Sum of ordinates mean value of functions

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SUMMARY

The discussion centers on the concept of the mean value of functions, specifically addressing the relationship between the sum of ordinates and the area under a curve defined by the function y = f(x) over the interval [a, b]. The sum of the ordinates, represented as ∫_a^c f(x) dx, corresponds to the area above the x-axis between x = a and x = c. The confusion arises from the terminology "sum of ordinates," which may mislead readers into thinking of discrete sums rather than the continuous nature of integration. The discussion highlights that the integral of a function represents the total area, which can be interpreted as the mean density over the interval.

PREREQUISITES
  • Understanding of integral calculus, specifically the Fundamental Theorem of Calculus.
  • Familiarity with the concept of mean value and average value of functions.
  • Knowledge of the relationship between area under a curve and definite integrals.
  • Basic grasp of mathematical terminology, including ordinates and density functions.
NEXT STEPS
  • Study the Fundamental Theorem of Calculus to solidify understanding of integration and its applications.
  • Learn about the concept of average value of a function using the formula (1/(b-a)) ∫_a^b f(x) dx.
  • Explore the relationship between density functions and integrals in physical contexts, such as mass density.
  • Review resources on integral calculus, such as the Khan Academy's section on average function value.
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Students of calculus, educators teaching integral calculus, and anyone seeking to clarify the relationship between integration and mean values of functions.

Appleton
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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.
 
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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.
 

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