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

  1. Sep 11, 2008 #1
    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.
     
  2. jcsd
  3. Sep 11, 2008 #2

    Dick

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    Science Advisor
    Homework Helper

    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.
     
  4. Sep 12, 2008 #3
    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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