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Thread: [SOLVED] Definite integral involving parts

  1. #1
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    [SOLVED] Definite integral involving parts

    $\displaystyle \int_2^{6} \sqrt {t} ln t dt$

    $\displaystyle u = ln t$
    $\displaystyle dv = \sqrt{t}$
    $\displaystyle du = 1/t$
    $\displaystyle v = 2t^{\frac{3}{2}}$
    $\displaystyle
    \frac{2 ln t*t^\frac{3}{2}}{3} - \int \frac {2t^\frac{3}{2}}{3t}$
    $\displaystyle \frac{2 ln t*t^\frac{3}{2}}{3} - \frac {-2}{3} \int \frac {t^\frac{3}{2}}{t}$
    $\displaystyle \frac{2 ln t*t^\frac{3}{2}}{3} - \frac {-2}{3}* \frac {2t^\frac{3}{2}}{3}$

    Could anyone point out the mistake I made? Continuing with this, my answer turns out wrong.

    EDIT: Nevermind guys, I have my answer. Sorry for the waste of topic. How do I delete my topic?
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Archduke01 View Post
    $\displaystyle \int_2^{6} \sqrt {t} ln t dt$

    $\displaystyle u = ln t$
    $\displaystyle dv = \sqrt{t}$
    $\displaystyle du = 1/t$
    $\displaystyle v = 2t^{\frac{3}{2}}$
    $\displaystyle
    \frac{2 ln t*t^\frac{3}{2}}{3} - \int \frac {2t^\frac{3}{2}}{3t}$
    $\displaystyle \frac{2 ln t*t^\frac{3}{2}}{3} - \frac {-2}{3} \int \frac {t^\frac{3}{2}}{t}$
    $\displaystyle \frac{2 ln t*t^\frac{3}{2}}{3} - \frac {-2}{3}* \frac {2t^\frac{3}{2}}{3}$

    Could anyone point out the mistake I made? Continuing with this, my answer turns out wrong.
    The fourth line of yours should be:
    $\displaystyle v = \frac{2}{3}t^{\frac{3}{2}}$

    and you should write $\displaystyle du = \frac{1}{t}dt$

    So,

    $\displaystyle uv - \int v du$

    $\displaystyle = \frac{2}{3}t^{\frac{3}{2}}.ln(t) - \int \frac{2}{3}t^{\frac{3}{2}} \times \frac{1}{t} dt$

    $\displaystyle = \frac{2}{3}t^{\frac{3}{2}}.ln(t) - \frac{2}{3}\int t^{\frac{3}{2}} \times \frac{1}{t} dt$

    $\displaystyle = \frac{2}{3}t^{\frac{3}{2}}.ln(t) - \frac{2}{3}\int t^{{\frac{3}{2}}-1} dt$
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