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 :)

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- May 1st 2009, 06:29 AMRedeemer_PieIntegral 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 :) - May 1st 2009, 06:34 AMMoo
- May 1st 2009, 06:41 AMRedeemer_Pie
thanks alot!!! :3

- May 1st 2009, 11:10 PMRedeemer_Pie
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 :) - May 1st 2009, 11:33 PMCaptainBlack
- May 1st 2009, 11:42 PMRedeemer_Pie
my apologies is that the final answer?(Speechless). how did you work that out?

My working out was this:

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

Thanks again - May 2nd 2009, 12:07 AMredsoxfan325
Working from here, we have $\displaystyle u=t$ and $\displaystyle dv=e^{at}\,dt$. Thus $\displaystyle du = dt$ and $\displaystyle v = \frac{1}{a}e^{at}$. Now we have:

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

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

$\displaystyle \frac{e^{at}}{a^2}(at-1) = \frac{2^{-2t}(-2t\ln(2)-1)}{4(\ln(2))^2} = \boxed{-\frac{2^{-2t}(2t\ln(2)+1)}{4(\ln(2))^2}}$ - May 2nd 2009, 12:25 AMRedeemer_Pie
thanks a lot man.

:)