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Math Help - integral with lne

  1. #1
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    integral with lne

    \int_{\ln e}^{\ln {e^3}} {\frac{{{e^{\frac{2}<br />
{{{x^2}}}}}}}<br />
{{{x^3}}}} dx<br />

    I'm not sure where to begin.
    Last edited by Chris L T521; October 4th 2009 at 06:28 PM.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by genlovesmusic09 View Post
    \int_{\ln e}^{\ln {e^3}} {\frac{{{e^{\frac{2}<br />
{{{x^2}}}}}}}<br />
{{{x^3}}}} dx<br />

    I'm not sure where to begin.
    It is a u sub

    u=\frac{2}{x^2} \implies du=-\frac{4}{x^3}dx

    This should get you started good luck.
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  3. #3
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    would this be right?
    \frac{{ - 1}}<br />
{4}\int_{\ln e}^{\ln {e^3}} { - 4{e^u}} du<br />
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  4. #4
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    so I figured: \frac{{ - 1}} {4}\int_{\ln e}^{\ln {e^3}} {{e^u}} du and \frac{{ - 1}} {4}{e^u} = \frac{{ - {e^{\frac{2} {{{x^2}}}}}}} {4}_{\ln e}^{\ln {e^3}} = \frac{{ - {e^{\frac{2} {{{{(\ln {e^3})}^2}}}}}}} {4} - \frac{{ - {e^{\frac{2} {{{{(\ln e)}^2}}}}}}} {4} = \frac{{ - {e^{\frac{2} {9}}}}} {4} + \frac{{{e^2}}} {4} = 1.535
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  5. #5
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    Hello, genlovesmusic09!

    \int_{\ln e}^{\ln e^3} \frac{e^{\frac{2}{x^2}}}{x^3}\,dx
    Note that: . \ln(e) = 1,\;\;\ln(e^3) = 3


    We have: . \int^3_1 e^{2x^{-2}}\left(x^{-3}\,dx\right)

    Let: . u \:=\:2x^{-2} \quad\Rightarrow\quad du \:=\:-4x^{-3}dx \quad\Rightarrow\quad x^{-3}dx \:=\:-\tfrac{1}{4}du


    Substitute: . \int e^u\left(-\tfrac{1}{4}\,du\right) \;=\;-\tfrac{1}{4}\int e^u\,du

    Got it?

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