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Integral calculus help

  1. Jul 25, 2010 #1
    Hello all,

    I wanted to know whether the following two relations are same.
    \int x\;f(x)\;{\rm d}x = {\rm A}
    \int (x-{\rm A})\;f(x)\;{\rm d}x = 0
    are same?
    'A' is some positive number, distribution function f(x) is normalized to 1. (i am trying to understand some details of first moment of f(x) and got this doubt.)
  2. jcsd
  3. Jul 25, 2010 #2
    Looks the same as long as the integration limits are such that the integration covers the entire range of normalization. I'm not a mathematician, but I'm bothered by the unspecified integration limits.
  4. Jul 25, 2010 #3
    Integration covers entire range in normalization. And for both equations the same integration limits.
    f(x) is a normalized function. In the above 2 equation the same f(x) is used.
    edit:Stevenb, I solved, both are same.
    Last edited: Jul 25, 2010
  5. Jul 26, 2010 #4
    Looks to me like the A in the 2nd [lower] equation needs to be at least d/dx A or the derivative of A.
    Then by the Fundamental Theorem of Calculus, the integral of the derivative of A is = A.

    Even then the algebra looks incorrect.

    The second equation should be
    Integral [ x f(x) ] - [ d/dx A ] dx NOT
    Integral ( x - A ) f(x) dx

    This is using the rule that the integral of a sum is the sum of the integrals due to
    the linearity of integration.
    Last edited: Jul 27, 2010
  6. Jul 27, 2010 #5
    Hi paulfr,

    do you mean this :

    \int \left( x-\frac{\rm dA}{{\rm d}{x}}\right)\;f(x)\;{\rm d}x = 0

  7. Jul 27, 2010 #6


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    Science Advisor

    If that is a definite integral and [math]\int_a^b f(x)dx= 1[/math] then it is true that
    [tex]\int_a^b (x- A)f(x) dx= \int_a^b xf(x)dx- A\int f(x)dx= \int_a^b xf(x)dx- A= 0[/tex]
    [tex]\int_a^b xf(x) dx= A[/tex].
  8. Jul 27, 2010 #7
    I never thought that it will be very simple as you show. thanks.
  9. Jul 27, 2010 #8
    No I did not mean that.

    My original post was wrong.
    Please see the correction.
    The A constant is not multiplied by f(x).
  10. Jul 27, 2010 #9
    Your first line here implies that the integral of f(x) dx = 1
    How do you arrive at that ?
  11. Jul 27, 2010 #10

    Gib Z

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    Homework Helper

    HallsofIvy didn't specify but he intended that [a,b] is the entire range of distribution, and since f(x) is normalized, by definition [tex]\int_a^b f(x) dx = 1[/tex]
  12. Jul 27, 2010 #11
    Hi Paulfr,
    I already informed that my function is normalized to 1. So integral of f(x)=1.
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