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Sum of ordinates mean value of functions

  1. Nov 18, 2014 #1
    I am having trouble deciphering the opening gambit of an explanation of mean values of functions. It begins as follows:

    "Consider the part of the curve y = f(x) for values of x in the range a ≤ x ≤ b."

    A graph is shown with a curve cutting the x axis at c with a shaded positive area bounded by the curve and the line x=a to the left of c and a shaded negative area bounded by the curve and the line x = b to the right of c.

    "The mean value of y in this range is the average value of y for that part of the curve.
    The sum of the ordinates (ie values of y) between x= a and x = c occupies the shaded area above the x axis and is positive.
    This area is ∫acy dx
    Hence the sum of the ordinates between x = a and x= c is ∫acy dx"

    I understand that an ordinate is the value of y. But are the ordinates taken at integer values of x or continuous values of x. I don't see how the sum of ordinates is equal to the value of the area under the curve. I must have misunderstood the definition of the sum of the ordinates.
    I can see how the sum of the continuous ordinates multiplied by change in x as change in x goes to 0 might equal the area under the curve.

    Sorry if my terminology and description of the graph leave a lot to be desired.
  2. jcsd
  3. Nov 18, 2014 #2

    Stephen Tashi

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    Science Advisor

    It's indeed meaningless to speak of the sum of all the y coordinates of the graph of a function that is defined on interval of real numbers. unless you define "sum" to be something besides an ordinary arithmetic sum.

    To argue the relation between an integral and a mean value in a better way, consider that "mean mass per unit length" is defined by a relation such as (total length of interval )(mean mass per unit length) = total mass

    Think of a f(x) as being "mass density" Then you just need to understand why the integral of a mass density over an interval is the total mass of the interval.

    You book could have said "Think of [itex] f(x) [/itex] as being a density of something. Then [itex] \int_a^b f(x) dx [/itex] is the total something in the interval [itex] [a,b] [/itex] and [itex] \frac{ \int_a^b f(x) dx}{ [b-a]} [/itex] is the mean something per unit length = the mean density.
  4. Nov 20, 2014 #3
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