Welcome to Calculus. I'm Professor Ghrist.

We're about to begin Lecture 39 on Averages.

What's an average? An average is something that's in the

middle. That doesn't sound very mathematical, but

that intuition will help you as we consider what it means to take the

average of a function. Averages are both ubiquitous and

intuitive. If someone were to ask, say, what is the

average unemployment rate. Well, you might get out the graph and

eyeball it. Try to pick a number that's sort of in

the middle. That's the way it feels like it ought to

be. Odd, of course, an average depends on a

couple of factors. First of all, it depends on the interval

over which you are looking. If in the case of unemployment you change

your time interval, then you might have very, very different result.

And in fact, the function that you are trying to average might look very, very

different and as the variation of that function increases it becomes harder to

see. What we actually mean when we say

average, and so, we need a definition. Now, our intuition says that the average

of a function f over an interval from a to b.

Should be the constant value where the area above and the area below balance out

or are equal. Let's say about there.

We would denote that average by the symbol f bar, the bar over the top

meaning the average. We can write out an integral form for

this intuitive definition, namely that the integral as x, goes from a to b of f

minus f bar dx equals 0. Now, let us solve this equation for f

bar. Integration is linear, so we can move the

integral of f bar over to the right-hand side.

But, by definition, f bar is supposed to be a constant, just a number, so we can

pull it outside the integral. And now, solving for f bar, we get the

integral of f dx over the integral of dx, both integrals running from a to b.

Now, you've probably seen what comes next.

Anyway, we evaluate the denominator to be x evaluated from a to b, that is b minus

a. And so, one typically writes f bar equal

to 1 over b minus a times the integral from a to b of f dx.

That is a great definition. It's fine, but it's not optimal.

The better definition is in terms, the ratio, the integral of f, to the integral

of 1. And why is this so much better?

Well, instead of your integration domain being the interval from a to b, we could

consider an arbitrary integration domain, d.

And still have a nice definition for the average fr.

For example, if D is a discrete set and your f is really just a sequence of

values, like say test scores, then you know, the classical formula for the

average. In this case, it's the sum of these f

values divided by n, the number of scored, but of course we could rewire it

that as the sum of the f values divided by the sum of 1.

As you're going from 1 to n that's giving you your denominator, and this really

fits into the same category of definition since a sum is really just an integral

for a discrete set. Now, let's look at a few examples.

Let's compute the average over the integral from zero to T of three

classical functions monomials, exponentials, and logarithms.

For the monomial, for x to the n, what we have to do is integrate x to the n dx

from 0 to t, and then divide by T. This is simple, you can really do it in

your head. What do you get?

You get T to the n. That is, the function evaluated at the

right-hand endpoint divided by n plus 1. That's the average value of x to the n

over this interval. What about the exponential, e to the x

goes beyond polynomial growth. Well, this too is an integral that we can

do in our head, but notice what you get. You get the right-hand endpoint, e to the

T minus 1 divided not by any n but by T. So it's as if the all the growth that

happens in the exponential function happens right at the end.

Now, lastly, for the logarithm, we're going to have to change our lower

integration value to 1 rather than 0. But notice what happens when we integrate

log of x. We get x log x minus x through

integration-like parts if you like, when we evaluate this.

We see that the average value for a log of x on the interval from 1 to T is

exactly log of T minus a little bit. This is again, telling you how slowly log

of x grows it's average value is almost equal to the right-hand endpoint.

That's pretty cool. Let's do another example.

What is the density of the earth? Well, by which I mean the average

density, since it changes. Recall, we know a little something about

the volumetric density function. Rho is a function of r radial distance.

Now, you might be tempted to look at this graph and, and try to eyeball it, and

figure out the average density there. But, be careful, we have to use this

integral formulation rho r is not the integral of rho of r dr over the integral

of dr. That is not what it is, because of

course, rho is a volumetric density. And so, our integration must be done with

respect to the volume form, rho bar is integral of rho dV over the integral of 1

time dV. Now, rho dV as we recall is really just

the mass element, and so, in retrospect, it's really kind of obvious that what we

would do to compute the average density is compute the mass, the integral of dM,

over the volume, the integral of dV. And you can write that out more

explicitly if you wish. Let's turn to another example.

This time involving blood flow through a tubular vessel.

There is something called Poiseuille's Law that tells you how the velocity of

the fluid varies with respect to location in the cylindrical vessel.

If we look at a cross section that is going to be a disc of radius capital R.

There's some maximum velocity, but the velocity is going to be a function of the

radial distance, little r. This is going to be a quadratic function,

Poiseuille's Law says that the velocity, V, is a function of radial distance of R

is P over 4 mu l times quantity big R squared minus little r squared.

In this case, P is a pressure, mu is a viscosity, l is the length of the vessel,

but don't worry about all that stuff, it's just constants in this case.

What we are going to worry about is the average velocity, vr.

Now, in this case, what do we integrate with respect to?

It's got to be with respect to the area element, since we're taking a cross

sectional area. v bar is the integral of vdA over the

integral of dA. In this case, dA is an annular strip with

constant little r. That is 2 pi rdr.

Now, plugging in our formula for V, what do we get?

Well, first of all, the denominator gives us a pi r squared, and that, we know.

And so, what we have left is the numerator.

That is the integral. As little r goes from 0 to big R of this

constant P over 4 mu l times quantity big R squared minus little r squared times

the area element 2 pi rdr. And we can simplify that integral quite a

bit pulling out p over 4 mu l constant, canceling the pi's, pulling the two

outside. And then, what do we have left?

some big constant times a simple integral..

We have to integrate big R squared times little rdr.

That gives big R squared, the little r squared over 2, and then we have to

subtract off the integral of little r cubed.

That's little r to the fourth over 4. Evaluate from 0 to big R and what do we

get? Well, we get something that looks a

little complicated at first, but is not so bad.

Some of the big Rs, and the coefficients cancel, and we're left with P over mu l

times big R squared over 2. If we consider our initial velocity

profile, we see that this is really just 1 half times the maximal velocity at the

center of the tube. That's a nice result.