Integral problem

**Redeemer_Pie** · May 1st 2009, 06:29 AM

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

**Moo** · May 1st 2009, 06:34 AM

Hello,

Originally Posted by Redeemer_Pie

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

**Redeemer_Pie** · May 1st 2009, 06:41 AM

thanks alot!!! :3

**Redeemer_Pie** · May 1st 2009, 11:10 PM

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

**CaptainBlack** · May 1st 2009, 11:33 PM

Originally Posted by Redeemer_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

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

**Redeemer_Pie** · May 1st 2009, 11:42 PM

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

**redsoxfan325** · May 2nd 2009, 12:07 AM

Originally Posted by Moo

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:

$\frac{e^{at}}{a^2}(at-1) = \frac{2^{-2t}(-2t\ln(2)-1)}{4(\ln(2))^2} = \boxed{-\frac{2^{-2t}(2t\ln(2)+1)}{4(\ln(2))^2}}$

**Redeemer_Pie** · May 2nd 2009, 12:25 AM

thanks a lot man.

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