# Average value of a function question

1. Jan 25, 2007

### drpizza

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?

2. Jan 25, 2007

### mathman

Why?? That seems to be an arbitrary restriction.

3. Jan 25, 2007

### drpizza

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."

4. Jan 25, 2007

### Crosson

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).

5. Jan 25, 2007

### drpizza

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.

6. Jan 25, 2007

### Crosson

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!

Last edited: Jan 25, 2007
7. Jan 25, 2007

### Gib Z

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 :)

8. Jan 25, 2007

### Crosson

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!

9. Jan 26, 2007

### HallsofIvy

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

10. Jan 26, 2007

### matt grime

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

11. Jan 26, 2007

### arunma

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.

12. Jan 26, 2007

### Gib Z

>.< 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..

13. Jan 26, 2007

### Crosson

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).

14. Jan 26, 2007

### Gib Z

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.

15. Jan 27, 2007

### HallsofIvy

Staff Emeritus
No, it wouldn't. The floor function is defined at all points but is not continuous at the integers.

16. Jan 27, 2007

### Crosson

Try drawing $$x Sin(\frac{1}{x})$$ on (0,1).

17. Jan 27, 2007

### Gib Z

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?

Last edited: Jan 27, 2007
18. Jan 27, 2007

### Crosson

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.

19. Jan 27, 2007

### Gib Z

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.

20. Jan 27, 2007

### Gib Z

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).