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PROFESSOR: Today we're going
to continue with integration.

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And we get to do the probably
the most important thing

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of this entire course.
 

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Which is appropriately named.
 

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It's called the fundamental
theorem of calculus.

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And we'll be abbreviating it
FTC and occasionally I'll put

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in a 1 here, because there
will be two versions of it.

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But this is the one that
you'll be using the

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most in this class.
 

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The fundamental theorem of
calculus says the following.

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It says that if F' = f, so F' (
x) = little f ( x), there's a

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capital F and a little f, then
the integral from a to b of

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f ( x) = F ( b) - F (a).
 

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That's it.
 

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That's the whole theorem.
 

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And you may recognize it.
 

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Before, we had the notation
that f was the antiderivative,

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that is, capital F was
the integral of f(x).

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We wrote it this way.
 

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This is this
indefinite integral.

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And now we're putting
in definite values.

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And we have a connection
between the two uses

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of the integral sign.
 

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But with the definite values,
we get real numbers out

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instead of a function.
 

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Or a function up to a constant.
 

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So this is it.
 

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This is the formula.
 

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And it's usually also written
with another notation.

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So I want to introduce that
notation to you as well.

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So there's a new notation here.
 

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Which you'll find
very convenient.

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Because we don't always
have to give a letter f to

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the functions involved.
 

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So it's an abbreviation.
 

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For right now there'll be
a lot of f's, but anyway.

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So here's the abbreviation.
 

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Whenever I have a difference
between a function at two

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values, I also can write this
as F ( x) with an a down

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here and a b up there.
 

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So that's the notation
that we use.

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And you can also, for emphasis,
and this sometimes turns out to

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be important, when there's more
than one variable floating

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around in the problem.
 

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To specify that the
variable is x.

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So this is the same
thing as x = a.

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And x = b.
 

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It indicates where you
want to plug in, what

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you want to plug in.
 

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And now you take the top value
minus the bottom value.

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So F ( b) - F(a).
 

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So this is just a notation, and
in that notation, of course,

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the theorem can be written with
this set of symbols here.

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Equally well.
 

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So let's just give a
couple of examples.

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The first example is the
one that we did last

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time very laboriously.
 

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If you take the function
capital F(x), which happens

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to be x^3 / 3, then if you
differentiate it, you get,

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well, the the factor
of 3 cancels.

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So you get x^2, that's
the derivative.

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And so by the fundamental
theorem, so this implies by the

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fundamental theorem, that the
integral from say, a to b of

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x^3 over - sorry, x^2 dx,
that's the derivative here.

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This is the function we're
going to use as f ( x) here

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= this function here.
 

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F ( b) - F ( a), that's here.
 

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This function here.
 

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So that's write F(b) -
F( a), and that's equal

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to b^3 / 3 - a^3 / 3.
 

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Now, in this new notation,
we usually don't have

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all of these letters.
 

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All we write is the following.
 

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We write the integral from a to
be, and I'm going to do the

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case 0 to b, because that was
the one that we actually

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did last time.
 

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So I'm going to set a = 0 here.
 

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And then, the problem we were
faced last time as this.

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And as I said we did
it very laboriously.

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But now you can see that
we can do it in ten

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seconds, let's say.
 

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Well, the antiderivative
of this is x^3 / 3.

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I'm going to evaluate it at
0 and at b and subtract.

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So that's going to be
b^3 / 3 - 0^3 / 3.

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Which of course is b^3 / 3.
 

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And that's the end,
that's the answer.

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So this is a lot faster
than yesterday.

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I hope you'll agree.
 

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And we can dispense with those
elaborate computations.

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Although there's a conceptual
reason, a very important one,

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for understanding the procedure
that we went through.

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Because eventually you're going
to be using integrals and these

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quick ways of doing things, to
solve problems like finding

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the volumes of pyramids.
 

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In other words, we're going
to reverse the process.

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And so we need to understand
the connection between the two.

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I'm going to give a
couple more examples.

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And then we'll go on.
 

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So the second example would be
one that would be quite

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difficult to do by this Riemann
sum technique that we

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described yesterday.
 

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Although it is possible.
 

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It uses much higher
mathematics to do it.

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And that is the area under one
hump of the sine curve, sine x.

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Let me just draw a
picture of that.

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The curve goes like
this, and we're talking

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about this area here.
 

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It starts out at
0, it goes to pi.

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That's one hump.
 

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And so the answer is,
it's the integral from

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0 to pi of sin x dx.
 

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And so I need to take the
antiderivative of that.

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And that's - cos x.
 

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That's the thing whose
derivative is sin x.

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Evaluating it at 0 and pi.
 

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Now, let's do this
one carefully.

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Because this is where I see a
lot of arithmetic mistakes.

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Even though this is the
easy part of the problem.

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It's hard to pay attention and
plug in the right numbers.

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And so, let's just pay
very close attention.

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I'm plugging in pi.
 

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That's - cos pi.
 

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That's the first term.
 

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And then I'm subtracting
the value at the bottom,

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which is - cos 0.
 

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There are already five
opportunities for you to make

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a transcription error or
an arithmetic mistake

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in what I just did.
 

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And I've seen all five of them.
 

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So the next one is
that this is - (- 1).

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Minus negative 1, if you like.
 

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And then this is minus,
and here's another - 1.

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So altogether we have 2.
 

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So that's it.
 

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That's the area.
 

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This area, which is hard
to guess, this is area 2.

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The third example is maybe
superfluous but I'm

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going to say it anyway.
 

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We can take the integral, say,
from 0 to 1, of x ^ 100.

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Any power, now, is
within our power.

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So let's do it.
 

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So here we have the
antiderivative is

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x ^ 101 / 101.
 

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Evaluate it at 0 and 1.
 

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And that is just 1 / 101.
 

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That's that.
 

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So that's the
fundamental theorem.

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Now this, as I say, harnesses
a lot of what we've already

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learned, all about
antiderivatives.

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Now, I want to give you an
intuitive interpretation.

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So let's try that.
 

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We'll talk about a proof of
the fundamental theorem

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a little bit later.
 

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It's not actually that hard.
 

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But we'll give an intuitive
reason, interpretation,

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if you like.
 

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Of the fundamental theorem.
 

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So this is going to be one
which is not related to

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area, but rather to
time and distance.

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So we'll consider x (t) is
your position at time t.

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And then x' (t), which is
dx/dt, is going to be what

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we know as your speed.
 

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And then what the theorem is
telling us, is the following.

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It's telling us the integral
from a to b of v ( t) dt.

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So, reading the relationship
is equal to x (b) - x ( a).

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And so this is some
kind of cumulative sum

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of your velocities.
 

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So let's interpret the
right-hand side first.

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This is the distance traveled.
 

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And it's also what you would
read on your odometer.

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Right, from the beginning
to the end of the trip.

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That's what you would
read on your odometer.

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Whereas this is what you would
read on your speedometer.

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So this is the interpretation.
 

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Now, I want to just go one
step further into this

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interpretation, to make the
connection with the Riemann

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sums that we had yesterday.
 

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Because those are very
complicated to understand.

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And I want you to understand
them viscerally on

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several different levels.
 

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Because that's how you'll
understand integration better.

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The first thing that I want to
imagine, so we're going to do a

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thought experiment now,
which is that you are

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extremely obsessive.
 

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And you're driving your car
from time a to time b, place

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q to place r, whatever.
 

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And you check your
speedometer every second.

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OK, so you've read your
speedometer in the i'th second,

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and you've read that you're
going at this speed.

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Now, how far do you
go in that second?

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Well, the answer is you go this
speed times the time interval,

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which in this case we're
imagining as 1 second.

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Alright?
 

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So this is how far you went.
 

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But this is the time interval.
 

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And this is the distance
traveled. in that a second

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number, i, in the i'th second.
 

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The distance traveled in the
i'th second, that's a total

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00:13:48 --> 00:13:49
distance you traveled.
 

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Now, what happens if you
go the whole distance?

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Well, you travel the sum
of all these distances.

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So it's some massive sum,
where n is some ridiculous

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number of seconds.
 

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3600 seconds or
something like that.

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00:14:04 --> 00:14:05
Whatever it is.
 

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And that's going to turn out to
be very similar to what you

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would read on your odometer.
 

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00:14:11 --> 00:14:13
Because during that
second, you didn't change

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velocity very much.
 

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00:14:14 --> 00:14:18
So the approximation that the
speed at one time that you

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spotted it is very similar
to the speed during

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the whole second.
 

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It doesn't change that much.
 

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00:14:24 --> 00:14:26
So this is a pretty good
approximation to how

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far you traveled.
 

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And so the sum is a very
realistic approximation

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00:14:33 --> 00:14:34
to the entire integral.
 

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Which is denoted this way.
 

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Which, by the fundamental
theorem, is exactly

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how far you traveled.
 

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So this is x ( b) - x (a).
 

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Exactly.
 

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The other one is approximate.
 

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OK, again this is
called a Riemann sum.

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Alright so that's the intro
to the fundamental theorem.

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And now what I need to do
is extend it just a bit.

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00:15:23 --> 00:15:29
And the way I'm going to
extend it is the following.

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00:15:29 --> 00:15:31
I'm going to do it on
this example first.

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00:15:31 --> 00:15:35
And then we'll do
it more formally.

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00:15:35 --> 00:15:39
So here's this example
where we went someplace.

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00:15:39 --> 00:15:44
But now I just want to draw you
an additional picture here.

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00:15:44 --> 00:15:49
Imagine I start here
and I go over to there

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and then I come back.
 

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And maybe even I
do a round trip.

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I come back to the same place.
 

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Well, if I come back to the
same place, then the position

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00:16:03 --> 00:16:06
is unchanged from the
beginning to the end.

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00:16:06 --> 00:16:08
In other words, the
difference is 0.

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00:16:08 --> 00:16:13
And the velocity, technically
rather than the speed.

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00:16:13 --> 00:16:15
It's the speed to the right and
the the speed to the left maybe

249
00:16:15 --> 00:16:17
are the same, but one of them
is going in the positive

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00:16:17 --> 00:16:19
direction and one of them is
going in the negative

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00:16:19 --> 00:16:22
direction, and they
cancel each other.

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00:16:22 --> 00:16:25
So if you have this kind
of situation, we want

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00:16:25 --> 00:16:26
that to be reflected.
 

254
00:16:26 --> 00:16:29
We like that interpretation and
we want to preserve it even

255
00:16:29 --> 00:16:35
when in the case when the
function v is negative.

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00:16:35 --> 00:16:47
And so I'm going to now extend
our notion of integration.

257
00:16:47 --> 00:17:02
So we'll extend integration
to the case f negative.

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00:17:02 --> 00:17:04
Or positive.
 

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In other words, it
could be any sign.

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Actually, there's no change.
 

261
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The formulas are all the same.
 

262
00:17:12 --> 00:17:15
We just, if this v is going
to be positive, we write

263
00:17:15 --> 00:17:16
in a positive number.
 

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If it's going to be negative,
we write in a negative number.

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And we just leave it alone.
 

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And the real, so here's, let
me carry out an example

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and show you how it works.
 

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I'll carry out the example
on this blackboard up here.

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00:17:32 --> 00:17:33
Of the sine function.
 

270
00:17:33 --> 00:17:36
But we're going to
try two humps.

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00:17:36 --> 00:17:38
We're going to try the first
hump and the one that

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00:17:38 --> 00:17:40
goes underneath.
 

273
00:17:40 --> 00:17:40
There.
 

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00:17:40 --> 00:17:44
So our example here is going
to be the integral from

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0 to 2pi of sin x dx.
 

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00:17:50 --> 00:17:54
And now, because the
fundamental theorem is so

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important, and so useful, and
so convenient, we just assume

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that it be true in
this case as well.

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00:18:01 --> 00:18:06
So we insist that this
is going to be - cos x.

280
00:18:06 --> 00:18:08
Evaluate it at 0 and 2pi.
 

281
00:18:08 --> 00:18:10
With the difference.
 

282
00:18:10 --> 00:18:14
Now, when we carry out that
difference, what we get here

283
00:18:14 --> 00:18:24
is - cos 2pi. - (- cos 0).
 

284
00:18:24 --> 00:18:33
Which is - 1 - (-
1), which is 0.

285
00:18:33 --> 00:18:39
And the interpretation of
this is the following.

286
00:18:39 --> 00:18:43
Here's our double hump,
here's pi and here's 2pi.

287
00:18:43 --> 00:18:46
And all that's happening
is that the geometric

288
00:18:46 --> 00:18:49
interpretation that we had
before of the area under

289
00:18:49 --> 00:18:52
the curve has to be taken
with a grain of salt.

290
00:18:52 --> 00:18:54
In other words, I lied to you
before when I said that the

291
00:18:54 --> 00:18:57
definite integral was the
area under the curve.

292
00:18:57 --> 00:18:59
It's not.
 

293
00:18:59 --> 00:19:02
The definite integral is the
area under the curve when it's

294
00:19:02 --> 00:19:05
above the curve, and it counts
negatively when it's

295
00:19:05 --> 00:19:08
below the curve.
 

296
00:19:08 --> 00:19:12
So yesterday, my geometric
interpretation was incomplete.

297
00:19:12 --> 00:19:19
And really just a plain lie.
 

298
00:19:19 --> 00:19:30
So the true geometric
interpretation of the definite

299
00:19:30 --> 00:19:46
integral is plus the area above
the axis, above the x

300
00:19:46 --> 00:20:00
axis, minus the area
below the x axis.

301
00:20:00 --> 00:20:02
As in the picture.
 

302
00:20:02 --> 00:20:05
I'm just writing it down in
words, but you should think

303
00:20:05 --> 00:20:08
of it visually also.
 

304
00:20:08 --> 00:20:12
So that's the setup here.
 

305
00:20:12 --> 00:20:17
And now we have the complete
definition of integrals.

306
00:20:17 --> 00:20:20
And I need to list for you a
bunch of their properties and

307
00:20:20 --> 00:20:21
how we deal with integrals.
 

308
00:20:21 --> 00:20:25
So are there any questions
before we go on?

309
00:20:25 --> 00:20:25
Yeah.
 

310
00:20:25 --> 00:20:32
STUDENT: [INAUDIBLE]
 

311
00:20:32 --> 00:20:39
PROFESSOR: Right.
 

312
00:20:39 --> 00:20:42
So the question was, wouldn't
the absolute value of the

313
00:20:42 --> 00:20:45
velocity function be involved?
 

314
00:20:45 --> 00:20:48
The answer is yes.
 

315
00:20:48 --> 00:20:51
That is, that's one question
that you could ask.

316
00:20:51 --> 00:20:54
One question you could
ask is what's the total

317
00:20:54 --> 00:20:57
distance traveled.
 

318
00:20:57 --> 00:21:01
And in that case, you would
keep track of the absolute

319
00:21:01 --> 00:21:05
value of the velocity
as you said.

320
00:21:05 --> 00:21:06
Whether it's positive
or negative.

321
00:21:06 --> 00:21:08
And then you would get
the total length of

322
00:21:08 --> 00:21:13
this curve here.
 

323
00:21:13 --> 00:21:18
That's, however, not what the
definite integral measures.

324
00:21:18 --> 00:21:21
It measures the net
distance traveled.

325
00:21:21 --> 00:21:24
So it's another thing.
 

326
00:21:24 --> 00:21:25
In other words, we can do that.
 

327
00:21:25 --> 00:21:27
We now have the
tools to do both.

328
00:21:27 --> 00:21:35
We could also, so if you like,
the total distance is equal

329
00:21:35 --> 00:21:39
to the integral of this.
 

330
00:21:39 --> 00:21:40
From a to b.
 

331
00:21:40 --> 00:21:48
But the net distance is
the one without the

332
00:21:48 --> 00:21:54
absolute value signs.
 

333
00:21:54 --> 00:21:57
So that's correct.
 

334
00:21:57 --> 00:22:03
Other questions?
 

335
00:22:03 --> 00:22:04
Alright.
 

336
00:22:04 --> 00:22:23
So now, let's talk about
properties of integrals.

337
00:22:23 --> 00:22:35
So the properties of integrals
that I want to mention

338
00:22:35 --> 00:22:39
to you are these.
 

339
00:22:39 --> 00:22:47
The first one doesn't
bear too much comment.

340
00:22:47 --> 00:22:53
If you take the cumulative
integral of a sum, you're just

341
00:22:53 --> 00:23:01
trying to get the sum of the
separate integrals here.

342
00:23:01 --> 00:23:03
And I won't say
much about that.

343
00:23:03 --> 00:23:06
That's because sums come
out, the because the

344
00:23:06 --> 00:23:07
integral is a sum.
 

345
00:23:07 --> 00:23:15
Incidentally, you know
this strange symbol here,

346
00:23:15 --> 00:23:17
there's actually a reason
for it historically.

347
00:23:17 --> 00:23:19
If you go back to old books,
you'll see that it actually

348
00:23:19 --> 00:23:21
looks a little bit
more like an s.

349
00:23:21 --> 00:23:24
This capital Sigma is a sum.
 

350
00:23:24 --> 00:23:27
S for sum, because everybody
in those days knew

351
00:23:27 --> 00:23:28
Latin and Greek.
 

352
00:23:28 --> 00:23:31
And this one is also an s,
but gradually it was such

353
00:23:31 --> 00:23:33
an important s that
they made a bigger.

354
00:23:33 --> 00:23:35
And then they stretched it out
and made it a little thinner,

355
00:23:35 --> 00:23:40
because it didn't fit into
one typesetting space.

356
00:23:40 --> 00:23:43
And so just for typesetting
reasons it got stretched.

357
00:23:43 --> 00:23:45
And got a little bit skinny.
 

358
00:23:45 --> 00:23:47
Anyway, so it's really an s.
 

359
00:23:47 --> 00:23:50
And in fact, in French
they call it sum.

360
00:23:50 --> 00:23:53
Even though we
call it integral.

361
00:23:53 --> 00:23:57
So it's a sum.
 

362
00:23:57 --> 00:24:00
So it's consistent with
sums in this way.

363
00:24:00 --> 00:24:09
And similarly, similarly we can
factor constants out of sums.

364
00:24:09 --> 00:24:20
So if you have an integral like
this, the constant factors out.

365
00:24:20 --> 00:24:25
But definitely don't try to
get a function out of this.

366
00:24:25 --> 00:24:27
That won't happen.
 

367
00:24:27 --> 00:24:30
OK, in other words, c
has to be a constant.

368
00:24:30 --> 00:24:41
Doesn't depend on x.
 

369
00:24:41 --> 00:24:44
The third property.
 

370
00:24:44 --> 00:24:46
What do I want to call
the third property here?

371
00:24:46 --> 00:24:51
I have sort of a preliminary
property, yes, here.

372
00:24:51 --> 00:24:52
Which is the following.
 

373
00:24:52 --> 00:24:53
And I'll draw a picture of it.
 

374
00:24:53 --> 00:24:59
I suppose you have three
points along a line.

375
00:24:59 --> 00:25:01
So then I'm going to
draw a picture that.

376
00:25:01 --> 00:25:03
And I'm going to use the
interpretation above the

377
00:25:03 --> 00:25:05
curve, even though that's
not the whole thing.

378
00:25:05 --> 00:25:07
So here's a, here's
b and here's c.

379
00:25:07 --> 00:25:11
And you can see that the area
of this piece, of the first

380
00:25:11 --> 00:25:14
two pieces here, when added
together, gives you the

381
00:25:14 --> 00:25:15
area of the whole.
 

382
00:25:15 --> 00:25:19
And that's the rule that
I'd like to tell you.

383
00:25:19 --> 00:25:24
So if you integrate from a to
b, and you add to that the

384
00:25:24 --> 00:25:37
integral from b to c, you'll
get the integral from a to c.

385
00:25:37 --> 00:25:39
This is going to be just a
little preliminary, because

386
00:25:39 --> 00:25:41
the rule is a little
better than this.

387
00:25:41 --> 00:25:47
But I will explain
that in a minute.

388
00:25:47 --> 00:25:52
The fourth rule is
a very simple one.

389
00:25:52 --> 00:26:00
Which is that the integral from
a to a of f ( x ) dx = 0.

390
00:26:00 --> 00:26:02
Now, that you can see
very obviously because

391
00:26:02 --> 00:26:04
there's no area.
 

392
00:26:04 --> 00:26:05
No horizontal movement there.
 

393
00:26:05 --> 00:26:08
The rectangle is infinitely
thin, and there's

394
00:26:08 --> 00:26:10
nothing there.
 

395
00:26:10 --> 00:26:12
So this is the case.
 

396
00:26:12 --> 00:26:17
You can also interpret
it a F ( a) - F ( a).

397
00:26:17 --> 00:26:21
So that's also consistent
with our interpretation.

398
00:26:21 --> 00:26:24
In terms of the fundamental
theorem of calculus.

399
00:26:24 --> 00:26:28
And it's perfectly reasonable
that this is the case.

400
00:26:28 --> 00:26:33
Now, the fifth property
is a definition.

401
00:26:33 --> 00:26:35
It's not really a property.
 

402
00:26:35 --> 00:26:38
But it's very important.
 

403
00:26:38 --> 00:26:46
The integral from a to b of f(
x) dx = minus the integral

404
00:26:46 --> 00:26:50
from b to a, of f( x) dx.
 

405
00:26:50 --> 00:26:56
Now, really, the right-hand
side here is an undefined

406
00:26:56 --> 00:26:58
quantity so far.
 

407
00:26:58 --> 00:27:05
We never said you could ever
do this where the a < b.

408
00:27:05 --> 00:27:09
Because this is working
backwards here.

409
00:27:09 --> 00:27:12
But we just have a convention
that that's the definition.

410
00:27:12 --> 00:27:14
Whenever we write down this
number, it's the same as

411
00:27:14 --> 00:27:17
minus what that number is.
 

412
00:27:17 --> 00:27:21
And the reason for all of these
is again that we want them to

413
00:27:21 --> 00:27:22
be consistent with the
fundamental theorem

414
00:27:22 --> 00:27:23
of calculus.
 

415
00:27:23 --> 00:27:26
Which is the thing that
makes all of this work.

416
00:27:26 --> 00:27:31
So if you notice the left-hand
side here is F ( b)

417
00:27:31 --> 00:27:34
- F ( a), capital F.
 

418
00:27:34 --> 00:27:36
The antiderivative of little f.
 

419
00:27:36 --> 00:27:40
On the other hand, the other
side is minus, and if we just

420
00:27:40 --> 00:27:42
ignore that, we say these are
letters, if we were a machine,

421
00:27:42 --> 00:27:44
we didn't know which one was
bigger than which, we just

422
00:27:44 --> 00:27:50
plugged them in, we would get
here F( a) - F( b), over here.

423
00:27:50 --> 00:27:53
And to make these two things
equal, what we want is to

424
00:27:53 --> 00:27:54
put that minus sign in.
 

425
00:27:54 --> 00:27:59
Now it's consistent.
 

426
00:27:59 --> 00:28:04
So again, these rules are
set up so that everything

427
00:28:04 --> 00:28:05
is consistent.
 

428
00:28:05 --> 00:28:11
And now I want to
improve on rule 3 here.

429
00:28:11 --> 00:28:13
And point out to you.
 

430
00:28:13 --> 00:28:16
So let me just go back
to rule 3 for a second.

431
00:28:16 --> 00:28:22
That now that we can evaluate
integrals regardless of the

432
00:28:22 --> 00:28:26
order, we don't have to have
a < b, b < c in order to

433
00:28:26 --> 00:28:28
make sense out of this.
 

434
00:28:28 --> 00:28:31
We actually have the
possibility of considering

435
00:28:31 --> 00:28:34
integrals where the a's and
the b's and the c's are

436
00:28:34 --> 00:28:36
in any order you want.
 

437
00:28:36 --> 00:28:38
And in fact, with this
definition, with this

438
00:28:38 --> 00:28:43
definition 5, 3 works no
matter what the numbers are.

439
00:28:43 --> 00:28:44
So this is much
more convenient.

440
00:28:44 --> 00:28:49
We don't, this is
not necessary.

441
00:28:49 --> 00:28:51
Not necessary.
 

442
00:28:51 --> 00:28:57
It just works using
convention 5.

443
00:28:57 --> 00:29:04
OK, with 5.
 

444
00:29:04 --> 00:29:09
Again, before I go on, let
me emphasize we really

445
00:29:09 --> 00:29:11
want to respect the
sign of this velocity.

446
00:29:11 --> 00:29:15
We really want the net
change in the position.

447
00:29:15 --> 00:29:18
And we don't want this
absolute value here.

448
00:29:18 --> 00:29:20
Because otherwise, all of our
formulas are going to mess up.

449
00:29:20 --> 00:29:22
We won't always be
able to check.

450
00:29:22 --> 00:29:25
Sometimes you have letters
rather than actual numbers

451
00:29:25 --> 00:29:28
here, and you won't know
whether a is bigger than b.

452
00:29:28 --> 00:29:31
So you'll want to know that
these formulas work and are

453
00:29:31 --> 00:29:36
consistent in all situations.
 

454
00:29:36 --> 00:29:39
OK, I'm going to
trade these again.

455
00:29:39 --> 00:29:47
In order to preserve the
ordering 1 through 5.

456
00:29:47 --> 00:29:54
And now I have a sixth property
that I want to talk about.

457
00:29:54 --> 00:30:02
This one is called estimation.
 

458
00:30:02 --> 00:30:14
And it says the following. if
f(x) <= g ( )x, then the

459
00:30:14 --> 00:30:21
integral from a to b of f(x)
dx <= the integral from

460
00:30:21 --> 00:30:28
a to b of g (x) dx.
 

461
00:30:28 --> 00:30:36
Now, this one says that if I'm
going more slowly than you,

462
00:30:36 --> 00:30:40
then you go farther than I do.
 

463
00:30:40 --> 00:30:41
OK.
 

464
00:30:41 --> 00:30:43
That's all it's saying.
 

465
00:30:43 --> 00:30:47
For this one, you'd
better have a < b.

466
00:30:47 --> 00:30:49
You need it.
 

467
00:30:49 --> 00:30:55
Because we flip the signs when
we flip the order of a and b.

468
00:30:55 --> 00:30:58
So this one, it's essential
that the lower limit be

469
00:30:58 --> 00:31:04
smaller than the upper limit.
 

470
00:31:04 --> 00:31:07
But let me just emphasize,
because we're dealing with

471
00:31:07 --> 00:31:08
the generalities of this.
 

472
00:31:08 --> 00:31:10
Actually if one of these is
negative and the other one is

473
00:31:10 --> 00:31:14
negative, then it also works.
 

474
00:31:14 --> 00:31:17
This one ends up being, if f
is more negative than g, then

475
00:31:17 --> 00:31:22
this added up thing is more
negative than that one.

476
00:31:22 --> 00:31:25
Again, under the
assumption that a < b.

477
00:31:25 --> 00:31:34
So as I wrote it's
in full generality.

478
00:31:34 --> 00:31:37
Let's illustrate this one.
 

479
00:31:37 --> 00:31:47
And then we have one more
property to learn after that.

480
00:31:47 --> 00:31:58
So let me give you an
example of estimation.

481
00:31:58 --> 00:32:01
The example is the same as
one that I already gave you.

482
00:32:01 --> 00:32:05
But this time, because we have
the tool of integration, we

483
00:32:05 --> 00:32:11
can just follow our
noses and it works.

484
00:32:11 --> 00:32:15
I start with the inequality,
so I'm trying to illustrate

485
00:32:15 --> 00:32:17
estimation, so I want to start
with an inequality which is

486
00:32:17 --> 00:32:19
what the hypothesis is here.
 

487
00:32:19 --> 00:32:21
And I'm going to integrate
the inequality to

488
00:32:21 --> 00:32:22
get this conclusion.
 

489
00:32:22 --> 00:32:25
And see what conclusion it is.
 

490
00:32:25 --> 00:32:28
The inequality that I
want to take is that e

491
00:32:28 --> 00:32:32
^ x >= 1, for x >= 0.
 

492
00:32:32 --> 00:32:37
That's going to be
our starting place.

493
00:32:37 --> 00:32:39
And now I'm going
to integrate it.

494
00:32:39 --> 00:32:40
That is, I'm going to
use estimation to

495
00:32:40 --> 00:32:42
see what that gives.
 

496
00:32:42 --> 00:32:45
Well, I'm going to integrate,
say, from 0 to b.

497
00:32:45 --> 00:32:50
I can't integrate below 0
because it's only true above 0.

498
00:32:50 --> 00:33:01
This is e ^ x dx >= the
integral from 0 to b of 1 dx.

499
00:33:01 --> 00:33:05
Alright, let's work out
what each of these is.

500
00:33:05 --> 00:33:13
The first one, e ^ x dx, is,
the antiderivative is e ^

501
00:33:13 --> 00:33:16
x, evaluated at 0 and b.
 

502
00:33:16 --> 00:33:18
So that's e ^ b - e ^ 0.
 

503
00:33:18 --> 00:33:23
Which is e ^ b - 1.
 

504
00:33:23 --> 00:33:29
The other one, you're supposed
to be able to get by

505
00:33:29 --> 00:33:32
the rectangle law.
 

506
00:33:32 --> 00:33:35
This is one rectangle of
base b and height 1.

507
00:33:35 --> 00:33:37
So the answer is b.
 

508
00:33:37 --> 00:33:44
Or you can do it by
antiderivatives, but it's b.

509
00:33:44 --> 00:33:49
That means that our inequality
says if I just combine these

510
00:33:49 --> 00:33:55
two things together,
that e ^ b - 1 >= b.

511
00:33:55 --> 00:34:02
And that's the same thing
as e ^ b >= 1 + b.

512
00:34:02 --> 00:34:05
Again, this only
works for b >= 0.

513
00:34:05 --> 00:34:10
Notice that if b were negative,
this would be a well

514
00:34:10 --> 00:34:13
defined quantity.
 

515
00:34:13 --> 00:34:18
But this estimation
would be false.

516
00:34:18 --> 00:34:22
We need that the b > 0 in
order for this to make sense.

517
00:34:22 --> 00:34:24
So this was used.
 

518
00:34:24 --> 00:34:25
And that's a good
thing, because this

519
00:34:25 --> 00:34:28
inequality is suspect.
 

520
00:34:28 --> 00:34:32
Actually, it turns out to be
true when b is negative.

521
00:34:32 --> 00:34:38
But we certainly
didn't prove it.

522
00:34:38 --> 00:34:42
I'm going to just
repeat this process.

523
00:34:42 --> 00:34:46
So let's repeat it.
 

524
00:34:46 --> 00:34:50
Starting from the inequality,
the conclusion, which

525
00:34:50 --> 00:34:51
is sitting right here.
 

526
00:34:51 --> 00:34:59
But I'll write it in a form
e ^ x >= 1 + x, for x >= 0.

527
00:34:59 --> 00:35:03
And now, if I integrate this
one, I get the integral from 0

528
00:35:03 --> 00:35:13
to b, e ^ x dx >= the integral
from 0 to b, (1 + x) dx and I

529
00:35:13 --> 00:35:16
remind you that we've already
calculated this one.

530
00:35:16 --> 00:35:19
This is e ^ b - 1.
 

531
00:35:19 --> 00:35:21
And the other one is
not hard to calculate.

532
00:35:21 --> 00:35:25
The antiderivative
is x + x^2 / 2.

533
00:35:25 --> 00:35:28
We're evaluating
that at 0 and b.

534
00:35:28 --> 00:35:34
So that comes out
to be b + b^2 / 2.

535
00:35:34 --> 00:35:42
And so our conclusion is that
the left side, which is e

536
00:35:42 --> 00:35:48
^ b - 1 >= b + b^2 / 2.
 

537
00:35:48 --> 00:35:51
And this is for b >= 0.
 

538
00:35:51 --> 00:36:00
And that's the same thing as
e ^ b >= 1 + b + b^2 / 2.

539
00:36:00 --> 00:36:07
This one actually is false for
b negative, so that's something

540
00:36:07 --> 00:36:15
that you have to be careful
with the b positives here.

541
00:36:15 --> 00:36:19
So you can keep on going
with this, and you

542
00:36:19 --> 00:36:20
didn't have to think.
 

543
00:36:20 --> 00:36:23
And you'll produce a very
interesting polynomial,

544
00:36:23 --> 00:36:30
which is a good
approximation to e^ b.

545
00:36:30 --> 00:36:34
So so that's it for
the basic properties.

546
00:36:34 --> 00:36:38
Now there's one tricky property
that I need to tell you about.

547
00:36:38 --> 00:36:47
It's not that tricky, but
it's a little tricky.

548
00:36:47 --> 00:37:07
And this is change
of variables.

549
00:37:07 --> 00:37:09
Change of variables in
integration, we've

550
00:37:09 --> 00:37:11
actually already done.
 

551
00:37:11 --> 00:37:14
We called that, the last
time we talked about it,

552
00:37:14 --> 00:37:23
we called it substitution.
 

553
00:37:23 --> 00:37:26
And the idea here, if you may
remember, was that if you're

554
00:37:26 --> 00:37:34
faced with an integral like
this, you can change it to, if

555
00:37:34 --> 00:37:38
you put in u = u (x) and you
have a du, which is equal

556
00:37:38 --> 00:37:42
to u' ( x) du, dx, sorry.
 

557
00:37:42 --> 00:37:45
Then you can change the
integral as follows.

558
00:37:45 --> 00:37:51
This is the same as g
(u( x)) u' (x) dx.

559
00:37:51 --> 00:37:58
This was the general
procedure for substitution.

560
00:37:58 --> 00:38:05
What's new today is that we're
going to put in the limits.

561
00:38:05 --> 00:38:10
If you have a limit here, u1,
and a limit here, u2, you want

562
00:38:10 --> 00:38:14
to know what the relationship
is between the limits here and

563
00:38:14 --> 00:38:18
the limits when you change
variables to the new variables.

564
00:38:18 --> 00:38:21
And it's the simplest
possible thing.

565
00:38:21 --> 00:38:26
Namely the two limits over here
are in the same relationship as

566
00:38:26 --> 00:38:29
u ( x) is to this
symbol u here.

567
00:38:29 --> 00:38:35
In other words, u1 = u
( x1), and u2 = u(x2).

568
00:38:35 --> 00:38:39
That's what works.
 

569
00:38:39 --> 00:38:44
Now there's only one danger
here, there's subtlety which

570
00:38:44 --> 00:39:02
is, this only works if
u' does not change sign.

571
00:39:02 --> 00:39:04
I've been worrying a little bit
about going backwards and

572
00:39:04 --> 00:39:07
forwards, and I allowed myself
to reverse and do all kinds of

573
00:39:07 --> 00:39:09
stuff, right, with
these integrals.

574
00:39:09 --> 00:39:11
So we're sort of free to do it.
 

575
00:39:11 --> 00:39:14
Well, this is one case where
you want to avoid it, OK?

576
00:39:14 --> 00:39:15
Just don't do it.
 

577
00:39:15 --> 00:39:18
It is possible, actually, to
make sense out of it, but it's

578
00:39:18 --> 00:39:21
also possible to get yourself
infinitely confused.

579
00:39:21 --> 00:39:25
So just make sure that -- now
it's OK is u' is always

580
00:39:25 --> 00:39:29
negative, or always going one
way, so OK if u' is always

581
00:39:29 --> 00:39:31
positive, you're always going
the other way, but if you mix

582
00:39:31 --> 00:39:39
them up you'll get
yourself mixed up.

583
00:39:39 --> 00:39:46
Let me give you an example.
 

584
00:39:46 --> 00:39:54
The example will be maybe close
to what we did last time.

585
00:39:54 --> 00:39:57
When we first did substitution.
 

586
00:39:57 --> 00:40:02
So the integral from 1 to 2,
this time I'll put in definite

587
00:40:02 --> 00:40:09
limits, of x^2 plus -- sorry,
maybe I call this x^3. x^3 +

588
00:40:09 --> 00:40:17
2, let's say, I don't know,
to the 5th power, x^2 dx.

589
00:40:17 --> 00:40:20
So this is an example of
an integral that we would

590
00:40:20 --> 00:40:25
have tried to handle by
substitution before.

591
00:40:25 --> 00:40:36
And the substitution we would
have used is u = x^3 + 2.

592
00:40:36 --> 00:40:38
And that's exactly what
we're going to do here.

593
00:40:38 --> 00:40:44
But we're just going to also
take into account the limits.

594
00:40:44 --> 00:40:47
The first step as in any
substitution or change

595
00:40:47 --> 00:40:54
of variables, is this.
 

596
00:40:54 --> 00:40:57
And so we can fill in the
things that we would

597
00:40:57 --> 00:40:58
have done previously.
 

598
00:40:58 --> 00:41:01
Which is that this is the
integral and this is u ^ 5.

599
00:41:01 --> 00:41:09
And then because this is
3x^2, we see that this is 3.

600
00:41:09 --> 00:41:13
Sorry, let's write
it the other way.

601
00:41:13 --> 00:41:17
1/3 du = x^2 dx.
 

602
00:41:17 --> 00:41:20
So that's what I'm going to
plug in for this factor here.

603
00:41:20 --> 00:41:26
So here's 1/3 du,
which replaces that.

604
00:41:26 --> 00:41:29
But now there's the
extra feature.

605
00:41:29 --> 00:41:31
The extra feature
is the limits.

606
00:41:31 --> 00:41:35
So here, really in disguise,
because, and now this is

607
00:41:35 --> 00:41:38
incredibly important.
 

608
00:41:38 --> 00:41:44
This is one of the reasons why
we use this notation dx and du.

609
00:41:44 --> 00:41:47
We want to remind ourselves
which variable is involved

610
00:41:47 --> 00:41:49
in the integration.
 

611
00:41:49 --> 00:41:52
And especially if you're the
one naming the variables, you

612
00:41:52 --> 00:41:54
may get mixed up
in this respect.

613
00:41:54 --> 00:41:59
So you must know which variable
is varying between 1 and 2.

614
00:41:59 --> 00:42:01
And the answer is, it's
x is the one that's

615
00:42:01 --> 00:42:04
varying between 1 and 2.
 

616
00:42:04 --> 00:42:07
So in disguise, even though
I didn't write it, it

617
00:42:07 --> 00:42:10
was contained in this
little symbol here.

618
00:42:10 --> 00:42:11
This reminded us
which variable.

619
00:42:11 --> 00:42:14
You'll find this amazingly
important when you get to

620
00:42:14 --> 00:42:16
multivariable calculus.
 

621
00:42:16 --> 00:42:18
When there are many
variables floating around.

622
00:42:18 --> 00:42:21
So this is an incredibly
important distinction to make.

623
00:42:21 --> 00:42:23
So now, over here
we have a limit.

624
00:42:23 --> 00:42:26
But of course it's supposed to
be with respect to u, now.

625
00:42:26 --> 00:42:29
So we need to calculate what
those corresponding limits are.

626
00:42:29 --> 00:42:33
And indeed it's just, I plug in
here u1 is going to be able to

627
00:42:33 --> 00:42:35
what I plug in for x = 1,
that's going to be 1

628
00:42:35 --> 00:42:38
^3 + 2, which is 3.
 

629
00:42:38 --> 00:42:46
And then u2 is 2^3 +
2, which = 10, right?

630
00:42:46 --> 00:42:47
8 + 2 = 10.
 

631
00:42:47 --> 00:42:57
So this is the integral from
3 to 10, of u ^ 5 (1/3 du).

632
00:42:57 --> 00:43:00
And now I can finish
the problem.

633
00:43:00 --> 00:43:06
This is 1/18 u ^
6, from 3 to 10.

634
00:43:06 --> 00:43:10
And this is where the most
common mistake occurs in

635
00:43:10 --> 00:43:12
substitutions of this type.
 

636
00:43:12 --> 00:43:16
Which is that if you ignore
this, and you plug in these 1

637
00:43:16 --> 00:43:18
and 2 here, you think, oh I
should just be putting

638
00:43:18 --> 00:43:20
it at 1 and 2.
 

639
00:43:20 --> 00:43:23
But actually, it should be, the
u value that we're interested

640
00:43:23 --> 00:43:27
in, and the lower limit is u
= 3 and u = 10 as

641
00:43:27 --> 00:43:29
the upper limit.
 

642
00:43:29 --> 00:43:31
So those are suppressed here.
 

643
00:43:31 --> 00:43:35
But those are the
ones that we want.

644
00:43:35 --> 00:43:37
And so, here we go.
 

645
00:43:37 --> 00:43:41
It's 1/18 times some ridiculous
number which I won't calculate.

646
00:43:41 --> 00:43:47
10 ^ 6 - 3 ^ 6.
 

647
00:43:47 --> 00:43:48
Yes, question.
 

648
00:43:48 --> 00:44:07
STUDENT: [INAUDIBLE]
 

649
00:44:07 --> 00:44:14
PROFESSOR: So, if you want to
do things with where you're

650
00:44:14 --> 00:44:19
worrying about the sign change,
the right strategy is,

651
00:44:19 --> 00:44:20
what you suggested works.
 

652
00:44:20 --> 00:44:22
And in fact I'm going
to do an example right

653
00:44:22 --> 00:44:24
now on this subject.
 

654
00:44:24 --> 00:44:30
But, the right strategy is
to break it up into pieces.

655
00:44:30 --> 00:44:34
Where u' has one sign
or the other, OK?

656
00:44:34 --> 00:44:37
Let me show you an example.
 

657
00:44:37 --> 00:44:40
Where things go wrong.
 

658
00:44:40 --> 00:44:47
And I'll tell you how
to handle it roughly.

659
00:44:47 --> 00:44:55
So here's our warning.
 

660
00:44:55 --> 00:45:00
Suppose you're integrating
for - 1 to 1, x^2 dx.

661
00:45:00 --> 00:45:02
Here's an example.
 

662
00:45:02 --> 00:45:09
And you have the temptation
to plug in u = x^2.

663
00:45:09 --> 00:45:11
Now, of course, we know
how to integrate this.

664
00:45:11 --> 00:45:16
But let's just pretend we
were stubborn and wanted

665
00:45:16 --> 00:45:19
to use substitution.
 

666
00:45:19 --> 00:45:26
Then we have du = 2x dx.
 

667
00:45:26 --> 00:45:30
And now if I try to make the
correspondence, notice that

668
00:45:30 --> 00:45:38
the limits are u1 = (- 1)^2,
that's the bottom limit.

669
00:45:38 --> 00:45:40
And u2 is the upper limit.
 

670
00:45:40 --> 00:45:43
That's 1 ^2, that's
also equal to 1.

671
00:45:43 --> 00:45:45
Both limits are 1.
 

672
00:45:45 --> 00:45:47
So this is going from 1 to 1.
 

673
00:45:47 --> 00:45:53
And no matter what it is, we
know it's going to be 0.

674
00:45:53 --> 00:45:55
But we know this is not 0.
 

675
00:45:55 --> 00:45:58
This is the integral of
a positive quantity.

676
00:45:58 --> 00:46:03
And the area under a curve is
going to be a positive area.

677
00:46:03 --> 00:46:04
So this is a positive quantity.
 

678
00:46:04 --> 00:46:07
It can't be 0.
 

679
00:46:07 --> 00:46:12
If you actually plug it in,
it looks equally strange.

680
00:46:12 --> 00:46:14
You put in here this
u and then, so that

681
00:46:14 --> 00:46:15
would be for the u^2.
 

682
00:46:16 --> 00:46:22
And then to plug in for dx, you
would write dx = (1 / 2x) du.

683
00:46:22 --> 00:46:27
And then you might
write that as this.

684
00:46:27 --> 00:46:31
And so what I should put in
here is this quantity here.

685
00:46:31 --> 00:46:33
Which is a perfectly
OK integral.

686
00:46:33 --> 00:46:37
And it has a value, I
mean, it's what it is.

687
00:46:37 --> 00:46:39
It's 0.
 

688
00:46:39 --> 00:46:45
So of course this is not true.
 

689
00:46:45 --> 00:46:55
And the reason is that u = x^2,
and u' ( x ) = 2x, which was

690
00:46:55 --> 00:47:00
positive for x positive, and
negative for x negative.

691
00:47:00 --> 00:47:03
And this was the sign change
which causes us trouble.

692
00:47:03 --> 00:47:08
If we break it off into its two
halves, then it'll be OK and

693
00:47:08 --> 00:47:09
you'll be able to use this.
 

694
00:47:09 --> 00:47:12
Now, there was a mistake.
 

695
00:47:12 --> 00:47:15
And this was essentially
what you were saying.

696
00:47:15 --> 00:47:19
That is, it's possible to see
this happening as you're doing

697
00:47:19 --> 00:47:21
it if you're very careful.
 

698
00:47:21 --> 00:47:24
There's a mistake in this
process, and the mistake

699
00:47:24 --> 00:47:26
is in the transition.
 

700
00:47:26 --> 00:47:28
This is a mistake here.
 

701
00:47:28 --> 00:47:33
Maybe I haven't used any
red yet today, so I get

702
00:47:33 --> 00:47:34
to use some red here.
 

703
00:47:34 --> 00:47:36
Oh boy.
 

704
00:47:36 --> 00:47:38
This is not true, here.
 

705
00:47:38 --> 00:47:39
This step here.
 

706
00:47:39 --> 00:47:40
So why isn't it true?
 

707
00:47:40 --> 00:47:43
It's not true for the
standard reason.

708
00:47:43 --> 00:47:50
Which is that really, x = plus
or minus square root of u.

709
00:47:50 --> 00:47:54
And if you stick to one side
or the other, you'll have

710
00:47:54 --> 00:47:55
a coherent formula for it.
 

711
00:47:55 --> 00:47:57
One of them will be the plus
and one of them will be the

712
00:47:57 --> 00:47:59
minus and it will work
out when you separate

713
00:47:59 --> 00:48:01
it into its pieces.
 

714
00:48:01 --> 00:48:02
So you could do that.
 

715
00:48:02 --> 00:48:04
But this is a can of worms.
 

716
00:48:04 --> 00:48:06
So I avoid this.
 

717
00:48:06 --> 00:48:10
And just do it in a place where
the inverse is well defined.

718
00:48:10 --> 00:48:11
And where the function
either moves steadily

719
00:48:11 --> 00:48:13
up or steadily down.
 

720
00:48:13 --> 00:48:14