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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Today we're moving
on from theoretical things, from the mean value theorem,
to the introduction to what's going to occupy us for the
whole rest of the course, which is integration. So, in order to
introduce that subject, I need to introduce for
you a new notation, which is called differentials. I'm going to tell you
what a differential is, and we'll get used to
using it over time. If you have a function
which is y = f(x), then the differential of y
is going to be denoted dy, and it's by definition f'(x) dx. So here's the notation. And because y is really
equal to f, sometimes we also call it the
differential of f. It's also called the
differential of f. That's the notation,
and it's the same thing as what happens if you
formally just take this dx, act like it's a number
and divide it into dy. So it means the same thing
as this statement here. And this is more or less
the Leibniz interpretation of derivatives. Of a derivative as a ratio of
these so called differentials. It's a ratio of what are
known as infinitesimals. Now, this is kind of a vague
notion, this little bit here being an infinitesimal. It's sort of like an
infinitely small quantity. And Leibniz perfected
the idea of dealing with these intuitively. And subsequently, mathematicians
use them all the time. They're way more effective than
the notation that Newton used. You might think that
notations are a small matter, but they allow you to think
much faster, sometimes. When you have the right
names and the right symbols for everything. And in this case it made
it very big difference. Leibniz's notation was adopted
on the continent and Newton dominated in Britain
and, as a result, the British fell
behind by one or two hundred years in the
development of calculus. It was really a serious matter. So it's really well
worth your while to get used to this idea of ratios. And it comes up all over the
place, both in this class and also in
multivariable calculus. It's used in many contexts. So first of all, just
to go a little bit easy. We'll illustrate it by its
use in linear approximations, which we've already done. The picture here, which we've
drawn a number of times, is that you have some function. And here's a value
of the function. And it's coming up like that. So here's our function. And we go forward
a little increment to a place which is
dx further along. The idea of this
notation is that dx is going to replace
the symbol delta x, which is the change in x. And we won't think
too hard about-- well, this is a small quantity,
this is a small quantity, we're not going to think too
hard about what that means. Now, similarly, if you see
how much we've gone up - well, this is kind of low, so
it's a small bit here. So this distance
here is, previously we called it delta y. But now we're just
going to call it dy. So dy replaces delta y. So this is the change in
level of the function. And we'll represent it
symbolically this way. Very frequently, this just
saves a little bit of notation. For the purposes
of this, we'll be doing the same things we did
with delta x and delta y, but this is the way that
Leibniz thought of it. And he would just have
drawn it with this. So this distance here is dx
and this distance here is dy. So for an example of
linear approximation, we'll say what's 64.1,
say, to the 1/3 power, approximately equal to? Now, I'm going to carry this
out in this new notation here. The function involved is x^1/3. And then it's a
differential, dy. Now, I want to use this
rule to get used to it. Because this is what we're going
to be doing all of today is, we're differentiating, or
taking the differential of y. So that is going to be
just the derivative. That's 1/3 x^(-2/3) dx. And now I'm just going to
fill in exactly what this is. At x = 64, which is the natural
place close by where it's easy to do the evaluations, we have
y = 64^(1/3), which is just 4. And how about dy? Well, so this is a little
bit more complicated. Put it over here. So dy = 1/3 64^(-2/3) dx. And that is 1/3 * 1/16
dx, which is 1/48 dx. And now I'm going
to work out what 64 to the, whatever it is
here, this strange fraction. I just want to be very careful
to explain to you one more thing. Which is that
we're using x = 64, and so we're thinking of x
+ dx is going to be 64.1. So that means that dx
is going to be 1/10. So that's the increment
that we're interested in. And now I can carry
out the approximation. The approximation says
that 64.1^(1/3) is, well, it's approximately what
I'm going to call y + dy. Because really, the dy
that I'm determining here is determined by this linear
relation. dy = 1/48 dx. And so this is only
approximately true. Because what's really
true is that this is equal to y + delta y. In our previous notation. So this is in disguise. What this is equal to. And that's the
only approximately equal to what the linear
approximation would give you. So, really, even though I wrote
dy is this increment here, what it really is if
dx is exactly that, is it's the amount
it would go up if you went straight
up the tangent line. So I'm not going to do
that because that's not what people write. And that's not even
what they think. They're really thinking
of both dx and dy as being infinitesimally small. And here we're going to the
finite level and doing it. So this is just something
you have to live with, is a little ambiguity
in this notation. This is the approximation. And now I can just calculate
these numbers here. y at this value is 4. And dy, as I said, is 1/48 dx. And that turns out to be 4
+ 1/480, because dx is 1/10. So that's approximately 4.002. And that's our approximation. Now, let's just compare it
to our previous notation. This will serve as
a review of, if you like, of linear approximation. But what I want to emphasize
is that these things are supposed to be the same. Just that it's really
the same thing. It's just a different
notation for the same thing. I remind you the basic formula
for linear approximation is that f(x) is approximately
f(a) + f'(a) (x-a). And we're applying it in the
situation that a = 64 and f(x) = x^(1/3). And so f(a), which is
f(64), is of course 4. And f'(a), which is 1/3
a^(-2/3), is in our case 1/16. No, 1/48. OK, that's the same
calculation as before. And then our relationship
becomes x^(1/3) is approximately equal to 4 plus
1/48 times x minus a, which is 64. So look, every single number
that I've written over here has a corresponding number
for this other method. And now if I plug in the
value we happen to want, which is the 64.1, this
would be 4 + 1/48 1/10, which is just the same
thing we had before. So again, same answer. Same method, new notation. Well, now I get to use this
notation in a novel way. So again, here's the notation. This notation of differential. The way I'm going to use it
is in discussing something called antiderivative Again,
this is a new notation now. But it's also a new idea. It's one that we
haven't discussed yet. Namely, the notation that
I want to describe here is what's called the
integral of g(x) dx. And I'll denote that by a
function capital G of x. So it's, you start
with a function g(x) and you produce a
function capital G(x), which is called
the antiderivative of g. Notice there's a
differential sitting in here. This symbol, this guy here,
is called an integral sign. Or an integral, or this whole
thing is called an integral. And another name for
the antiderivative of g is the indefinite integral of g. And I'll explain to you why
it's indefinite in just-- very shortly here. Well, so let's carry
out some examples. Basically what I'd like
to do is as many examples along the lines of
all the derivatives that we derived at the
beginning of the course. In other words, in
principle you want to be able to integrate as
many things as possible. We're going to start out with
the integral of sin x dx. That's a function whose
derivative is sin x. So what function would that be? Cosine x, minus, right. It's -cos x. So -cos x differentiated
gives you sin x. So that is an
antiderivative of sine. And it satisfies this property. So this function,
G(x) = - cos x, has the property that
its derivative is sin x. On the other hand, if you
differentiate a constant, you get 0. So this answer is what's
called indefinite. Because you can also
add any constant here. And the same thing will be true. So, c is constant. And as I said, the integral
is called indefinite. So that's an explanation
for this modifier in front of the "integral". It's indefinite because
we actually didn't specify a single function. We don't get a single answer. Whenever you take the
antiderivative of something it's ambiguous up to a constant. Next, let's do some
other standard functions from our repertoire. We have the integral of x^a dx. Some power, the
integral of a power. And if you think about it, what
you should be differentiating is one power larger than that. But then you have to
divide by 1/(a+1), in order that the
differentiation be correct. So this just is the fact
that d/dx of x^(a+1), or maybe I should
even say it this way. Maybe I'll do it in
differential notation. d(x^(a+1)) = (a+1) x^a dx. So if I divide that
through by a+1, then I get the relation above. And because this is
ambiguous up to a constant, it could be any
additional constant added to that function. Now, the identity that I
wrote down below is correct. But this one is
not always correct. What's the exception? Yeah. a equals-- STUDENT: 0. PROFESSOR: Negative 1. So this one is OK for all a. But this one fails because
we've divided by 0 when a = -1. So this is only true when
a is not equal to -1. And in fact, of course,
what's happening when a = 0, you're getting 0 when you
differentiate the constant. So there's a third case
that we have to carry out. Which is the exceptional case,
namely the integral of dx/x. And this time, if
we just think back to what our-- So
what we're doing is thinking backwards here,
which a very important thing to do in math at all stages. We got all of our formulas, now
we're reading them backwards. And so this one, you
may remember, is ln x. The reason why I want to do
this carefully and slowly now, is right now I also want to
write the more standard form which is presented. So first of all, first we
have to add a constant. And please don't put
the parentheses here. The parentheses go there. But there's another formula
hiding in the woodwork here behind this one. Which is that you can also
get the correct formula when x is negative. And that turns out
to be this one here. So I'm treating the case, x
positive, as being something that you know. But let's check the
case, x negative. In order to check
the case x negative, I have to differentiate the
logarithm of the absolute value of x in that case. And that's the
same thing, again, for x negative as the derivative
of the logarithm of negative x. That's the formula,
when x is negative. And if you carry
that out, what you get, maybe I'll put
this over here, is, well, it's the chain rule. It's 1/(-x) times the derivative
with respect to x of -x. So see that there
are two minus signs. There's a -x in the
denominator and then there's the derivative
of -x in the numerator. That's just -1. This part is -1. So this -1 over
-x, which is 1/x. So the negative signs cancel. If you just keep track of this
in terms of ln(-x) and its graph, that's a function
that looks like this. For x negative. And its derivative
is 1/x, I claim. And if you just look at
it a little bit carefully, you see that the slope
is always negative. Right? So here the slope is negative. So it's going to
be below the axis. And, in fact, it's getting
steeper and steeper negative as we go down. And it's getting less and less
negative as we go horizontally. So it's going like
this, which is indeed the graph of this
function, for x negative. Again, x negative. So that's one other
standard formula. And very quickly, very often,
we won't put the absolute value signs. We'll only consider the
case x positive here. But I just want you to
have the tools to do it in case we want to
use, we want to handle, both positive and negative x. Now, let's do two more examples. The integral of sec^2 x dx. These are supposed to
get you to remember all of your differentiation
formulas, the standard ones. So this one, integral of
sec^2 dx is what? tan x. And here we have + c, all right? And then the last one of, a
couple of, this type would be, let's see. I should do at least this one
here, square root of 1 - x^2. This is another
notation, by the way, which is perfectly acceptable. Notice I've put the
dx in the numerator and the function in
the denominator here. So this one turns
out to be sin^(-1) x. And, finally, let's see. About the integral
of dx / (1 + x^2). That one is tan^(-1) x. For a little while, because
you're reading these things backwards and
forwards, you'll find this happens to you on exams. It gets slightly worse
for a little while. You will antidifferentiate when
you meant to differentiate. And you'll differentiate
when you're meant to antidifferentiate. Don't get too
frustrated by that. But eventually, you'll
get them squared away. And it actually helps
to do a lot of practice with antidifferentiations,
or integrations, as they're sometimes called. Because that will
solidify your remembering all of the
differentiation formulas. So, last bit of
information that I want to emphasize before
we go on some more complicated examples is this. It's obvious because the
derivative of a constant is 0. That the antiderivative is
ambiguous up to a constant. But it's very
important to realize that this is the only
ambiguity that there is. So the last thing that
I want to tell you about is uniqueness of antiderivatives
up to a constant. The theorem is the following. The theorem is if F' =
G', then F equals G-- so F(x) equals G(x)
plus a constant. But that means, not only that
these are antiderivatives, all these things with these
plus c's are antiderivatives. But these are the only ones. Which is very reassuring. And that's a kind of uniqueness,
although its uniqueness up to a constant, it's
acceptable to us. Now, the proof of
this is very quick. But this is a fundamental fact. The proof is the following. If F' = G', then if you take
the difference between the two functions, its derivative,
which of course is F' - G', is equal to 0. Hence, F(x) - G(x)
is a constant. Now, this is a key fact. Very important fact. We deduced it last time
from the mean value theorem. It's not a small matter. It's a very, very
important thing. It's the basis for calculus. It's the reason why
calculus make sense. If we didn't have the fact
that the derivative is 0 implied that the function
was constant, we would be done. We would have-- Calculus
would be just useless for us. The point is, the
rate of change is supposed to determine
the function up to this starting value. So this conclusion
is very important. And we already checked it
last time, this conclusion. And now just by
algebra, I can rearrange this to say that F(x) is
equal to G(x) plus a constant. Now, maybe I should leave
differentials up here. Because I want to
illustrate-- So let's go on to some trickier,
slightly trickier, integrals. Here's an example. The integral of, say,
x^3 (x^4 + 2)^5 dx. This is a function
which you actually do know how to integrate,
because we already have a formula for all powers. Namely, the integral of
x^a is equal to this. And even if it were a negative
power, we could do it. So it's OK. On the other hand, to expand the
5th power here is quite a mess. And this is just a
very, very bad idea. There's another trick for
doing this that evaluates this much more efficiently. And it's the only
device that we're going to learn now for integrating. Integration actually is much
harder than differentiation, symbolically. It's quite difficult. And
occasionally impossible. And so we have to
go about it gently. But for the purposes
of this unit, we're only going
to use one method. Which is very good. That means whenever
you see an integral, either you'll be able to divine
immediately what the answer is, or you'll use this method. So this is it. The trick is called the
method of substitution. And it is tailor-made for
notion of differentials. So tailor-made for
differential notation. The idea is the following. I'm going to to
define a new function. And it's the messiest
function that I see here. It's u = x^4 + 2. And then, I'm going to take
its differential and what I discover, if I
look at its formula, and the rule for differentials,
which is right here. Its formula is what? 4x^3 dx. Now, lo and behold with
these two quantities, I can substitute, I can
plug in to this integral. And I will simplify
it considerably. So how does that happen? Well, this integral is the
same thing as, well, really I should combine
it the other way. So let me move this over. So there are two pieces here. And this one is u^5. And this one is 1/4 du. Now, that makes it the
integral of u^5 du / 4. And that's relatively
easy to integrate. That is just a power. So let's see. It's just 1/20 u to
the-- whoops, not 1/20. The antiderivative
of u^5 is u^6. With the 1/6, so
it's 1/24 u^6 + c. Now, that's not the
answer to the question. It's almost the answer
to the question. Why isn't it the answer? It isn't the answer
because now the answer's expressed in terms of u. Whereas the problem was posed
in terms of this variable x. So we must change back
to our variable here. And we do that just
by writing it in. So it's 1/24 (x^4 + 2)^6 + c. And this is the
end of the problem. Yeah, question. STUDENT: [INAUDIBLE] PROFESSOR: The question is,
can you see it directly? Yeah. And we're going to talk about
that in just one second. OK. Now, I'm going to
do one more example and illustrate this method. Here's another example. The integral of x dx over
the square root of 1 + x^2. Now, here's another example. Now, the method of substitution
leads us to the idea u = 1 + x^2. du = 2x dx, etc. It takes about as long as
this other problem did. To figure out what's going on. It's a very similar
sort of thing. You end up integrating u^(-1/2). It leads to the
integral of u^(-1/2) du. Is everybody seeing
where this...? However, there is a
slightly better method. So recommended method. And I call this method
advanced guessing. What advanced guessing
means is that you've done enough of these
problems that you can see two steps ahead. And you know what's
going to happen. So the advanced
guessing leads you to believe that here
you had a power -1/2, here you have the
differential of the thing. So it's going to
work out somehow. And the advanced guessing allows
you to guess that the answer should be something like
this. (1 + x^2)^(1/2). So this is your advanced guess. And now you just differentiate
it, and see whether it works. Well, here it is. It's 1/2 (1 + x^2)^(-1/2) 2x,
that's the chain rule here. Which, sure enough, gives you
x over square root of 1 + x^2. So we're done. And so the answer is square
root of (1 + x^2) + c. Let me illustrate this
further with another example. I strongly recommend
that you do this, but you have to get used to it. So here's another example. e^(6x) dx. My advanced guess is e^(6x). And if I check, when
I differentiate it, I get 6e^(6x). That's the derivative. And so I know that
the answer, so now I know what the answer is. It's 1/6 e^(6x) + c. Now, OK, you could,
it's also OK, but slow, to use a substitution,
to use u = 6x. Then you're going to get
du = 6dx, dot, dot, dot. It's going to work, it's
just a waste of time. Well, I'm going to give
you a couple more examples. So how about this one. x e^(-x^2) dx. What's the guess? Anybody have a guess? Well, you could also correct. So I don't want you to
bother - yeah, go ahead. STUDENT: [INAUDIBLE] PROFESSOR: Yeah, so you're
already one step ahead of me. Because this is too easy. When they get more
complicated, you just want to make this guess here. So various people have said
1/2, and they understand that there's 1/2 going here. But let me just show
you what happens, OK? If you make this guess
and you differentiate it, what you get here is
e^(-x^2) times the derivative of negative 2x, so that's -2x. x^2, so it's -2x. So now you see that you're off
by a factor of not 2, but -2. So a number of you
were saying that. So the answer is
-1/2 e^(-x^2) + c. And I can guarantee
you, having watched this on various problems, that
people who don't write this out make arithmetic mistakes. In other words, there
is a limit to how much people can think ahead
and guess correctly. Another way of doing
it, by the way, is simply to write
this thing in and then fix the coefficient by doing
the differentiation here. That's perfectly OK as well. All right, one more example. We're going to integrate
sin x cos x dx. So what's a good
guess for this one? STUDENT: [INAUDIBLE] PROFESSOR: Someone
suggesting sin^2 x. So let's try that. Over 2 - well, we'll get the
coefficient in just a second. So sin^2 x, if I differentiate,
I get 2 sin x cos x. So that's off by a factor of 2. So the answer is 1/2 sin^2 x. But now I want to
point out to you that there's another way
of doing this problem. It's also true that if
you differentiate cos^2 x, you get 2 cos x (-sin x). So another answer is that the
integral of sin x cos x dx is equal to -1/2 cos^2 x + c. So what is going on here? What's the problem with this? STUDENT: [INAUDIBLE] PROFESSOR: Pardon me? STUDENT: [INAUDIBLE] PROFESSOR: Integrals
aren't unique. That's part of the-- but
somehow these two answers still have to be the same. STUDENT: [INAUDIBLE] PROFESSOR: OK. What do you think? STUDENT: If you add them
together, you just get c. PROFESSOR: If you add
them together you get c. Well, actually,
that's almost right. That's not what you
want to do, though. You don't want to add them. You want to subtract them. So let's see what happens
when you subtract them. I'm going to ignore the
c, for the time being. I get sin^2 x, 1/2 sin^2
x - (-1/2 cos^2 x). So the difference between
them, we hope to be 0. But actually of
course it's not 0. What it is, is it's 1/2 sine
squared plus cosine squared, which is 1/2. It's not 0, it's a constant. So what's really going on here
is that these two formulas are the same. But you have to understand
how to interpret them. The two constants, here's
a constant up here. There's a constant, c_1
associated to this one. There's a different constant,
c_2 associated to this one. And this family of functions
for all possible c_1's and all possible c_2's, is
the same family of functions. Now, what's the relationship
between c_1 and c_2? Well, if you do the
subtraction, c_1 - c_2 has to be equal to 1/2. They're both constants,
but they differ by 1/2. So this explains,
when you're dealing with families of things, they
don't have to look the same. And there are lots
of trig functions which look a little different. So there can be several formulas
that actually are the same. And it's hard to check that
they're actually the same. You need some trig
identities to do it. Let's do one more example here. Here's another one. Now, you may be thinking,
and a lot of people are, thinking ugh,
it's got a ln in it. If you're experienced,
you actually can read off the
answer just the way there were several people
who were shouting out the answers when we were doing
the rest of these problems. But, you do need to relax. Because in this case, now
this is definitely not true in general when
we do integrals. But, for now, when
we do integrals, they'll all be manageable. And there's only one method. Which is substitution. And in the substitution
method, you want to go for the
trickiest part. And substitute for that. So the substitution
that I proposed to you is that this should
be, u should be ln x. And the advantage that that
has is that its differential is simpler than itself. So du = dx / x. Remember, we use that in
logarithmic differentiation, too. So now we can express this
using this substitution. And what we get is,
the integral of, so I'll divide the
two parts here. It's 1 / ln x, and
then it's dx / x. And this part is 1 /
u, and this part is du. So it's the integral of du / u. And that is ln u + c. Which altogether, if I put back
in what u is, is ln (ln x) + c. And now we see
some uglier things. In fact, technically
speaking, we could take the
absolute value here. And then this would be
absolute values there. So this is the type of
example where I really would recommend that you
actually use the substitution, at least for now. All right, tomorrow
we're going to be doing differential equations. And we're going to
review for the test. I'm going to give you a handout
telling you just exactly what's going to be on the test. So, see you tomorrow.
