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Thread: Integrals

  1. #1
    Nas
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    Exclamation Integrals

    I'm having a problem trying to figure out how to integrate the following integral: ln(y)/y with respect to dy with limits of e and 1.

    Please can someone assist me

    Thanks
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  2. #2
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    let $\displaystyle u = \ln y $

    then $\displaystyle du = \frac{dy}{y} $

    so the integral becomes $\displaystyle \int u \ du = \frac{u^{2}}{2} = \frac{(\ln y)^{2}}{2} \Big|^{e}_{1} $

    $\displaystyle = \frac{1}{2} \Big((\ln e)^{2} - (\ln 1)^{2}\Big) = \frac{1}{2} \Big(1-0\Big) = \frac{1}{2}$
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  3. #3
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    Quote Originally Posted by Nas View Post
    I'm having a problem trying to figure out how to integrate the following integral: ln(y)/y with respect to dy with limits of e and 1.

    Please can someone assist me

    Thanks
    $\displaystyle \int{\frac{\ln{y}}{y}\,dy} = \int{\ln{y}\left(\frac{1}{y}\right)\,dy}$.

    Now make the substitution $\displaystyle u = \ln{y}$ so that $\displaystyle \frac{du}{dy} = \frac{1}{y}$, the integral becomes

    $\displaystyle \int{u\,\frac{du}{dy}\,dy}$

    $\displaystyle = \int{u\,du}$

    $\displaystyle = \frac{1}{2}u^2 + C$

    $\displaystyle = \frac{1}{2}(\ln{y})^2 + C$.



    Therefore:

    $\displaystyle \int_1^e{\frac{\ln{y}}{y}\,dy}= \left[\frac{1}{2}(\ln{y})^2\right]_1^e$

    $\displaystyle = \left[\frac{1}{2}(\ln{e})^2\right] - \left[\frac{1}{2}(\ln{1})^2\right]$

    $\displaystyle = \frac{1}{2}$.
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  4. #4
    Nas
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    Thank you
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