Length of the curve integral, can't solve the integral

  • Thread starter Prometheos
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  • #1
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This question was on my test I have no idea how to do the middle work.

Find the length of the curve
[tex] y = \frac{1}{2}(e^x + e^{-x}) , 0 \leq x \leq 2 [/tex]

Problem set up was easy enough
[tex] L= \int_0^2 \sqrt{ 1 + \frac{1}{4}( e^{2x} -2 + e^{-2x} ) } dx [/tex]

Looking back in my notes I see that the answer is
[tex] \frac{1}{2}( e^2 - e^{-2} ) [/tex]

But, how do you get there? I think my main problem is probably the algebra behind combing the 1 and derivative of y squared.
 

Answers and Replies

  • #2
Dick
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Combine the 1 and (1/4)(-2) constants under the radial. You can now rearrange into a perfect square. Remember e^(2x)=(e^x)^2.
 
  • #3
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Ah, so you get [tex] \frac{1}{4}e^
{2x} + \frac{1}{2} + \frac{1}{4}e^{-2x} [/tex]
after distributing the .25 and adding the one which is equal to
[tex] \frac{1}{4} ( e^{x} + e^{-x} )^2 [/tex]
Wow, high school algebra comes back to haunt me again.

Thanks for the help, I may have missed this problem on the test, but hopefully it won't happen again now lol
 

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