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Math Help - Integral problem

  1. #1
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    Integral problem

    Hi guys i've got a problem with an integral question:

    'integral sign' t/(2^(2t)). Can someone please help me with this. I dont know what to do. Thanks in advance
    Last edited by mr fantastic; May 2nd 2009 at 03:39 AM. Reason: Restored original post.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Redeemer_Pie View Post
    Hi guys i've got a problem with an integral question:

    'integral sign' t/(2^(2t)). Can someone please help me with this. I dont know what to do. Thanks in advance
    This can be rewritten :
    \int t \cdot 2^{-2t} ~dt=\int t \left[e^{\ln(2)}\right]^{-2t} ~dt=\int t e^{-2t \ln(2)} ~dt=\int t e^{at} ~dt

    where a=-2\ln(2)

    and now, do an integration by parts
    Last edited by CaptainBlack; May 1st 2009 at 11:28 PM.
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  3. #3
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    thanks alot!!! :3
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  4. #4
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    I was wondering is the final answer:

    t(2^(-2t))/(-2ln2t) - 2^(-2t)/(-2ln2t)^2

    can someone confirm for me? It doesnt look quite right.

    thanks guys
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  5. #5
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    Quote Originally Posted by Redeemer_Pie View Post
    I was wondering is the final answer:

    t(2^(-2t))/(-2ln2t) - 2^(-2t)/(-2ln2t)^2

    can someone confirm for me? It doesnt look quite right.

    thanks guys
    What you have typed is almost impossible to decipher, it should be:

    -{{\left(2\,\ln(2)\,t+1\right)\,2^{-2t}}\over{4\,(\ln(2))^2}}

    CB
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  6. #6
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    my apologies is that the final answer?. how did you work that out?

    My working out was this:

    t 'multiply' (e^(-2ln2t)/(-2ln2t)) - 'integral sign' (e^(-2ln2t)/(-2ln2t))

    Thanks again
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  7. #7
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    Quote Originally Posted by Moo View Post
    Hello,

    This can be rewritten :
    \int t \cdot 2^{-2t} ~dt=\int t \left[e^{\ln(2)}\right]^{-2t} ~dt=\int t e^{-2t \ln(2)} ~dt=\int t e^{at} ~dt

    where a=-2\ln(2)

    and now, do an integration by parts
    Working from here, we have u=t and dv=e^{at}\,dt. Thus du = dt and v = \frac{1}{a}e^{at}. Now we have:

    \frac{t}{a}e^{at}-\frac{1}{a}\int e^{at} = \frac{t}{a}e^{at}-\frac{1}{a^2}e^{at} = \frac{e^{at}}{a^2}(at-1)

    Since a=-2\ln(2) and e^{-2t\ln(2)} = 2^{-2t}, we have:

    \frac{e^{at}}{a^2}(at-1) = \frac{2^{-2t}(-2t\ln(2)-1)}{4(\ln(2))^2} = \boxed{-\frac{2^{-2t}(2t\ln(2)+1)}{4(\ln(2))^2}}
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  8. #8
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    thanks a lot man.

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