Integrate?

Apr 2010
58
2
Riverside, CA
(t^2 + e^(2t) + e^(-2t))^1/2

anyone got any ideas, I suck at integration so its hard for me to see what to do, if someone can start me in the right track I am sure I can finish.

thanks,

Warren.
 
Nov 2009
130
24
Hello!

Is the original integral that you want to solve:

\(\displaystyle \int \sqrt{t^2 + e^{2t} + e^{-2t}} dt\)

Please use LaTeX.
 
Apr 2010
58
2
Riverside, CA
Hello!

Is the original integral that you want to solve:

\(\displaystyle \int \sqrt{t^2 + e^{2t} + e^{-2t}} dt\)

Please use LaTeX.
Yes that is it. I am sorry I did not use Latex, have headache right now and couldn't think
 
Apr 2010
58
2
Riverside, CA
Dear Warrenx,

This integration could not be evaluated using elementary functions. You can get the answer using, Function calculator
hmm... I think I wrote the problem write:

It is asking for the arc length:

so s(t) =\(\displaystyle \int_0^1 |r'(t)|\)

r(t) = \(\displaystyle \sqrt2t , e^t , e^{-t}\)

so I get r'(t) = \(\displaystyle t , e^t , e^{-t}\)

then lr'(t)l = \(\displaystyle \sqrt{(t)^2 + (e^t)^2 + (-e^{-t})^2}\)

(sorry it was -e for the third term)

and then I am supposed to integrate it, but I suck at integratin' so I are stuck :(
 

simplependulum

MHF Hall of Honor
Jan 2009
715
427
hmm... I think I wrote the problem write:

It is asking for the arc length:

so s(t) =\(\displaystyle \int_0^1 |r'(t)|\)

r(t) = \(\displaystyle \sqrt2t , e^t , e^{-t}\)

so I get r'(t) = \(\displaystyle t , e^t , e^{-t}\)

then lr'(t)l = \(\displaystyle \sqrt{(t)^2 + (e^t)^2 + (-e^{-t})^2}\)

(sorry it was -e for the third term)

and then I am supposed to integrate it, but I suck at integratin' so I are stuck :(

Oh , the derivative of \(\displaystyle \sqrt{2} t\) is \(\displaystyle \sqrt{2} \)

so what the integral you are going to evaluate should be :

\(\displaystyle \int \sqrt{ \sqrt{2}^2 + e^{2t} + e^{-2t} }~dt \)

Trick is also found

\(\displaystyle 2 + e^{2t} + e^{-2t} = (e^t)^2 + 2 (e^t)(e^{-t}) + (e^{-t})^2 = (e^t + e^{-t} )^2 \) (Happy)

the heaven is just in front of you now .
 
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Apr 2010
58
2
Riverside, CA
Oh , the derivative of \(\displaystyle \sqrt{2} t\) is \(\displaystyle \sqrt{2} \)

so what the integral you are going to evaluate should be :

\(\displaystyle \int \sqrt{ \sqrt{2}^2 + e^{2t} + e^{-2t} }~dt \)

Trick is also found

\(\displaystyle 2 + e^{2t} + e^{-2t} = (e^t)^2 + 2 (e^t)(e^{-t}) + (e^{-t})^2 = (e^t + e^{-t} )^2 \) (Happy)

the heaven is just in front of you now .
If you don't tell anyone about this, I wont tell anyone... :)
but thank you!