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Thread: a to the u integral

  1. #1
    MHF Contributor Jason76's Avatar
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    a to the u integral

    Does look right? As far as correct form, etc..?

    $\displaystyle \int_{1}^{5} 3^{2x} dx$.

    Using $\displaystyle \int a^{u} du = \dfrac{a^{u}}{\ln a} + C$ and $\displaystyle \dfrac{d}{dx} (a^{u}) = a^{u} * du * \ln a$.

    $\displaystyle u = 2x$.

    $\displaystyle du = 2$.

    $\displaystyle a = 3$.

    $\displaystyle \dfrac{1}{2} \int_{1}^{5} 3^{u} du$.

    $\displaystyle \int_{1}^{5} \dfrac{1}{2} 3^{u} du $.

    $\displaystyle \int_{1}^{5} \dfrac{3^{u}}{2} du $.

    $\displaystyle \dfrac{3^{2(u)}}{2\ln 3} |_{1}^{5} $.

    $\displaystyle \dfrac{3^{2(5)}}{2\ln 3}$ - $\displaystyle \dfrac{3^{2(1)}}{2\ln 3}$.

    $\displaystyle \dfrac{3^{10}}{2\ln 3}$ - $\displaystyle \dfrac{3^{2}}{2\ln 3}$.

    $\displaystyle \dfrac{59049}{2\ln 3}$ - $\displaystyle \dfrac{9}{2\ln 3} = \dfrac{59040}{2\ln 3} = \dfrac{29520}{\ln 3}$
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  2. #2
    Junior Member FelixFelicis28's Avatar
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    Re: a to the u integral

    Yes, that's correct, but there are a few mistakes that I feel the need to point out.

    Using $\displaystyle \int a^{u} du = \dfrac{a^{u}}{\ln a} + C$ and $\displaystyle \dfrac{d}{dx} (a^{u}) = a^{u} * du * \ln a$.
    This should be $\displaystyle \frac{d}{dx} a^u = \frac{d}{du} \cdot \frac{du}{dx} a^u = a^u \ln a \cdot \frac{du}{dx}$ by the chain rule.

    $\displaystyle u = 2x$

    $\displaystyle du = 2$
    This should be $\displaystyle u = 2x \implies \frac{du}{dx} = 2$

    $\displaystyle \dfrac{3^{2(u)}}{2\ln 3} \bigg|_{1}^{5} $.
    I'm not quite sure if this is a typo or not but you have to either re-write your integral after you've integrated in terms of $\displaystyle x$ again if you're going to evaluate it with those limits (which I think you've done by putting the $\displaystyle 2$ back in but it got a bit confusing by keeping it in terms of $\displaystyle u$) OR change your limits for $\displaystyle u$ i.e. $\displaystyle u = 2x \implies 1 \to 2, \ 5 \to 10$.

    $\displaystyle \dfrac{59049}{2\ln 3}$ - $\displaystyle \dfrac{9}{2\ln 3} = \dfrac{59040}{2\ln 3} = \dfrac{29520}{\ln 3}$
    That's correct.
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  3. #3
    MHF Contributor Jason76's Avatar
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    Re: a to the u integral

    Now, it's all coming together.

    Pretty sure, this is right:

    $\displaystyle \int_{5}^{1} 3^{2x} dx $.

    $\displaystyle u = 2x$.

    $\displaystyle du = 2 dx \rightarrow du(\dfrac{1}{2}) = dx \rightarrow \dfrac{du}{2} = dx$.

    $\displaystyle \dfrac{1}{2} \int_{5}^{1} 3^{u} du \rightarrow \int_{5}^{1} (\dfrac{1}{2}) 3^{u} du \rightarrow \dfrac{3^{u}}{2}$.

    $\displaystyle = \dfrac{3^{u}}{2\ln 3} |_{5}^{1} \rightarrow \dfrac{3^{2x}}{2\ln 3} |_{5}^{1}$.

    $\displaystyle \dfrac{3^{2(5)}}{2\ln 3} - \dfrac{3^{2(1)}}{2\ln 3} \rightarrow \dfrac{3^{10}}{2\ln 3} - \dfrac{3^{2}}{2\ln 3}$.

    $\displaystyle = \dfrac{59049}{2\ln 3} \rightarrow \dfrac{29520}{\ln 3}$.
    Last edited by Jason76; Sep 4th 2013 at 11:07 PM.
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