke. Anyone experienced enough would see that the domain of integration is a subset of the positive real numbers, therefore the absolute values is not necessary.

    Moreover, when doing an integral as apposed to an antederivative one need not use a C.
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    Re: Can anyone help with this integration thanks

    -\int_{20}^{10} -1/x dx= -[\int_{10}^{20} 1/x dx]= \left[ln|x|\right]_{10}^{20}= [ln(20)- ln(10)]= ln(20/10)= ln(2)
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    Re: Can anyone help with this integration thanks

    Quote Originally Posted by Plato View Post
    $ \begin{align*}\int_{20}^{10} -1/x dx&=\int_{10}^{20} 1/x dx \\&=\left. {\log(x)} \right|_{10}^{20}\\&=\log(2) \end{align*}$


    It was by no means a fluke. Anyone experienced enough would see that the domain of integration is a subset of the positive real numbers, therefore the absolute values is not necessary.

    Moreover, when doing an integral as apposed to an antederivative one need not use a C.
    You don't know what you're talking about. The *correct* antiderivative formula is as I stated with the absolute value bars. That is a separate issue from continuing into the next part
    of the problem. Anyone experienced enough knows that. Therefore, *after* noticing that, then the qualification may be stated that parentheses are sufficient where it is placed into
    the definite integral. So, yes, it was by means a fluke, because the argument without the absolute value didn't change anything. Please be careful next time with your logic.


    Moreover, in my haste to correct him, I left off the "+ C" for the indefinite integral. It should be:

    $\displaystyle \int -\dfrac 1 x~dx = -\ln|x| + C$
    Last edited by greg1313; Jul 6th 2017 at 09:50 AM.
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