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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: So we're going
on to the third unit here. So we're getting
started with Unit 3. And this is our
intro to integration. It's basically the
second half of calculus after differentiation. Today what I'll talk
about is what are known as definite integrals. Actually, it looks
like, are we missing a bunch of overhead lights? Is there a reason for that? Hmm. Let's see. Aah. All right. OK, that's a little
brighter now. All right. So the idea of
definite integrals can be presented in
a number of ways. But I will be consistent with
the rest of the presentation in the course. We're going to start with
the geometric point of view. And the geometric
point of view is, the problem we want to solve is
to find the area under a curve. The other point of
view that one can take, and we'll mention that at
the end of this lecture, is the idea of a cumulative sum. So keep that in mind that
there's a lot going on here. And there are many
different interpretations of what the integral is. Now, so let's draw
a picture here. I'll start at a place
a and end at a place b. And I have some curve here. And what I have in mind
is to find this area here. And, of course, in
order to do that, I need more information
than just where we start and where we end. I also need the
bottom and the top. By convention, the bottom
is the x axis and the top is the curve that we've
specified, which is y = f(x). And we have a notation
for this, which is the notation using
calculus for this as opposed to some geometric notation. And that's the
following expression. It's called an
integral, but now it's going to have what are
known as limits on it. It will start at a and end at b. And we write in the
function f(x) dx. So this is what's known
as a definite integral. And it's interpreted
geometrically as the area under the curve. The only difference between
this collection of symbols and what we had before
with indefinite integrals is that before we didn't
specify where it started and where it ended. Now, in order to understand
what to do with this guy, I'm going to just describe
very abstractly what we do. And then carry out
one example in detail. So, to compute this
area, we're going to follow initially three steps. First of all, we're going
to divide into rectangles. And unfortunately, because
it's impossible to divide a curvy region into rectangles,
we're going to cheat. So they're only
quote-unquote rectangles. They're almost rectangles. And the second thing we're going
to do is to add up the areas. And the third thing
we're going to do is to rectify this problem
that we didn't actually hit the answer on the nose. That we were missing
some pieces or were choosing some extra bits. And the way we'll
rectify that is by taking the limit as
the rectangles get thin. Infinitesimally thin, very thin. Pictorially, again,
that looks like this. We have a and our b, and
we have our guy here, this is our curve. And I'm going to chop it up. First I'm going to chop up the
x axis into little increments. And then I'm going to
chop things up here. And I'll decide on
some rectangle, maybe some staircase pattern here. Like this. Now, I don't care so much. In some cases the rectangles
overshoot; in some cases they're underneath. So the new area that
I'm adding up is off. It's not quite the same as
the area under the curve. It's this region here. But it includes these
extra bits here. And then it's missing
this little guy here. This little bit
there is missing. And, as I say, these
little pieces up here, this a little bit
up here is extra. So that's why we're
not really dividing up the region into rectangles. We're just taking rectangles. And then the idea is that as
these get thinner and thinner, the little itty-bitty amounts
that we miss by are going to 0. And they're going
to be negligible. Already, you can see it's
kind of a thin piece of area, so we're not missing by much. And as these get thinner and
thinner, the problem goes away and we get the answer on
the nose in the limit. So here's our first example. I'll take the first interesting
curve, which is f(x) = x^2. I don't want to do anything more
complicated than one example, because this is a
real labor here, what we're going to go through. And to make things
easier for myself, I'm going to start at a = 0. But in order to see
what the pattern is, I'm going to allow
b to be arbitrary. Let's draw the graph and
start breaking things up. So here's the
parabola, and there's this piece that we
want, which is going to stop at this place, b, here. And the first step is
to divide into n pieces. That means, well, graphically,
I'll just mark the first three. And maybe there are
going to be many of them. And then I'll draw
some rectangles here, and I'm going to choose
to make the rectangles all the way from the right. That is, I'll make us
this staircase pattern here, like this. That's my choice. I get to choose
whatever level I want, and I'm going to
choose the right ends as the shape of the staircase. So I'm overshooting
with each rectangle. And now I have to
write down formulas for what these areas are. Now, there's one big advantage
that rectangles have. And this is the starting place. Which is that it's easy
to find their areas. All you need to know is
the base and the height, and you multiply,
and you get the area. That's the reason why we can
get started with rectangles. And in this case,
these distances, I'm assuming that they're
all equal, equally spaced, intervals. And I'll always be doing that. And so the spacing, the bases,
the base length, is always b/n. All equal intervals. So that's the base length. And next, I need the heights. And in order to keep
track of the heights, I'm going to draw a little
table here, with x and f(x), and plug in a few values just
to see what the pattern is. The first place here,
after 0, is b/n. So here's b/n,
that's an x-value. And the f(x) value
is the height there. And that's just, I
evaluate f(x), f(x) is x^2. And that's (b/n)^2. And similarly, the
next one is 2b/n. And the value here is (2b/n)^2. That's this. This height here is 2b/n. That's the second rectangle. And I'll write down one more. 3b/n, that's the third one. And the height is (3b/n)^2. And so forth. Well, my next job is
to add up these areas. And I've already prepared
that by finding out what the base and the height is. So the total area, or the
sum of the areas, let's say, of these rectangles, is-- Well,
the first one is (b/n) (b/n)^2. The second one is 2b/n --
I'm sorry, is (b/n) (2b/n)^2. And it just keeps on going. And the last one is
(b/n) (nb / n)^2. So it's very important
to figure out what the general formula is. And here we have a base. And here we have a
height, and here we have the same kind of base,
but we have a new height. And so forth. And the pattern is that the
coefficient here is 1, then 2, then 3, all the way up to n. The rectangles are
getting taller and taller, and this one, the last
one is the biggest. OK, this is a very
complicated gadget. and the first thing I
want to do is simplify it and then I'm actually
going to evaluate it. But actually I'm not going
to evaluate it exactly. I'm just going to
evaluate the limit. Turns out, limits
are always easier. The point about calculus here is
that these rectangles are hard. But the limiting value
is an easy value. So what we're heading for is
the simple formula, as opposed to the complicated one. Alright, so the first
thing I'm going to do is factor out all
these b/n factors. There's a b/n, here, and
there's a (b/n)^2, So all told, we have a (b/n)^3. As a common factor. And then the first term
is 1, and the second term, what's left over, is 2^2. 2^2. And then the third
term would be 3^2, although I haven't written it. In the last term, there's
an extra factor of n^2. In the numerator. OK, is everybody with me here? Now, what I'd like to do is
to eventually take the limit as n goes to infinity here. And the quantity that's
hard to understand is this massive quantity here. And there's one change
that I'd like to make, but it's a very modest one. Extremely minuscule. Which is that I'm
going to write 1, just to see that there's
a general pattern here. Going to write 1 as 1^2. And let's put in
the 3 here, why not. And now I want to use a trick. This trick is not
completely recommended, but I will say a
lot more about that when we get through to the end. I want to understand how
big this quantity is. So I'm going to use a
geometric trick to draw a picture of this quantity. Namely, I'm going
to build a pyramid. And the base of the pyramid
is going to be n by n blocks. So imagine we're in Egypt
and we're building a pyramid. And the next layer is
going to be n-1 by n-1. So this next layer
in is n-1 by n-1. So the total number of blocks
on the bottom is n squared. That's this rightmost term here. But the next term, which I
didn't write in but maybe I should, the next-to-the-last
term was this one. And that's the second
layer that I've put on. Now, this is, if you
like, the top view. But perhaps we should also
think in terms of a side view. So here's the same picture,
we're starting at n and we build up this layer here. And now we're going to put
a layer on top of it, which is a little shorter. So the first layer
is of length n. And the second layers is of
length n-1, and then on top of that we have something
of length n-2, and so forth. And we're going to pile them up. So we pile them up. All the way to the top, which is
just one giant block of stone. And that's this last one, 1^2. So we're going
backwards in the sum. And so I have to build
this whole thing up. And I get all the way up
in this staircase pattern to this top block, up there. So here's the trick
that you can use to estimate the size
of this, and it's sufficient in the limit
as n goes to infinity. The trick is that I can
imagine the solid thing underneath the
staircase, like this. That's an ordinary pyramid,
not a staircase pyramid. Which is inside. And this one is inside. And so, but it's an
ordinary pyramid as opposed to a staircase pyramid. And so, we know the formula
for the volume of that. Because we know the formula
for volumes of cones. And the formula for the volume
of this guy, of the inside, is 1/3 base times height. And in that case, the
base here-- so that's 1/3, and the base is n by n, right? So the base is n^2. That's the base. And the height, it goes all
the way to the top point. So the height is n. And what we've discovered
here is that this whole sum is bigger than 1/3 n^3. Now, I claimed that - this
line, by the way has slope 2. So you go 1/2 over
each time you go up 1. And that's why you
get to the top. On the other hand, I can
trap it on the outside, too, by drawing a
parallel line out here. And this will go down 1/2
more on this side and 1/2 more on the other side. So the base will be n+1 by
n+1 of this bigger pyramid. And it'll go up 1 higher. So on the other end, we get that
this is less than 1/3 (n+1)^3. Again, (n+1)^2 times n+1, again,
this is a base times a height. Of this bigger pyramid. Yes, question. STUDENT: [INAUDIBLE] and
then equating it to volume. PROFESSOR: The
question is, it seems as if I'm adding up areas
and equating it to volume. But I'm actually
creating volumes by making these honest
increments here. That is, the base is
n but the height is 1. Thank you for pointing that out. Each one of these
little staircases here has exactly height 1. So I'm honestly
sticking blocks there. They're sort of square blocks,
and I'm lining them up. And I'm thinking of n
by n cubes, if you like. Honest cubes, there. And the height is 1. And the base is n^2. Alright, so I claim that I've
trapped this guy in between two quantities here. And now I'm ready
to take the limit. If you look at what
our goal is, we want to have an
expression like this. And I'm going to-- This was the
massive expression that we had. And actually, I'm going
to write it differently. I'll write it as b^3 times
1^2 plus 2^2 plus... plus n^2, divided by n^3. I'm going to combine
all the n's together. Alright, so the
right thing to do is to divide what
I had up there. Divide by n^3 in this set
of inequalities there. And what I get here is 1/3 is
less than 1 plus 2^2 plus 3^2 plus... plus n^2 divided by n^3
is less than 1/3 (n+1)^3 / n^3. And that's 1/3 + (1 + 1/n)^3. And now, I claim we're done. Because this is
1/3, and the limit, as n goes to infinity,
of this quantity here, is easily seen to be, this
is, as n goes to infinity, this goes to 0. So this also goes to 1/3. And so our total here, so
our total area, under x^2, which we sometimes might write
the integral from 0 to b x^2 dx, is going to be
equal to - well, it's this 1/3 that I've got. But then there was
also a b^3 there. So there's this
extra b cubed here. So it's 1/3 b^3. That's the result from
this whole computation. Yes, question. STUDENT: [INAUDIBLE] PROFESSOR: So that was
a very good question. The question is, why did
we leave the b/n^3 out, for this step. And a part of the answer
is malice aforethought. In other words, we know
what we're heading for. We know, we understand,
this quantity. It's all one thing. This thing is a sum, which
is growing larger and larger. It's not what's
called a closed form. So, the thing that's not
known, or not well understood, is how big is this
quantity here. 1^2 + 2^2, the sum
of the squares. Whereas, this is
something that's quite easy to understand. So we factor it out. And we analyze carefully the
piece which we don't know yet, how big it is. And we discovered that it's
very, very similar to n^3. But it's more
similar to 1/3 n^3. It's almost
identical to 1/3 n^3. This extra piece here. So that's what's going on. And then we match that. Since this thing is very
similar to 1/3 n^3 we cancel the n^3's and we
have our result. Let me just mention that
although this may seem odd, in fact this is what
you always do if you analyze these kinds of sum. You always factor
out whatever you can. And then you still are
faced with a sum like this. So this will happen
systematically, every time you're faced with such a sum. OK, now I want to say one
more word about notation. Which is that this notation
is an extreme nuisance here. And it's really sort of too
large for us to deal with. And so, mathematicians
have a shorthand for it. Unfortunately, when you
actually do a computation, you're going to end up with
this collection of stuff anyway. But I want to just show you
this summation notation in order to compress it a little bit. The idea of summation
notation is the following. So this thing tends-- The
ideas are the following. I'll illustrate it
with an example first. So, the general notation is
the sum of a_i, i = 1 to n, is a_1 plus a_2 plus
dot dot dot plus a_n. So this is the abbreviation. And this is a capital sigma. And so, this quantity
here, for instance, is 1/n^3 times the
sum i^2, i = 1 to n. So that's what this
thing is equal to. And what we just showed
is that that tends to 1/3 as n goes to infinity. So this is the way the
summation notation is used. There's a formula for each
of these coefficients, each of these entries
here, or summands. And then this is
just an abbreviation for what the sum is. And this is the reason
why I stuck in that 1^2 at the beginning, so that you
could see that the pattern worked all the
way down to i = 1. It isn't an exception
to the rule. It's the same as
all of the others. Now, over here, in
this board, we also had one of these
extremely long sums. And this one can be written
in the following way. And I hope you agree, this
is rather hard to scan. But one way of writing it is,
it's the sum from i = 1 to n of - now I have to write down the
formula for the general term. Which is b/n (ib/n)^2. So that's a way of abbreviating
this massive formula into one which is just a lot shorter. And now, the manipulation
that I performed with it, which is to factor
out this (b/n)^3, is something that I'm perfectly
well allowed to do also over here. This is the distributive law. This, if I factor out b^3 /
n^3, I'm left with the sum, i = 1 to n, of i^2, right? So these notations make it
a little bit more compact. What we're dealing with. The conceptual phenomenon
is still the same. And the mess is really still
just hiding under the rug. But the notation is-- at
least fits with fewer symbols, anyway. So let's continue here. I've given you one calculation. And now I want to fit
it into a pattern. And here's the thing that
I'd like to calculate. So, first of all let's
try the case-- S I'm going to do two more examples. I'll do two more
examples, but they're going to be much, much easier. And then things are going to
get much easier from now on. So, the second example is going
to be the function f(x) = x. If I draw that, that's
this function here, that's the line with slope 1. And here's b. And so this area
here is the same as the area of the triangle
with base b and height b. So the area is equal to 1/2
b * b, so this is the base. And this is the height. We also know how to find
the area of triangles. And so, the formula is 1/2 b^2. And the third example--
Notice, by the way, I didn't have to do this
elaborate summing to do that, because we happen
to know this area. The third example is going
to be even easier. f(x) = 1. By far the most
important example. Remarkably, when you get to
18.02, multivariable calculus, you will forget
this calculation. Somehow. And I don't know why, but
it happens to everybody. So, the function is just
horizontal, like this. Right? It's the constant 1. And if we stop it at b, then
the area we're interested in is just this, from 0 to b. And we know that this is
height 1, so this is area is base, which is b, times 1. So it's b. Let's look now at the pattern. We're going to look at the
pattern of the function, and it's the area
under the curve, which is this-- has this
elaborate formula in terms of-- so this is just the
area under the curve. Between 0 and b. And we have x^2, which
turned out to be b^3 / 3. And we have x, which
turned out to be-- well, let me write them over just a
bit more to give myself some room. x, which turns
out to be b^2 / 2. And then we have 1,
which turned out to be b. So this, I claim, is suggestive. If you can figure
out the pattern, one way of making it a little
clearer is to see that x is x^1. And 1 is x^0. So this is the case, 0, 1 and 2. And b is b^1 / 1. So, if you want to guess what
happens when f(x) is x^3, well if it's 0, you do b^1 /
1; if it's 1, you do b^2 / 2; if it's 2, you do b^3 / 3. So it's reasonable to guess
that this should be b^4 / 4. That's a reasonable
guess, I would say. Now, the strange thing is that
in history, Archimedes figured out the area under a parabola. So that was a long time ago. It was after the pyramids. And he used, actually, a
much more complicated method than I just described here. And his method, which is
just fantastically amazing, was so brilliant that it may
have set back mathematics by 2,000 years. Because people were
so-- it was so difficult that people couldn't
see this pattern. And couldn't see that, actually,
these kinds of calculations are easy. So they couldn't
get to the cubic. And even when they
got to the cubic, they were struggling
with everything else. And it wasn't until
calculus fit everything together that people were
able to make serious progress on calculating these areas. Even though he was the expert on
calculating areas and volumes, for his time. So this is really a great
thing that we now can have easy methods of doing it. And the main thing that I
want to tell you is that's we will not have to labor to
build pyramids to calculate all of these quantities. We will have a way
faster way of doing it. This is the slow, laborious way. And we will be able to do it
so easily that it will happen as fast as you differentiate. So that's coming up tomorrow. But I want you to know that
it's going to be-- However, we're going to go through just
a little pain before we do it. And I'll just tell you one
more piece of notation here. So you need to have a
little practice just to recognize how much
savings we're going to make. But never again will
you have to face elaborate geometric
arguments like this. So let me just add a
little bit of notation for definite integrals. And this goes under the
name of Riemann sums. Named after a mathematician
from the 1800s. So this is the general procedure
for definite integrals. We divide it up into pieces. And how do we do that? Well, so here's our
a and here's our b. And what we're going to do is
break it up into little pieces. And we're going to give
a name to the increment. And we're going to
call that delta x. So we divide up into these. So how many pieces are there? If there are n pieces,
then the general formula is always the delta x is
1/n times the total length. So it has to be (b-a) / n. We will always use
these equal increments, although you don't
absolutely have to do it. We will, for these Riemann sums. And now there's only
one bit of flexibility that we will allow ourselves. Which is this. We're going to pick any height
of f between-- in the interval, in each interval. So what that means is,
let me just show it to you on the picture here. Is, I just pick any
value in between, I'll call it c_i,
which is in there. And then I go up here. And I have the level,
which is f(c_i). And that's the
rectangle that I choose. In the case that
we did, we always chose the right-hand,
which turned out to be the largest one. But I could've chosen
some level in between. Or even the left-hand end. Which would have meant
that the staircase would've been quite a bit lower. So any of these staircases
will work perfectly well. So that means were picking
f(c_i), and that's a height. And now we're just going
to add them all up. And this is the sum of the
areas of the rectangles, because this is the height. And this is the base. This notation is
supposed to be, now, very suggestive of the
notation that Leibniz used. Which is that in the limit,
this becomes an integral from a to b of f(x) dx. And notice that the delta
x gets replaced by a dx. So this is what
happens in the limit. As the rectangles get thin. So that's as delta x goes to 0. And these gadgets are
called Riemann sums. This is called a Riemann sum. And we already worked
out an example. This very complicated guy was
an example of a Riemann sum. So that's a notation. And we'll give you
a chance to get used to it a little
more when we do some numerical work at the end. Now, the last
thing for today is, I promised you an example
which was not an area example. I want to be able to show
you that integrals can be interpreted as cumulative sums. Integrals as cumulative sums. So this is just an example. And, so here's the way it goes. So we're going to
consider a function f, we're going to consider a
variable t, which is time. In years. And we'll consider
a function f(t), which is in dollars per year. Right, this is a
financial example here. That's the unit here,
dollars per year. And this is going to
be a borrowing rate. Now, the reason why I
want to put units in here is to show you that
there's a good reason for this strange dx, which
we append on these integrals. This notation. It allows us to
change variables, it allows this to be
consistent with units. And allows us to develop
meaningful formulas, which are consistent across the board. And so I want to
emphasize the units in this when I set up this
modeling problem here. Now, you're borrowing
money, let's say, every day. So that means delta t = 1/365. That's almost 1 /
infinity, from the point of view of various purposes. So this is how much
you're borrowing. In each time increment
you're borrowing. And let's say that you borrow--
your rate varies over the year. I mean, sometimes you need more
money sometimes you need less. Certainly any business
would be that way. And so here you are,
you've got your money. And you're borrowing
but the rate is varying. And so how much did you borrow? Well, in day 45, which
corresponds to t is 45/365, you borrowed the
following amount. Here was your borrowing
rate times this quantity. So, dollars per year. And so this is, if
you like-- I want to emphasize the scaling
that comes about here. You have dollars per year. And this is this
number of years. So that comes out
to be in dollars. This final amount. This is the amount that
you actually borrow. So you borrow this amount. And now, if I want to
add up how much you get-- you've borrowed in
the entire year. That's this sum. i = 1 to 365
of f of, well, it's (i / 365) delta t. Which I'll just leave
as delta t here. This is total amount borrowed. This is kind of a messy sum. In fact, your bank probably
will keep track of it and they know how to do that. But when we're modeling things
with strategies, you know, trading strategies,
of course, you're really some kind of
financial engineer and you want to cleverly
optimize how much you borrow. And how much you spend,
and how much you invest. This is going to be very,
very similar to the integral from 0 to 1 of f(t) dt. At the scale of 1/35,
it's probably-- 365, it's probably enough
for many purposes. Now, however,
there's another thing that you would want to model. Which is equally important. This is how much you
borrowed, but there's also how much you owe the back
at the end of the year. And the amount that you owe the
bank at the end of the year, I'm going to do
it in a fancy way. It's, the interest, we'll say,
is compounded continuously. So the interest rate,
if you start out with P as your principal, then after
time t you owe-- So borrow P, after time t, you owe P e^(rt),
where r is your interest rate. Say .05 per year. That would be an example
of an interest rate. And so, if you want to
understand how much money you actually owe at
the end of the year. At the end of the
year what you owe is, well, you borrowed
these amounts here. But now you owe more
at the end of the year. You owe e^r times the amount
of time left in the year. So the amount of time left
in the year is 1 - i / 365. Or 365 - i days left. So this is 1 - i / 365. And this is what you have to
add up, to see how much you owe. And that is essentially
the integral from 0 to 1. The delta t comes out. And you have here e^(r(1-t)),
so the t is replacing this i / 365, f(t) dt. And so when you start computing
and thinking about what's the right strategy, you're faced
with integrals of this type. So that's just an example. And see you next time. Remember to think
about questions that you'll ask next time.
