# is this correct?

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• July 1st 2008, 09:58 PM
10164064
is this correct?
{4 - 5e^ - 5t + (e^2t)/3}dt

I got
4t + e^-5t + (e^2t)/6
• July 1st 2008, 10:06 PM
Mathstud28
Quote:

Originally Posted by 10164064
{4 - 5e^ - 5t + (e^2t)/3}dt

I got
4t + e^-5t + (e^2t)/6

(Yes)
• July 1st 2008, 11:00 PM
CaptainBlack
Quote:

Originally Posted by 10164064
{4 - 5e^ - 5t + (e^2t)/3}dt

I got
4t + e^-5t + (e^2t)/6

You can check this yourself by differentiating $4t + e^{-5t} + e^{2t}/6$, you should get your integrand $4 - 5e^{- 5t} + e^{2t}/3$.

Also since this is an indefinite integralb there should be an arbitary constant added so you should have:

$4t + e^{-5t} + e^{2t}/6+c$

RonL