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Thread: Integral of e^(2x-1) as an infinite series

  1. #1
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    Integral of e^(2x-1) as an infinite series

    I am trying to figure out what I am even being asked to do here. I'm guessing that I'm supposed to integrate the Mclaurin Series? If so, is my answer at the bottom left correct?
    Integral of e^(2x-1) as an infinite series-revu_2018-12-08_20-59-58.png
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Integral of e^(2x-1) as an infinite series

    When you are at this point:

    $\displaystyle \int e^{2x-1}\,dx=\sum_{n=0}^{\infty}\frac{(2x-1)^{n+1}}{2(n+1)!}$

    You might then observe that we can write:

    $\displaystyle \int e^{2x-1}\,dx=\frac{1}{2}\left(\sum_{n=1}^{\infty}\frac{( 2x-1)^{n}}{n!}\right)=\frac{1}{2}\left(e^{2x-1}-1\right)$
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: Integral of e^(2x-1) as an infinite series

    As an addendum, we should probably write:

    $\displaystyle \int e^{2x-1}\,dx=\sum_{n=0}^{\infty}\frac{(2x-1)^{n+1}}{2(n+1)!}+C$

    And then we'd end up with:

    $\displaystyle \int e^{2x-1}\,dx=\frac{1}{2}\left(e^{2x-1}-1\right)+C$

    And it we define:

    $\displaystyle C_1=C-\frac{1}{2}$

    We could write:

    $\displaystyle \int e^{2x-1}\,dx=\frac{1}{2}e^{2x-1}+C_1$
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