Operations with negative sign

  1. While working on an integration problem I found that I will arrive at two different solutions depending on how I approach it.

    I'm finding the arc length of y=ln(1-x2) on the interval [0,0.5]

    The formula for finding the arc length is ∫sqrt[1+[f'(x)]2]dx

    So f'(x) = -2x / ( 1-x2 )

    Here I first simplify this to 2x / ( x2 - 1 ) and squaring gives

    4x2 / ( x2 -1 )2

    Working from here I end up integrating from 0 to 0.5

    ∫ [1 + 1/(x-1) - 1/(x+1)] dx = 0.5 - ln3

    On the other hand if I leave f'(x) as it is without simplifying, when I squared f'(x) I get

    4x2 / ( 1-x2 )2

    and end up integrating from 0 to 0.5

    ∫ [1 + 1/(1+x) + 1/(1-x)] dx = -0.5 - ln3

    Should both have the same solution or is this simply a possible effect from squaring numbers?

    Thank you
  2. jcsd
  3. arildno

    arildno 11,265
    Science Advisor
    Homework Helper
    Gold Member

    Note that when taking the square root of the perfect square in your denominator, you must use the ABSOLUTE value, |x^2-1| as your new denominator.
    1 person likes this.
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