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Thread: Arc Length of Polynomials

  1. #1
    Mar 2014
    Riverside, CA

    Question Arc Length of Polynomials

    I need to find the arc length of ((x^2)/2 -ln(x)/4) from 2 to 4.

    Using the arc length formula, I get that L = ∫ds = ∫ sqrt(1+(f'(x))^2) dx =

    ∫ sqrt (1 + ((x^3)/6 -1/(4x))^2) dx from x=2 to x=4,

    but I cannot solve it any further.

    I cannot use u-substitution and I have tried using trig-substitution but it does not seem to work either.

    The answer given by WolframAlpha is (1/4)(24 +ln(2)) which is correct, but I want to know how it computed that. I understand that the ln(2) part came from ln(4) - ln(2) = ln(4/2) but beyond that I am lost.
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  2. #2
    MHF Contributor
    Prove It's Avatar
    Aug 2008

    Re: Arc Length of Polynomials

    If $\displaystyle \begin{align*} f(x) = \frac{x^2}{2} - \frac{\ln{(x)}}{4} \end{align*}$ then $\displaystyle \begin{align*} f'(x) = x - \frac{1}{4x} \end{align*}$. It looks like you integrated instead of differentiating...
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