Operations with negative sign

  1. Mar 18, 2014 #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. Mar 18, 2014 #2

    arildno

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    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.
     
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