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Integral Average Value?

  1. Apr 30, 2010 #1
    Hey peoplez o:)

    In the Fundamental Theorem of Calculus;

    [tex] \Delta F \ = \ F(b) \ - \ F(a) \ and \ \Delta x \ = \ b \ - \ a [/tex]

    we can rewrite this as;

    [tex] \Delta F \ = \ \int_{a}^{b} f (x)\,dx [/tex]

    Then if we multiply both sides by 1/Δx we get;

    [tex] \frac{\Delta F}{ \Delta x} \ = \ \frac{1}{b \ - \ a} \int_{a}^{b} f (x)\,dx [/tex]

    This is called the Average of the function f.

    What does this mean?

    I was always extremely bad at any form of statistics because I didn't understand it but maybe now I'll get it.

    Would this be like those graphs of average rainfall throughout the year where they showed the 12 months and the rainfall in each month and you had to find the average for the year?

    It just makes very little sense to me and I don't know what it's good for.
     
  2. jcsd
  3. Apr 30, 2010 #2

    mathman

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    It is the average in the ordinary sense for something which has a continuum of values, rather than just a discrete set.
     
  4. Apr 30, 2010 #3
    So if [itex]x[/itex] is time, say in days, and at each time [itex]x[/itex] we write [itex]f(x)[/itex] for the instantaneous rainfall rate, say in inches per day, a=midnight preceding January 1, b=midnight following December 31. Then your formula tells the average rainfall for the year, in inches per day. So the integral is the total rainfall for the year, and [itex]b-a[/itex] is the number of days in the year.
     
  5. Apr 30, 2010 #4
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