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Thread: stuck on integration by parts

  1. #1
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    stuck on integration by parts

    hello,

    im stuck on an integration by parts problem:

    $\displaystyle \int{s \times 2^s}$

    here is what i have so far

    $\displaystyle u = s$, $\displaystyle du = ds$
    $\displaystyle dv = 2^{s}ds$, $\displaystyle v = \frac{2^{s+1}}{s+1}$

    i know that

    $\displaystyle \int{s \times 2^s} = uv - \int{vdu}$
    $\displaystyle = s \times \frac{2^{s+1}}{s+1} - \int{\frac{2^{s+1}}{s+1}du}$

    but im not sure how to continue...

    could someone show me?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by absvalue View Post
    hello,

    im stuck on an integration by parts problem:

    $\displaystyle \int{s \times 2^s}$

    here is what i have so far

    $\displaystyle u = s$, $\displaystyle du = ds$
    $\displaystyle dv = 2^{s}ds$, $\displaystyle v = \frac{2^{s+1}}{s+1}$

    i know that

    $\displaystyle \int{s \times 2^s} = uv - \int{vdu}$
    $\displaystyle = s \times \frac{2^{s+1}}{s+1} - \int{\frac{2^{s+1}}{s+1}du}$

    but im not sure how to continue...

    could someone show me?
    Be careful!! $\displaystyle \displaystyle\int 2^s\,ds\neq \frac{2^{s+1}}{s+1}$!! You can't apply the power rule when your base is a number and your exponent is a variable!!

    However, observe that $\displaystyle 2^s=e^{s\ln 2}$.

    Can you proceed?
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  3. #3
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    Ok, I think I understand...

    Is this correct then?

    $\displaystyle \int \!s2^s\, ds$

    Note that $\displaystyle 2^s = e^{s\ln{2}}$. Thus

    $\displaystyle \int \!s2^s\, ds = \int \!se^{s\ln{2}}\,ds$

    Let $\displaystyle u=s$ and $\displaystyle dv = e^{s\ln{2}}ds$.

    Then $\displaystyle du = ds$ and $\displaystyle v = \frac{e^{s\ln{2}}}{\ln{2}}$.

    Using integration by parts:

    $\displaystyle \int \!se^{s\ln{2}}\,ds = uv - \int vdu$

    $\displaystyle = s\frac{e^{s\ln{2}}}{\ln{2}} - \int \!\frac{e^{s\ln{2}}}{\ln{2}}\,ds$

    $\displaystyle = s\frac{e^{s\ln{2}}}{\ln{2}} - \frac{e^{s\ln{2}}}{\ln^2{2}} + c$

    $\displaystyle = s\frac{2^s}{\ln{2}} - \frac{2^s}{\ln^2{2}} + c$
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  4. #4
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    Yes, that is correct.

    It is also useful to learn as general rules: $\displaystyle \frac{d a^x}{dx}= ln(a) a^x$ and $\displaystyle \int a^x dx= \frac{a^x}{ln(a)}+ C$.

    Notice that if a= e, ln(a)= ln(e)= 1 so that becomes the "usual" derivative and anti-derivative of $\displaystyle e^x$.
    Last edited by Chris L T521; Jun 28th 2010 at 07:56 PM. Reason: fixed LaTeX.
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