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Math Help - curve length with ln function

  1. #1
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    curve length with ln function

    Find the length of the curve:

     y = \frac{x^2}{4} - \ln \sqrt x  for  1\leq x\leq 2<br /> <br />

    If I understand correctly, Length = integral from 1 to 2 of \sqrt {1+(\frac{dy}{dx})^2} dx

    So in this case, I need to find y' and substitute it in the formula and then find the integral.

    y =  \frac{x}{2} - \frac{1}{2}\frac{1}{x}  ==>  \frac{x^2-1}{2x}

    Then I plug in y' into the formula to get (integral from 1 to 2)
    L = \int \sqrt{1 + (\frac{x^2-1}{2x})^2} dx<br /> <br /> <br />
I'm not sure where to go from here to make the integration work. I've been stuck on this for a while, so if anyone could help, I'd appreciate it!
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  2. #2
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    After you get your derivative, square it and all that.

    \sqrt{1+\left(\frac{d}{dx}\left[\frac{x^{2}}{4}-ln(\sqrt{x})\right]\right)^{2}}=\frac{x^{2}+1}{2x}

    which is not too bad to integrate...no radical.

    Most of these things are made to simplify down with some algebra.
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  3. #3
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    am i right that dy/dx =  \frac{x^2-1}{2x}?
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  4. #4
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    Continuing on.. for the integration of x^2+1 / 2x I got:

     \frac{1}{2}\ln\mid\frac{x^2+1}{2x}\mid + C

    I did this by making u = 2x du = 2 and dx = 1/2

    Is this even close?

    If so, then since the curve length is from 1 to 2, would I input 2 and 1 and subtract to find the difference? Is the "C" ignored in this sense?

    length = .1115717757 ??
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