Sum of ordinates mean value of functions

[Appleton]

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 ∫_a^cy dx
Hence the sum of the ordinates between x = a and x= c is ∫_a^cy 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.

 

[Stephen Tashi]

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 f(x) as being a density of something. Then \int_a^b f(x) dx is the total something in the interval [a,b] and \frac{ \int_a^b f(x) dx}{ [b-a]} is the mean something per unit length = the mean density.

 

[Appleton]

Thanks for your explanation. I found this explanation pretty straightforward:

https://www.khanacademy.org/math/in...alue/v/average-function-value-closed-interval

Which leaves me quite puzzled as to why my book deployed the rather confusing "sum of ordinates" term.