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Average value for integrals

  1. Sep 19, 2013 #1
    1. The problem statement, all variables and given/known data
    I have a question can the average value for an integral be negative. I don't see why not just checking.


    You know this evalutation f_ave = (1/b-a) ∫ f(x) dx


    2. Relevant equations

    thx

    3. The attempt at a solution
     
  2. jcsd
  3. Sep 19, 2013 #2

    pasmith

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    Homework Helper

    That is the average value of the function f on [a,b]. It can of course be negative, and will be if f(x) < 0 for all x in [a,b].
     
  4. Sep 19, 2013 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    If the average value of the function is negative, of course!

    More correctly f_ave = (1/(b-a)) ∫ f(x) dx. What you wrote would normally be interpreted
    f_ave = ((1/b)-a) ∫ f(x) dx


    Of course. Take the simplest example: f(x)= -1 for all x.
    Then [tex]\int_0^1 f(x)dx= -\int_0^1 dx= -(1- 0)= -1[/tex]
    Slightly more complicated, if f(x)= -x,
    [tex]\int_0^1 f(x)dx= -\int_0^1 xdx= -\frac{1}{2}[/tex].
     
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