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MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Today we're going to
hold off just a little bit on boiling water. And talk about another
application of integrals, and we'll get to the witches'
cauldron in the middle. The that I'd like to start with
today is average value. This is something that I
mentioned a little bit earlier, and there was a
misprint on the board, so I want to make sure that we have
the definitions straight. And also the reasoning
straight. This is one of the most
important applications of integrals, one of the most
important examples. If you take the average
of a bunch of numbers, that looks like this. And we can view this as
sampling a function. As we would with
Riemann's sum. And what I said last week was
that this tends to this expression here, which is called
the continuous average. So this guy is the continuous
average. Or just the average of f. And I want to explain that, just
to make sure that we're all on the same page. In general, if you have a
function and you want to interpret the integral, our
first interpretation was that it's something like the
area under the curve. But average value is another
reasonable interpretation. Namely, if you take equally
spaced points here, starting with x0, x1, x2, all the way
up to xn, which is the left point b, and then we have values
y1, which = f ( x1), y2, which = f ( x2), all the way
up to yn, which = f ( xn). And again, the spacing here
that we're talking about is b - a / n. So remember that spacing,
that's going to be the connection that we'll draw. Then the Riemann sum is y1
through yn, the sum of (y1 ... yn) delta x. And that's what tends, as delta
x goes to 0, to the integral. . The only change in point of view
if I want to write this limiting property, which is
right above here, the only change between here and here
is that I want to divide by the length of the
interval. b - a. So I will divide
by b - a here. And divide by b - a over here. And then I'll just check what
this thing actually is. Delta x / b - a, what
is that factor? Well, if we look over here to
what delta x is, if you divide by b - a, it's 1 / n. So the factor delta
x / b - a = 1 / n. That's what I put over here, the
sum of y1 through yn / n. And as this tends to
0, it's the same as n going to infinity. Those are the same things. The average value and
the integral are very closely related. There's only this difference
that we're dividing by the length of the interval. I want to give an example
which is an incredibly simpleminded one, but it'll
come into play later on. So let's take the example
of a constant. And this is, I hope, will make
you slightly less confused about what I just wrote. As well as making you think that
this is as simpleminded and reasonable as
it should be. If I check what the average
value of this constant is, it's given by this relatively
complicated formula here. That is, I have to integrate
the function c. Well, it's just the
constant c. And however you do this, as an
antiderivative I was thinking of it as a rectangle, the answer
that you're going to get is c here. So work that out. The answer is c. And so the fact that the average
of c = c, which had better be the case for averages,
explains the denominator. Explains the 1 / b - a there. That's cooked up exactly so that
the average of a constant is what it's supposed to be. Otherwise we have the wrong
normalizing factor. We've clearly got a piece of
nonsense on our hands. And incidentally, it also
explains the 1 / n in the very first formula that
I wrote down. The reason why this is called
the average, or one reason why it's the right thing, is that if
you took the same constant c, for y all the way across
there n times, if you divide it by n, you get back c. That's what we mean by average
value and that's why the n is there. So that was an easy example. Now none of the examples that we
are going to give are going to be all that complicated. But they will get sort of
steadily more sophisticated. The second example is going to
be the average height of a point on a semicircle. And maybe I'll draw a picture of
the semicircle first here. And we'll just make
it the standard circle, the unit circle. So maybe I should have called
it a unit semicircle. This is the point negative
1, this is the point 1. And we're picking a point over
here and we're going to take the typical, or the average,
height here. Integrating with
respect to dx. So sort of continuously
with respect to dx. Well, what is that? Well, according to the rule,
it's 1 / b - a times - sorry, it's up here in the box. 1 / b - a, the interval from
a to b, f ( x) dx. That's 1 / + 1 - (- 1). The integral from - 1 to 1,
square root of 1 - x ^2 dx. Right, because the height is y
= , this is y = the square root of 1 - x^2. And to evaluate this is not as
difficult as it seems. This is 1/2 times this quantity
here, which we can interpret as an area. It's the area of
the semicircle. So this is the area of the
semicircle, which we know to be half the area
of the circle. So it's pi / 2. And so the answer, here the
average height, is pi/ 4. Now, later in the class and
actually not in this unit, we'll actually be able
to calculate the antiderivative of this. So in other words, we'll be
able to calculate this analytically. For right now we just have the
geometric reason why the value of this is pi / 2. And we'll do that in the fourth
unit when we do a lot of techniques of integration. So here's an example. Turns out, the average height
of this is pi / 4. Now, the next example that I
want to give introduces a little bit of confusion. And I'm not going to resolve
this confusion completely, but I'm going to try to get
you used to it. I'm going to take the average
height again. But now, with respect
to arc length. Which is usually
denoted theta. Now, this brings
up an extremely important feature of averages. Which is that you have to
specify the variable with respect to which the average
is taking place. And the answer will
be different depending on the variable. So it's not going
to be the same. Wow, can't spell the
word length here. Just like the plural of
witches the last time. We'll work on that. We'll fix all of our,
that's an ancient Gaelic word, I think. Lengh. So now, let me show you that
it's not quite the same here. It's especially exaggerated if
maybe I shift this little interval dx over to the
right-hand end. And you can see that the
little portion that corresponds to it, which is
the d theta piece, has a different length from
the dx piece. And indeed, as you come down
here, these very short portions of dx length
have much longer portions of theta length. So that the average that we're
taking when we do it with respect to theta, is going to
emphasize the low values more. They're going to be
more exaggerated. And the average should
be lower than the average that we got here. So we should expect a
different number. And it's not going to be
pi / 4, it's going to be something else. Whatever it is, it should
be smaller than pi / 4. Now, let's set up
the integral. The integral follows
the same rule. It's just 1 over the length
of the interval times the integral over the interval
of the function. That's the integral, but now
where does theta range? This time, theta goes
from 0 to pi. So the integral is
from 0 to pi. And the thing we divide
by is pi. And the integration requires
us to know the formula for the height. Which is sin theta. In terms of theta, of course. It's the same as square root of
- x^2, but it's expressed in terms of theta. So it's this. And here's our average. I'll put this up here. So that's the formula
for the height. So let's work it out. This one, we have the advantage
of being able to work out because we know the
antiderivative of sin theta. It happens with this factor
of pi, it's - cos theta. And so, that's - 1 / pi ( cos
pi), sorry. (cos pi - cos) 0. Which is - 1 / pi ( -
2), which is 2 / pi. And sure enough, if you check
it, you'll see that 2 / pi < pi / 4, because pi^2 > 8. Yeah, question. STUDENT: [INAUDIBLE] PROFESSOR: The question is
how do I get sin theta. And the answer is, on this
diagram, if theta is over here then this height is this, and
this is the angle theta, then the height is the sine. OK. Another question. STUDENT: [INAUDIBLE] PROFESSOR: The question is,
what is the first one, the first one is an average of
height, of a point on a semicircle and this one
is with respect to x. So what this reveals is that
it's ambiguous to say what the average value of something is,
unless you've explained what the underlying averaging
variable is. STUDENT: [INAUDIBLE] PROFESSOR: The next question
is how should you interpret this value. That is, what came out
of this calculation? And the answer is only
sort of embedded in this calculation itself. So here's a way of thinking of
it which is anticipating our next subject. Which is probability. Which is, suppose you
picked a number at random in this interval. With equal likelihood, one
place and another. And then you saw what height
was above that. That would be the interpretation
of this first average value. And the second one is, I picked
something at random on this circle. And equally likely, any possible
point on this circle according to its length. And then I ask what the height
of that point is. And those are just
different things. Another question. STUDENT: [INAUDIBLE] PROFESSOR: Cos pi, shouldn't
it be 0? No. cos of, it's - 1. Cos pi = - 1. Cosine, sorry. No, cos 0 = 1. cos pi = - 1. And so they cancel. That is, they don't cancel. It's - 1 - 1, which = - 2. Key point. Yeah. STUDENT: [INAUDIBLE] PROFESSOR: All right,
let me repeat. So the question was to repeat
the reasoning by which I guessed in advance that probably
this was going to be the relationship between the
average value with respect to arc length versus the average
value with respect to this horizontal distance. And it had to do with
the previous way this diagram was drawn. Which is comparing an
interval in dx with an interval in theta. A little section in theta. And when you're near the top,
they're nearly this same. That is, it's more
or less balanced. It's a little curved here,
a little different. But here it becomes
very exaggerated. The d theta lengths are much
longer than the dx lengths. Which means that importance
given by the theta of variable to these parts of the
circle is larger, relative to these parts. Whereas if you look at this
section versus this section for the dx, they give
equal weights to these two equal lengths. But here, with respect to theta,
this is relatively short and this is much larger. So, as I say, the theta
variable's emphasizing the lower parts of the
semicircle more. That's because this length
is shorter and this length is longer. Whereas these two
are the same. It's a balancing act of
the relative weights. I'm going to say that again
in a different way, and maybe this will. The lower part is more
important for theta. STUDENT: [INAUDIBLE] PROFESSOR: So the question is,
but shouldn't it have a bigger value because it's
a longer length. Never with averages. Whatever the length is,
we're always dividing. We're always compensating
by the total. We have the integral from 0 to
pi, but we're dividing by pi. Here we had the integral
from - 1 to 1, but we're dividing by 2. So we divide by something
different each time. And this is very,
very important. It's that the average of a
constant is that same constant regardless of which
one we did. So if it were a constant, we
would always compensate for the length. So the length never matters. If it's the integral from 0 to
1,000,000, or 100, let's say, 1/100 cdx, it's just the same. It's always that, it doesn't
matter how long it is. Because we compensate. That's really the difference
between an integral and an average, is that we're dividing
by the total. Now I want to introduce another
notion, which is actually what's underlying these
two examples that I just wrote down. And this is by far the one which
you should emphasize the most in your thoughts. Because it is much more
flexible, and is much more typical of real life problems.
So the idea of a weighted average is the following. You take the integral, say from
a to b, of some function. But now you multiply
by a weight. And you have to divide
by the total. And what's the total
going to be? It's the integral from a to b
of this total weighting that we have. Now, why is this
the correct notion? I'm going to explain it
to you in two ways. The first is this very
simpleminded thing that I wrote on the board there,
with the constants. What we want is the average
value of c to be c. Otherwise this makes no
sense as an average. Now, let's just look at
this definition here. And see that that's correct. If you integrate c, from a to
b, w ( x) dx, and you divide by the integral from a to b, w (
x) dx, not surprisingly, the c factors out. It's a constant. So this is c times the integral
a to b, w (x), dx, divided by the same thing. And that's why we picked it. We picked it so that these
things would cancel. And this would give c. So in the previous case, this
property explains the denominator. And again over here, it explains
the denominator. And let me just give you
one more explanation. Which is maybe a real life
pretend real life example. You have a stock which you
bought for $10 one year. And then six months later you
brought some more for $20. And then you bought
some more for $30. Now, what's the average
price of your stock? Well, it depends on how many
shares you bought. If you bought this many shares
the first time, and this many shares the second time, and
this many shares the third time, this is the total
amount that you spent. And the average price is the
total price divided by the total number of shares. And this is the discrete analog
of this continuous averaging process here. The function f now, so I use w
for weight, the function f now is the function whose values
are 10, 20 and 30. And the weightings are the
relative importance of the different purchases. So again, these wi's
are weights. There was another question. Out in the audience,
at some point. Over here, yes. STUDENT: [INAUDIBLE] PROFESSOR: Very, very
good point. So in this numerator here, the
statement is, in this example, we factored out c. But here we cannot factor
out f ( x). That's extremely important and
that is the whole point. So, in other words, the weighted
average is very interesting you have to do two
different integrals to figure it out in general. When it happens that
this is c, it's an extremely boring integral. Which in fact because, it's an
average, you don't even have to calculate at all. Factor it out and cancel these
things and never bother to calculate these two numbers. So these massive numbers
just cancel. So it's a very special property
of a constant, that it factors out. That was our first discussion,
and now with this example I'm going to go back to the heating
up of the witches' cauldron and we'll use average
value to illustrate the integral that we get in
that context as well. I remind you, let's see. The situation with the witches'
cauldron was this. The first important thing is
that there were, so this is the big cauldron here. This is the one whose height
is 1 meter and whose width is 2 meters. And it's a parabola of
revolution here. And it had about approximately
1600 liters in it. And this curve was y = x^2. And the situation that I
described at the end of last time was that the initial
temperature was T = 0 degrees Celsius. And the final temperature,
instead of being a constant temperature, we were heating
this guy up from the bottom. And it was hotter on the
bottom than on the top. And the final temperature was
given by the formula T = 100 - 30 times the height y. So at y = 0, at the
bottom, it's 100. And at the top, T
= 70 degrees. OK, so this is the final
configuration for the temperature. And the question was how
much energy do we need. So, the first observation here,
and this is the reason for giving this example, is that
it's important to realize that you want to use the method
of disks in this case. The reason, so it doesn't have
to do with, you shouldn't think of the disks first. But
what you should think of is the horizontal. We must use horizontals because
T is constant on horizontals. It's not constant
on verticals. If we set things up with shells,
as we did last time, to compute the volume
of this, then T will vary along the shell. And we will still have an
averaging problem, an integral problem along the vertical
portion. But if we do it this way,
T is constant on this whole level here. And so there's no more
calculus involved in calculating what the
contribution is of any given level. So t is constant
on horizontals. Actually, in disguise, this
is that same trick that we have here. We can factor constants
out of integrals. You could view it as an
integral, but the point is that it's more elementary
than that. Now I have to set
it up for you. And in order to do that, I
need to remember what the equation is. Which is y = x ^2. And the formula for the total
amount of energy is going to be volume times the
number of degrees. That's going to be equal to the
energy that we need here. And so let's add it up. It's the integral from 0 to 1,
and this is with respect to y. So the y level goes
from 0 to 1. This top level's y = 1, this
bottom level's y = 0. And the disk that we get, this
is the point (x, y) here, is rotated around. And its radius is x. So the thickness is dy,
and the area of the disk is pi x^2. And the thing that we're
averaging is T. Well, we're not yet averaging, we're
just integrating it. We're just adding
up the total. Now I'm just going to plug in
the various values for this. And what I'm going to get is
T, again, = 100 - 30y. And this radius is measured
up to this very end. So x^2 = y. So this is pi y dy. And this is the integral that
we'll be able to evaluate. Yeah, question. STUDENT: [INAUDIBLE] PROFESSOR: All right. Well, let's carry this out. Let's finish off the
calculation here. Let's see. This is equal to, what
does it equal to. Well, I'll put it over here. It's equal to 50 pi y ^2 -
right, because this is 100 pi y, and then there's a 30, this
is 100 pi y - 30 pi y ^2, and I have to take the
antiderivative of that. So I get 50 pi y ^2, and
I get 10 pi y ^3. Evaluate it at 0 and 1. And that is 40 pi. Now, I spent a tremendous
amount of time last time focusing on units. Because I want to tell you
how to get a realistic number out of this. And there's a subtle point here
that I pointed out last time that had to do with
changing meters to centimeters. I claim that I've treated
those correctly. So, what we have here is that
the answer is in degrees, that is Celsius, times
cubic meters. These are the correct units. And now, I can translate this
into Celsius is spelled with a C. That's interesting. Celsius. I can translate this into
units that you're more familiar with. So let's try 40 pi deg
* m ^3, and then do the conversion factors. First of all there's one calorie
per degree times a milliliter. That's one conversion. And then let's see. I'm going to have to translate
from centimeters so I have here (100 cm / m)^ 3. So these are the two conversion factors that I need. And so, I get 40 pi ( 10
^ 6), that's 100^3. And this is in calories. So how much is this? Well, it's a little better,
maybe, to do it in 40 pi * 1,000 kilocalories, because
these are the ones that you actually see on your nutrition
labels of foods. And so this number is
around 125 or so. Let's see, is that
about right? Let's make sure I've got
these numbers right. Yeah, this is about 125. 40 pi. And so one candy bar, this is
a Halloween example, so. One candy bar is about
250 kilocalories. So this is half a candy bar. So the answer to our question
is that it takes 500 candy bars to heat up this thing. OK, so that's our example. Now, yeah. Question. STUDENT: [INAUDIBLE] PROFESSOR: What does the
integral give us? This integral is, the
integral represents the following things. So the question is, what does
this integral give us. So here's the integral. Here it is, rewritten so that
it can be calculated. And what this integral
is giving us is the following thing. You have to imagine the
following idea. You've got a little chunk
of water in here. And you're going to raise is
from 0 degrees all the way up to whatever the target
temperature is. And so that little milliliter of
water, if you like, has to be raised from 0 to some number
which is a function of the height. It's something between
70 and 100 degrees. And the one right above it also
has to be raised to a temperature, although
a slightly different temperature. And what we're doing with the
integral is we're adding up all of those degrees and the
calorie represents how much it takes, one calorie represents
how much it takes to raise by 1 degree 1 milliliter
of water. One cubic centimeter of water. That's the definition
of a calorie. And we're adding it up. So in other words, each of
these cubes is one thing. And now we have to add it up
over this massive thing, which is 1600 liters. And we have a lot of different
little cubes. And that's what we did. When we glommed them
all together. That's what the integral
is doing for us. Other questions. Now I want to connect this
with weighted averages before we go on. Because that was the reason why
I did weighted averages first. I'm going to compute
also the average final temperature. So, final because this is the
interesting one, the average starting temperature's
very boring, it's 0. The average final temperature
is, individually the temperatures are different. And the answer here is it's the
integral from 0 to 1 of T pi y dy divided by the
integral from 0 to 1 of pi y dy. So this is the total
temperature, weighted appropriately to the volume of
water that's involved at that temperature, divided by the
total volume of water. And we computed these
two numbers. The number in the numerator
is what we call 40 pi. And the number in the
denominator, actually this is easier than what we did last
time with shells you can just look at this and see that it's
the area under a triangle. It's pi / 2. And so the answer here
is 80 degrees. This is the average
temperature. Note that this is a
weighted average. The weighting here is different according to the height. The weighting factor is pi y. That's the weighting factor. And that's not surprising. When y is small, there's
less volume down here. Up above, those are more
important volumes, because there's more water up at the top
of the cauldron than there is down at the bottom
of the cauldron. If you compare this to the
ordinary average, if you take the maximum temperature plus
the minimum temperature, divided by 2, that would
be 100 + 70 / 2. You would get 85 degrees. And that's bigger. Why? Because the cooler
water is on top. And the actual average, the
correct weighted average, is lower than this fake average. Which is not the true average
in this context. All right so the weighting is
that the thing is getting fatter near the top. So now I'm going to do another
example of weighted average. And this example is also very
much worth your while. It's the other incredibly
important one in interpreting integrals. And it's a very, very simple
example of a function, f. The weightings will be
different, but the functions, f, will be of a very
particular kind. Namely, the function f will
be practically a constant. But not quite. It's going to be a constant on
one interval, and then 0 on the rest. So we'll do those
weighted averages now. And this subject is called
probability. In probability, what we do, so
I'm just going to give some examples here. I'm going to pick a point to in
quotation marks at random. In the region y <
x < 1 - x ^2. That's this shape here. Well, let's draw it
right down here. For now. So, somewhere in here. Some point, (x, y). And then I need to tell you,
according to what this random really means. This is proportional to
area, if you like. So area inside of
this section. And then the question that we're
going to answer right now is, what is the chance
that, it's usually called probability, that x > 1/2. Let me show you what's
going on here. And this is always the case with
things in probability. So, first of all, we have
a name for this. This is called P ( x > 1/2). And so that's what it's called
in our notation here. And what it is, is the
probability is always equal to the part / the whole. It's a ratio just like
the one over there. And which is the part and
which is the whole. Well, in this picture, the whole
is the whole parabola. And the part is the
section x > 1/2. And it's just the ratio
of those two areas. Let's write that down. That's the integral from 1/2 to
1 of (1 - x ^2) dx, divided by the integral from - 1
to 1, (1 - x ^2 ) dx. And again, the weighting
factor here is 1 - x^2. And to be a little bit more
specific here, the starting point a = - 1 and the
endpoint b = + 1. So this is P(x < 1/2). And if you work it out,
it turns out to be 5/18, we won't do it. Yeah. STUDENT: [INAUDIBLE] PROFESSOR: What we're trying
to do with probability. So I can't repeat
your question. But I can try say, because
it was a little bit too complicated. But it was not correct, OK. What we're taking is,
we have two possible things that could happen. Either, let's put it this way. Let's make it a gamble. Somebody picks a point
in here at random. And we're trying to figure
out what your chances are of winning. In other words, the chances the
person picks something in here versus something
in there. And the interesting thing
is, so what percent of the time do you win. The answer is it's some
fraction of 1. And in order to figure that out,
I have to figure out the total area here. Versus the total of the entire,
all the way from - 1 to 1, the beginning
to the end. So in the numerator, I put
success, and in the denominator I put all
possibilities. So that, right? STUDENT: [INAUDIBLE] PROFESSOR: And that's the
interpretation of this. So maybe I didn't understand
your question. STUDENT: [INAUDIBLE] PROFESSOR: Ah, why is 1 - x
^2 the weighting factor. That has to do with how you
compute areas under curves. The curve here is
y = 1 - x ^2. And so, in order to calculate
how much area is between 1/2 and 1, I have to integrate. That's the interpretation
of this. This is the area under
that curve. This integral. And the denominator's the area
under the whole thing. OK, yeah. Another question. STUDENT: [INAUDIBLE] PROFESSOR: Ah. Yikes. It was supposed to be the same
question as over here. Thank you. STUDENT: [INAUDIBLE] PROFESSOR: This has something to
do with weighting factors. Here's the weight factor. Well, it's the relative
importance from the point of view of this probability of
these places versus those. That is, so this is a weighting
factor because it's telling me that in some sense
this number 5/18, actually that makes me think that this
number is probably wrong. Well, I'll let you
calculate it out. It looks like it should be less
than 1/4 here, because this is 1/4 of the total
distance and there's a little less in here than there
is in the middle. So in fact it probably should be
less than 1/4, the answer. STUDENT: [INAUDIBLE] PROFESSOR: The equation of
the curve is 1 - x^2. The reason why it's the
weighting factor is that we're interpreting, the question
has to do with the area under that curve. And so, this is showing us how
much is relatively important versus how much is not. This is, these parts are
relatively important, these parts are less important. According to area. Because we've said that
area is the way we're making the choice. So I don't have quite enough
time to tell you about my next example. Instead, I'm just going to
tell you what the general formula is. And we'll do our example
next time. I'll tell you what
it's going to be. So here's the general formula
for probability here. We're going to imagine that we
have a total range which is maybe going from a to b, and
we have some intermediate values x1 and x2, and then we're
going to try to compute the probability that some
variable that we picked at random occurs between
x1 and x2. And by definition, we're saying
that it's an integral. It's the integral from x1 to x2
of the weight dx, divided by the integral all the
way from a to b. Of the weight. So, again, this is the part
divided by the whole. And the relationship between
this and the weighted average that we had earlier was that the
function f ( x) is kind of a strange function. It's 0 and 1. It's just the picture, if you
like, is that you have this weighting factor. And it's going from a to b. But then in between there, we
have the part that we're interested in. Which is between x1 and x2. And it's the ratio of this inner
part to the whole thing that we're interested in. Tomorrow I'm going to try to
do a realistic example. And I'm going to tell
you what it is, but we'll take it up tomorrow. I told you it was going to be
tomorrow, but we still have a whole minute, so I'm going to
tell you what the problem is. So this is going to be a target
practice problem. You have a target here
and you're throwing darts at this target. And so you're throwing
darts at this target. And somebody is standing
next to the dartboard. Your little brother is
standing next to the dartboard here. And the question is, how likely
you are to hit your little brother. So this will, let's see. You'll see whether you
like that or not. Actually, I was the
little brother. So, I don't know which
way you want to go. We'll go either way. We'll find out next time.
