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Average value of a curve

  1. Jan 22, 2006 #1
    If I had a sinusoid, how would I find the average value of it over a given interval. Say -pi/5 to pi/5 for instance. Thanks everybody.
     
  2. jcsd
  3. Jan 22, 2006 #2

    benorin

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    The average value of a function, say [tex]f(x),[/tex] over the interval [a,b] is given by the the formula

    [tex]f_{\mbox{ave}}=\frac{1}{b-a}\int_{x=a}^{b}f(x)dx[/tex]

    where I have assumed that [tex]f(x),[/tex] is properly integrable over [a,b].
     
  4. Jan 23, 2006 #3

    HallsofIvy

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    The point is: if you had a constant function, f(x)= c, the "area under the curve" from a to b would f(x)(b-a)= c(b-a). With a variable function, that area is [tex]\int_a^b f(x)dx[/tex].

    If fave is the average of the function we must have
    [tex]\int_a^b f(x)dx= f_{ave}(b-a)[/tex]
     
  5. Jan 23, 2006 #4

    mathman

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    The sine function is odd. Therefore the average over an interval symmetric around 0 will be 0.
     
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