# Average value of a function question

## Main Question or Discussion Point

This question popped up in my high school class today. I couldn't think of an explanation that worked for the students, other than "the definition of the average value states that the function has to be continuous on the interval."

Their intuition helped them understand the definition of the average value of f(x): To find the average height of an area, take the area and divide by the width of the area. Thus, the average value of a function f(x) is the integral of f(x) from a to b, divided by b-a, or $$\frac{1}{b-a}\displaystyle\int^b_a f(x)\,dx$$

No problem with this... yet. Then, came improper integrals.
They calculated the following integral:
$$\displaystyle\int^4_1 \frac{dx}{(x-2)^\frac{2}{3}} = 3+3\sqrt[3]{2}$$

This was followed by the question, "well, even though there's an infinite discontinuity, the area over the interval is finite. Why can't we divide by the width of the interval (3) and get the average value of f(x) over that interval?"

Anyone have a better answer than I had, that they (relatively bright students) might be able to comprehend?

mathman
"the definition of the average value states that the function has to be continuous on the interval."
Why?? That seems to be an arbitrary restriction.

It didn't seem that arbitrary to me, because I recognize the problem that would exist with asymptotes. I had glanced at each of the calculus texts that I have in my classroom; all had essentially the same definition:
If f is continuous on [a,b], then the average value (or mean value) of f on [a,b] is defined to be: ...

This has simply left me scratching my head, but I know it shouldn't. I'm leaning toward explaining this way, "if you take the function f(x) = 1/x, from 1 to infinity, and you generate a solid by rotating it about the x-axis, then the volume of such a solid is finite, but the surface area is infinite. It's kind of the same idea: there is no average y-value although the area is finite."

In this case the students are absolutely correct, your result divided by the measure (width) of the interval does indeed give the average value of the function.

Think of it as a weighted average: even though the function is unbounded it is only "very large" over an interval that is "very small" i.e. the greater values of the function acount for less of the width of the domain. That this function would have a (weighted by width) average value should come as no surpise, since integration is all about getting finite information (numbers) from the infinite.

Note: since your function was undefined at x = 2, it is incorrect to say that the function is not continuous (it is continuous at every point in its domain).

Thanks, Crosson, that really puts my mind at ease. I wanted to agree with the students, but at the same time, I was having trouble wrapping my mind around infinity.

But, I don't agree with the statement that it's incorrect to say that the function is not continuous on that interval.

But, I don't agree with the statement that it's incorrect to say that the function is not continuous on that interval.
Please, do not confuse your students! We do not discuss continuity at points where the function fails to be defined! (Just say the function is unbounded on [1,4]-{2} , and that any attempt to define the function at the value x = 2 will produce a discontinuous function).

The reason it is incorrect to say the function is 'not continuous' is because the function is continuous everywhere that it is defined!

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Gib Z
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Ahh yes That is something I forgot looking at this question as well, just because the function is not defined somewhere, does not mean it isn't continuous :) Because you could still draw a graph over a limited domain, without taking your pencil off the paper :)

Or if you prefer, for every x in the domain we have f(x) is equal to 'the limit from the left' and the 'limit from the right'. Hence I say the function is continuous on its entire domain!

HallsofIvy
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The textbook I am looking at, Anton's Calculus, only requires that f be integrable on an interval in order that its "average" exist.

matt grime
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Ahh yes That is something I forgot looking at this question as well, just because the function is not defined somewhere, does not mean it isn't continuous :) Because you could still draw a graph over a limited domain, without taking your pencil off the paper :)
PLEASE don't use that description of continuity. It is SOOO wrong it is making me use CAPS and extra letters in a word.

This was followed by the question, "well, even though there's an infinite discontinuity, the area over the interval is finite. Why can't we divide by the width of the interval (3) and get the average value of f(x) over that interval?"

Anyone have a better answer than I had, that they (relatively bright students) might be able to comprehend?
It might help to look briefly at the proof of the fundamental theorem of calculus (which is fairly straightforward). There is a certain point at which you need to assume that the integrand has a maximum and minimum value on the interval defined by the limits of integration. In fact you need to multiply the integral by both of these values. Obviously it doesn't work when the maximum value of the function is infinity. Therefore, when an integral's integrand has an infinite discontinuity over the interval, the fundamental theorem of calculus doesn't apply.

Gib Z
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PLEASE don't use that description of continuity. It is SOOO wrong it is making me use CAPS and extra letters in a word.

>.< Sorry about that, simplistic ways for me to understand, and surprisingly it says that in alot of textbooks...I probably wouldn't understand any other definition anyway, with my simplistic mind...unless its Crossons about the limits, i get that..

The limit definition is considered completely rigorous (provided limits had been defined, which is difficult to do but yields no more understanding then does the intuitive understanding). So if you understand the limit definition, you understand continuity (of real functions of a single real variable).

If you read the textbook carefully they probably say something like "Any function you can draw without taking the pen of the paper is continuous" (true) but are careful not to say "Any continuous function can be drawn without taking your pen of the paper" (false).

Gib Z
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O I see, that seems correct. A continuous function that can't be drawn without taking your pen off the paper would be, eg, The Floor Function? Its defined at all points, but you have to take your pen off.

HallsofIvy
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O I see, that seems correct. A continuous function that can't be drawn without taking your pen off the paper would be, eg, The Floor Function? Its defined at all points, but you have to take your pen off.
No, it wouldn't. The floor function is defined at all points but is not continuous at the integers.

A continuous function that can't be drawn without taking your pen off the paper
Try drawing $$x Sin(\frac{1}{x})$$ on (0,1).

Gib Z
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Im going to go and cry now >.< I'm not good at this..

So it is continuous where it is defined, but not defined at zero?

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If you want, define as a special case that function is 0 at x = 0 (then the function is even continuous at the point 0). The point is that we can't draw the function because it is oscillating infinitely rapidly near the origin, yet it is continuous every where.

Gib Z
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We'll, say i used the pencil definition, Couldn't it apply if I had an infinite length of time to do it? I know even that long time couldn't draw a line of infinite length...but you get what i mean.

Would it be correct to say the function is not differentiable near zero? It looks like it going to be sort of like a fractal.

Gib Z
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Or at least would it be correct to say the derivative function will also oscillate rapidly near zero? It happens to be x(cos (1/x))-1/(x^2) + sin (1/x).

We'll, say i used the pencil definition, Couldn't it apply if I had an infinite length of time to do it?
If you start at 0, then as soon as you move the pencil either up or down the drawing is already in accurate.

Would it be correct to say the function is not differentiable near zero?
Yes that is correct. Interestingly $$x^2 Sin(\frac{1}{x})$$ is differentiable at 0.

Gib Z
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I wonder what it is about the extra factor of x that stabilizes it...Near zero it looks like the line y=0.

matt grime
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@Gib Z.

Firstly, what makes you think you can 'draw' any function? What about functions from R^6 to R^8? What about functions from Q to R?

Anyway, the trivial example of why 'pencil and paper' is silly as a definition, is the example:

f(x)=0 if x^2<2
f(x)=1 if x^2>2

as a function from Q to Q.

let's assume that we draw the graph as s subset of R^2. There's a whopping great jump at sqrt(2), but that ain't in Q, so there's no where where it is discontinuous.

Alright, back to continuity.
I've always used the definitions in the textbook I have in class, which are identical to the definitions I learned 20 years ago.
Definition: A function f is said to be continuous at a point c if the following conditions are satisfied:
1. f(c) is defined
2. lim (as x approaches c) f(x) exists
3. lim (as x approaches c) f(x) = f(c)

Later, in the same book, Theorem: A rational function is continuous everywhere except at the points where the denominator is zero.

I have no problem saying that a function is continuous over its domain; however, I still don't find it wrong to say that the function I mentioned a few posts up, is *not* continuous over that interval. i.e. f(x) = (x^2 -4)/ (x-2); it seems that you're saying "well, 2 isn't in the domain of f(x). Exactly! That's why the function isn't continuous at 2. You could call f(x) a continuous function, because it's continuous over its entire domain, but it's not continuous on the interval [0,4], because 2 isn't in its domain. Am I missing something??

Try drawing $$x Sin(\frac{1}{x})$$ on (0,1).
But, if it makes you feel better, that's one of my favorite functions to get the students really thinking (or to really twist their minds); especially when I use a piecewise definition to fill in the point f(0)=0