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Question about integral and Abs function

  1. Feb 8, 2010 #1

    I want to find


    I know the method for separating the integral on two integrals, but I want to try second method.

    I know that


    But in this case I got F(2)-F(0)=-2 and on the first method 5/2.

    Why the second method does not work?

    Thanks in advance.
  2. jcsd
  3. Feb 8, 2010 #2


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    What makes you think that the last equation is true?
    I don't see where the x(x - 3) comes from, and how taking the derivative of sgn(2x-3) x (x - 3) is |2x - 3| :S
  4. Feb 8, 2010 #3
    It is.

    Let |2x-3|=[itex]\sqrt{(2x-3)^2}[/itex]


    Now taking the derivative:

    [tex]\frac{(\sqrt{(2x-3)^2}(x-3)x)'*(2x-3) - \sqrt{(2x-3)^2}(x-3)x*(2x-3)'}}{(2x-3)^2}[/tex]

    Also you can check it by first

    x>= 3/2

    [tex](x-3)x=x^2 -3x[/tex]




    x< 3/2




    This is a lot work to do here, so I check it again with Mathematica 5.0 and it is true. But still can't see the problem.
  5. Feb 8, 2010 #4


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    You can't ignore the fact that there is a discontinuity at x = 3/2. The proper computation is

    [tex]\left.\mbox{sgn}(2x-3)(x-3)x\right|^{2}_{3/2^+} + \left.\mbox{sgn}(2x-3)(x-3)x\right|^{3/2^-}_{0},[/tex]
    where 3/2^+ means infinitesimally above 3/2 and 3/2^- means infinitesimally below 3/2, such that in the first term sgn(2*x-3) = 1, while in the second it is -1. You then get

    [itex]-2 - (-3/2)(3/2) - ((-3/2)(3/2) - 0) = 5/2[/itex], which is your desired result.
  6. Feb 9, 2010 #5
    Thanks for the reply. I think I understand.

    [tex]A_1(x)=(x-3)x[/tex] when x>=3/2


    [tex]A_2(x)=-x(x-3)[/tex] when x<3/2

    Where A1(x) and A2(x) represent the area of the interval [0,x]

    But if I want to find the area on the interval [0,2]

    Now I find the area of [0,3/2] + (area [0,2] - area [0, 3/2])

    First I must find A2(3/2)+(A1(2) - A2(3/2))=A1(2)
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