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Welcome to Calculus. I'm Professor Ghrist.

Â We're about to begin Lecture 39 on Averages.

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

Â middle. That doesn't sound very mathematical, but

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

Â average of a function. Averages are both ubiquitous and

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

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

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

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

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

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

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

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

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

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

Â see. What we actually mean when we say

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

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

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

Â or are equal. Let's say about there.

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

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

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

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

Â bar. Integration is linear, so we can move the

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

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

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

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

Â Now, you've probably seen what comes next.

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

Â a. And so, one typically writes f bar equal

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

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

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

Â of 1. And why is this so much better?

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

Â consider an arbitrary integration domain, d.

Â And still have a nice definition for the average fr.

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

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

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

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

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

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

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

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

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

Â classical functions monomials, exponentials, and logarithms.

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

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

Â your head. What do you get?

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

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

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

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

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

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

Â happens in the exponential function happens right at the end.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â This time involving blood flow through a tubular vessel.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â constant little r. That is 2 pi rdr.

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

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

Â And so, what we have left is the numerator.

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

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

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

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

Â outside. And then, what do we have left?

Â some big constant times a simple integral..

Â We have to integrate big R squared times little rdr.

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

Â subtract off the integral of little r cubed.

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

Â get? Well, we get something that looks a

Â little complicated at first, but is not so bad.

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

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

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

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

Â 11:18

For our last example consider what happens when you plug something into the

Â wall with alternating currents, the voltage is going to be sinusoidal.

Â The voltage is a function of time is going to be some constant, Vp times sine

Â omega t. This constant, Vp, is the amplitude or

Â peak voltage. Omega is giving you some sort of

Â frequency of oscillation. The period is going to be defined as 2 pi

Â over omega. Now, the question is what is the average

Â voltage. Well, the problem is over every period if

Â we compute the average. We wind up integrating a sine function

Â over an entire period from 0 to 2 pi, as it were, I'll let you do the computation

Â to see that the average voltage is 0. That is totally useless for our purposes.

Â So, what we do is define a different type of average.

Â This is called the root mean square in some circles.

Â Sometimes, it's called the quadratic mean.

Â The root mean square, it is defined as follows.

Â Square your function, f, take the average of the square and then take the square

Â root of that. So, this square root, the average of the

Â square will give us something that is non-zero, because that squaring makes all

Â of the negative terms positive. In the case that we looked at before, the

Â root mean square voltage, V RMS is what? We square the voltage, Vp squared times

Â sin squared omega t. Then we average that, taking the integral

Â from 0 to 2 pi over omega, and dividing by that length, 2 pi over omega then

Â square rooting. Now, this integral is not so bad, we can

Â use the double angle formula and take advantage of the fact that we're

Â integrating of a full period so that that cosine term integrates to 0.

Â Then, we're left with the square root of omega over 2 pi times Vp squared times 1

Â half t as t goes from 0 to 2pi over omega.

Â Well, that evaluates simply enough, and wonderful to say, cancels the period, and

Â we're left with Vp over the square root of 2.

Â Now, in practice that works out to about 71% of the peak voltage.

Â That's what the route mean square is. So if you see something that is labeled

Â at a 120 volts, what you're really getting is that root mean square voltage.

Â The peak voltage is actually bigger at almost 170 volts.

Â This lesson should lead you to wonder what other kinds of things can be

Â averaged. Can you average locations or points on a

Â map? What happens when birds fly in a flock?

Â They seem to do so with a great deal of coordination.

Â What they are often doing is simply averaging the orientations or the

Â directions of their neighbors? Fish do similar things.

Â What other kinds of things can be averaged?

Â Can we average things like faces? Can we take a collection of human faces

Â and compute an average? In what way would that be like an

Â integral? These are all excellent questions that

Â with enough mathematics, you can answer. We now know how to average a function

Â over a domain and we've seen hints that there are more interesting types of

Â averages out there. In our next lesson, we'll look at what it

Â means to average a collection of locations or positions.

Â Some of the integrals involved are going to be a bit intricate.

Â You might want to take a look at the Lecture 31 Bonus Material before

Â beginning. [BLANK_AUDIO]

Â