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PROFESSOR: One correction
from last time.
Sorry to say, I forgot a very
important factor when I was
telling you what an
average value is.
If you don't put in that factor,
it's only half off on
the exam problem that will
be given on this.
So I would have gotten half off
for missing out on this
factor, too.
So remember you have to divide
by n here, certainly when
you're integrating over 0 to
n, the Riemann sum is the
numerator here.
And if I divide by n on that
side, I've got to divide by n
on the other side.
This was meant to illustrate
this idea that we're dividing
by the total here.
And we are going to be
talking about average
value in more detail.
Not today, though.
So this has to do with
average value.
And we'll discuss it in
considerable detail in a
couple of days, I guess.
Now, today I want to continue.
I didn't have time to finish
my discussion of the
Fundamental Theorem
of Calculus 2.
And anyway it's very important
to write it down on the board
twice, because you want to
see it at least twice.
And many more times as well.
So let's just remind you, the
second version of the
Fundamental Theorem of Calculus
says the following.
It says that the derivative of
an integral gives you the
function back again.
So here's the theorem.
And the way I'd like to use
it today, I started this
discussion last time.
But we didn't get into it.
And this is something that's on
your problem set along with
several other examples.
Is that we can use this to solve
differential equations.
And in particular, for example,
we can solve the
equation y' = 1 / x
with this formula.
Namely, using an integral.
L ( x ) is the integral
from 1 to x of dt / t.
The function f ( t
) is just 1 / t.
Now, that formula can be taken
to be the starting place for
the derivation of all
the properties of
the logarithm function.
So what we're going to do right
now is we're going to
take this to be the definition
of the logarithm.
And if we do that, then I claim
that we can read off the
properties of the logarithm
just about as easily as we
could before.
And so I'll illustrate
that now.
And there are a few other
examples of this where
somewhat more unfamiliar
functions come up.
This one is one that in theory
we know something about.
The first property of this
function is the one that's
already given.
Namely, its derivative
is 1 / x.
And we get a lot of information
just out of the
fact that its derivative
is 1 / x.
The other thing that we need
in order to nail down the
function, besides its
derivative, is one value of
the function.
Because it's really not
specified by this equation,
only specified up to a constant
by this equation.
But we nail down that constant
when we evaluate it at this
one place, L ( 1).
And there we're getting the
integral from 1 to 1 of dt /
t, which is 0.
And that's the case with all
these definite integrals.
If you evaluate them at their
starting places, the
value will be 0.
And together these two
properties specify this
function L (x), uniquely.
Now, the next step is to try
to think about what its
properties are.
And the first approach to that,
and this is the approach
that we always take, is to maybe
graph the function, to
get a feeling for it.
And so I'm going to take
the second derivative.
Now, notice that when you have
a function which is given as
an integral, its first
derivative is
really easy to compute.
And then its second derivative,
well, you have to
differentiate whatever
you get.
So it may or may not be easy.
But anyway, it's a lot harder in
the case when I start with
a function to get to the
second derivative.
Here it's relatively easy.
And these are the properties
that I'm going to use.
I won't really use very much
more about it than that.
And qualitatively, the
conclusions that we can draw
from this are, first of all,
from this, for example we see
that this thing is concave
down every place.
And then to get started with the
graph, since I see I have
a value here, which is L( 1) =
0, I'm going to throw in the
value of the slope.
So L' ( 1), which I know is 1 of
1, that's reading off from
this equation here,
so that's 1.
And now I'm ready to sketch at
least a part of the curve.
So here's a sketch
of the graph.
Here's the point (1, 0);
that is, x = 1, y = 0.
And the tangent line,
I know, has slope 1.
And the curve is concave down.
So it's going to look
something like this.
Incidentally, it's
also increasing.
And that's an important
property, it's strictly
increasing.
That's because L' (
x) is positive.
And so, we can get from this
the following important
definition.
Which, again, is working
backwards from this
definition.
We can get to where we started
with a log in our previous
discussion.
Namely, if I take the level
here, which is y = 1, then
that crosses the
axis someplace.
And his point is what we're
going to define as e.
So the definition of e is
that it's the value such
that L ( e) = 1.
And again, the fact that there's
exactly one such place
just comes from the fact that
this L' is positive, so that L
is increasing.
No, there's just one other
feature of this graph that I'm
going to emphasize to you.
There's one other thing which
I'm not going to check, which
you would ordinarily
do with graphs.
Once it's increasing there are
no critical points, so the
only other interesting
thing is the n's.
And it turns out that the limit
as you go down to 0 is
minus infinity.
As you go over to the right
here it's plus infinity.
It does get arbitrarily high;
it doesn't level off.
But I'm not going to
discuss that here.
Instead, I'm going to just
remark on one qualitative
feature of the graph, which is
this remark that the part
which is to the left
of 1 is below 0.
So I just want to remark, why
is L (x) negative for x < 1.
Maybe I don't have room
for that, so I'll just
put in here x < 1.
I want to give you
two reasons.
Again, we're only working from
very first principles here.
Just that the property that
L' = 1 / x, and L (1) = 0.
So our first reason is that,
well, I just said it.
L ( 1) = 0.
And L is increasing.
And if you read that backwards,
if it gets up to 0
here, it must have been
negative before 0.
So this is one way of seeing
that L ( x) is negative.
There's a second way
of seeing it,
which is equally important.
And it has to do with just
manipulation of integrals.
Here I'm going to start out
with L (x), and its
definition.
Which is the integral
from 1 to x dt / t.
And now I'm going to reverse
the order of integration.
This is the same, by our
definition of our properties
of integrals, as the integral
from x to 1 with a
minus sign dt / t.
Now, I can tell that this
quantity is negative.
And the reason that I can tell
is that this chunk of it here,
this piece of it, is
a positive number.
This part is positive.
And this part is positive
because x < 1.
So the lower limit is less than
the upper limit, and so
this is interpreted.
The thing in the green box is
interpreted as an area.
It's an area.
And so negative a positive
quantity is negative, minus a
positive quantity's negative.
So both of these work
perfectly well as
interpretations.
And it's just to illustrate
what we can do.
Now, there's one more
manipulation of integrals that
gives us the fanciest
property of the ln.
And that's the last one
that I'm going to do.
And you have a similar thing
on your homework.
So I'm going to prove that
this is, as I say, the
fanciest property of the log.
On your homework, by the way,
you're going to check that L(
1 / x) = - L ( x).
But we'll do this one.
The idea is just to plug
in the formula
and see what it gives.
On the left-hand side, I
have 1 to ab, dt / t.
That's L ( ab).
And then that's certainly equal
to the left-hand side.
And then I'm going to now split
this into two pieces.
Again, this is a property
of integrals.
That if you have an integral
from one place to another, you
can break it up into pieces.
So I'm going to start at
1 but then go to a.
And then I'm going to continue
from a to ab.
So this is the question
that we have. We
haven't proved this.
Well, this one is
actually true.
If we want this to be
true, we know by
definition L ( ab) is this.
We know, we can see it,
that L(a) is this.
So the question that this boils
down to is, we want to
know that these two
things are equal.
We want to know that L ( b) is
that other integral there.
So let's check it.
I'm going to rewrite
the integral.
It's the integral from - sorry,
from lower limit a to
upper limit ab of dt / t.
And now, again, to illustrate
properties of integrals, the
key property here that we're
going to have to use is change
of variables.
This is a kind of a scaled
integral where everything is
multiplied by a factor of a from
what we want to get to
this L ( b) quantity.
And so this suggests that
we write down t = a u.
That's going to be our trick.
And if I use that new variable
u, then the change
in t, dt, is adu.
And as a result, I can write
this as equal to an integral
from, let's see, dt = adu.
And t = au.
So I've now substituted
in for the integrand.
But on top of this, with
definite integrals, we also
have to check the limits.
And the limits work
out as follows.
When t = a, that's
the lower limit.
Let's just take a
look. t = a u.
So that means that u
is equal to, what?
It's 1.
Because a * 1 = a.
So if t = a, this is
if and only if.
So this lower limit, which
really in disguise was where t
= a, becomes where u = 1.
And similarly, when
t = ab, u = b.
So the upper limit here is b.
And now, if you notice, we're
just going to cancel these two
factors here.
And now we recognize that this
is just the same as the
definition of L ( b).
Because L ( x) is over
here in the box.
And the fact that I use the
letter t there is irrelevant;
it works equally well
with the letter u.
So this is just L (b).
Which is what we
wanted to show.
So that's an example, and you
have one in your homework,
which is a little similar.
Now, the last example, that I'm
going to discuss of this
type, I already mentioned
last time.
Which is the function F ( x),
which is the integral from 0
to x of e ^ -t^2 dt.
This one is even more exotic
because unlike the logarithm
it's a new function.
It really is not any function
that you can express in terms
of the functions that
we know already.
And the approach, always, to
these new functions is to
think of what their
properties are.
And the way we think of
functions in order to
understand them is to
maybe sketch them.
And so I'm going to do
exactly the same
thing I did over here.
So, what is it that I
can get out of this?
Well, immediately I can figure
out what the derivative is.
I read it off from the
fundamental theorem.
It's this.
I also can figure out the value
at the starting place.
In this case, the starting
place is 0.
And the value is 0.
And I should check the second
derivative, which is also not
so difficult to compute.
The second derivative
is - 2x e ^ x^2.
And so now I can see that this
function is increasing,
because this derivative is
positive, it's always
increasing.
And it's going to be concave
down when x is positive and
concave up when x is negative.
Because there's a minus sign
here, so the sign is negative.
This is less than 0 when x is
positive and greater than 0
when x is negative.
And maybe to get started I'll
remind you F( 0), 0.
It's also true that F' ( 0),
that just comes right out of
this, F' ( 0) = e ^
- 0^2 which is 1.
That means the tangent line
again has slope 1.
We do this a lot
with functions.
We normalize them so that their
slopes of their tangent
lines are 1 at convenient
spots.
So here's the tangent
line of slope 1.
We know this thing is concave
down to the right and concave
up to the left.
And so it's going to look
something like this.
With an inflection point.
Right?
Now, I want to say one more,
make one more remark about
this function, or maybe two
more remarks about this
function, before we go on.
Really, you want to know this
graph as well as possible.
And so there are just a
couple more features.
And one is enormously helpful
because it cuts in half all of
the work that you have. and
that is the property that
turns out that this
function is i.
Namely, F( - x) = - F(x).
That's what's known as
an odd function.
Now, the reason why it's
odd is that it's the
antiderivative of something
that's even.
This function in here is even.
And we nailed it down so
that it was 0 at 0.
Another way of interpreting
that, and let me show it to
you underneath, is
the following.
When we look at its derivative,
its derivative,
course, is the function e ^ x.
Sorry, e ^ - x ^2.
So that's this shape here.
And you can see the slope is 0,
but fairly close to 0, but
positive along here.
It's getting, this is
its steepest point.
This is the highest
point here.
And then it's leveling off
again; the slope is going
down, always positive.
This is the graph of
F' = e ^ -x^2.
Now, the interpretation of the
function that's up above is
that the value here is
the area from 0 to x.
So this is area F(x).
Maybe I'll color it in, decorate
it a little bit.
So this area here is F ( x).
Now, I want to show you
this odd property,
by using this symmetry.
The graph here is even, so in
other words, what's back here
is exactly the same
as what's forward.
But now there's a reversal.
Because we're keeping track
of the area starting
from 0 going forward.
That's positive.
If we go backwards, it's
counted negatively.
So if we went backwards to - x,
we'd get exactly the same
as that green patch
over there.
We'd get a red patch
over here.
But it would be counted
negatively.
And that's the property
that it's odd.
You can also check this
by properties
of integrals directly.
That would be just like
this process here.
So it's completely analogous
to checking this
formula over there.
So that's one of the comments
I wanted to make about this.
And why does this save us
a lot of time, if we
know this is odd?
Well, we know that the shape of
this branch is exactly the
reverse, or the reflection,
if you like, of the
shape of this one.
What we want to do is flip it
under the axis and then
reflect it over that way.
And that's the symmetry property
of the graph of F(x).
Now, the last property that I
want to mention is what's
happening with the ends.
And at the end there's
an asymptote,
there's a limit here.
So this is an asymptote.
And the same thing down here,
which will be exactly because
of the odd feature, this'll
be exactly negative.
The opposite value over here.
And you might ask yourself, what
level is this, exactly.
Now, that level turns out to be
a very important quantity.
It's interpreted down here as
the area under this whole
infinite stretch.
It's all the way out
to infinity.
So, let's see.
What do you think it is?
You're all clueless.
Well, maybe not all of you,
you're just afraid to say.
So it's obvious.
It's the square root of pi/2.
That was right on the tip of
your tongue, wasn't it?
STUDENT: Ah, yes.
PROFESSOR: Right, so this is
actually very un-obvious, but
it's a very important
quantity.
And it's an amazing fact
that this thing
approaches this number.
And it's something that people
worried about for many years
before actually nailing down.
And so what I just claimed here
is that the limit as x
approaches infinity of F ( x)
= the square root of pi / 2.
And similarly, if you do it to
minus infinity, you'll get
minus square root of pi/2.
And for this reason, people
introduce a new function
because they like
the number 1.
This function is erf, short
for error function.
And it's 2 / the square root of
pi times the integral from
0 to x, e ^ - t ^2 dt.
In other words, it's just our
original, our previous
function multiplied by 2 /
the square root of pi.
And that's the function which
gets tabulated quite a lot.
You'll see it on the internet
everywhere, and it's a very
important function.
There are other normalizations
that are used, and the
discussions of the other
normalizations are in your
problems. This is one
of them, and another
one is in your exercises.
The standard normal
distribution.
There are tons of functions
like this, which are new
functions that we can
get at once we
have the tool of integrals.
And I'll write down just one
or two more, just so that
you'll see the variety.
Here's one which is called
a Fresnel integral.
On your problem set next week,
we'll do the other Fresnel
integral, we'll look
at this one.
These functions cannot be
expressed in elementary terms.
The one on your homework for
this week was this one.
This one comes up in
Fourier analysis.
And I'm going to just tell you
maybe one more such function.
There's a function which is
called Li ( x), logarithmic
integral of x, which
is this guy.
The reciprocal of the logarithm,
the natural log.
And the significance of this
one is that Li ( x) is
approximately equal to the
number of primes < x.
And, in fact, if you can make
this as precise as possible,
you'll be famous for millennia,
because this is
known as the Riemann
hypothesis.
Exactly how closely this
approximation occurs.
But it's a hard problem, and
already a century ago the
prime number theorem, which
established this connection
was extremely important
to progress in math.
Yeah, question.
STUDENT: [INAUDIBLE]
PROFESSOR: Is this stuff you're
supposed to understand.
That's a good question.
I love that question.
The answer is, this is,
so we launched off
into something here.
And let me just explain
it to you.
I'm going to be talking a fair
amount more about this
particular function, because
it's associated to the normal
distribution.
And I'm going to let you
get familiar with it.
What I'm doing here is
purely cultural.
Well, after this panel, what I'm
doing is purely cultural.
Just saying there's a lot
of other beasts out
there in the world.
And one of them is called
C of [INAUDIBLE].
So we'll have a just a very
passing familiarity with one
or two of these functions.
But there are literally dozens
and dozens of them.
The only thing that you'll need
to do with such functions
is things like understanding
the derivative, the second
derivative, and tracking
what the function does.
Sketching the same way you
did with any other tool.
So we're going to do this type
of thing with these functions.
And I'll have to lead
you through.
If I wanted to ask you a
question about one of these
functions, I have to tell you
exactly what I'm aiming for.
Yeah, another question.
STUDENT: [INAUDIBLE]
PROFESSOR: Yeah, I did.
I called these guys
Fresnel integrals.
The guy's name is Fresnel.
It's just named after
a person.
But, and this one, Li's
logarithmic integral, it's not
named after a person.
Logarithm is not somebody's
name.
So look, in fact this will be
mentioned also on a problem
set, but I don't expect you
to remember these names.
In particular, that you
definitely don't want to try
to remember.
Yes, another question.
STUDENT: [INAUDIBLE]
PROFESSOR: The question is,
will we prove this limit.
And the answer is yes,
if we have time.
It'll be in about
a week or so.
We're not going to do it now.
It takes us quite a bit
of work to do it.
OK.
I'm going to change years now,
I'm going to shift gears.
And we're going to go back to
a more standard thing which
has to do with just setting
up integrals.
And this has to do with
understanding where integrals
play a role, and they play a
role in cumulative sums, in
evaluating things.
This is much more closely
associated with the first
Fundamental Theorem.
That is, we'll take, today
we were talking about how
integrals are formulas
for functions.
Or solutions to differential
equations.
We're going to go back and talk
about integrals as being
the answer to a question
as opposed to.
what we've done now.
So in other words, and the first
example, or most of the
examples for now, are going
to be taken from geometry.
Later on we'll get
to probability.
And the first topic is just
areas between curves.
Here's the idea.
If you have a couple of curves
that look like this and maybe
like this, and you want to start
at a place a and you
want to end at a place b, then
you can chop it up the same
way we did with Riemann
sums. And take a chunk
that looks like this.
And I'm going to write the
thickness of that chunk.
Well, let's give these
things names.
Let's say the top curve is
f(x), and the bottom
curve is g ( x).
And then this thickness
is going to be dx.
That's the thickness.
And what is the height?
Well, the height is the
difference between the top
value and the bottom value.
So here we have (f (
x) - g ( x)) dx.
This is, if you like, base times
- whoops, backwards.
This is height, and this is
the base of the rectangle.
And these are approximately
correct.
But of course, only in limit
when this is an infinitesimal,
is it exactly right.
In order to get the
whole area, I have
add these guys up.
So I'm going to integrate
from a to b.
That's summing them, that's
adding them up.
And that's going
to be my area.
So that's the story here.
Now, let me just say two
things about this.
First of all, on a very abstract
level before we get
started with details of more
complicated problems. The
first one is that every problem
that I'm going to be
talking about from now on for,
several days, involves the
following collection of,
the following goals.
I want to identify something
to integrate.
That's called an integrand.
And I want to identify what
are known as the limits.
The whole game is simply to
figure out what a, b, and this
quantity is here.
Whatever it is.
And the minute we have that, we
can calculate the integral
if we like.
We have numerical procedures
or maybe we have analytic
procedures, but anyway we
can get at the integral.
The goal here is
to set them up.
And in order to set
them up, you must
know these three things.
The lower limit, the upper
limit, and what we're
integrating.
If you leave one of these out,
it's like the following thing.
I ask you what the area
of this region is.
If I left out this end, how
could I possibly know?
I don't even know where it
starts, so how can I figure
out what this area is if I
haven't identified what the
left side is.
I can't leave out the bottom.
It's sitting here,
in this formula.
Because I need to know
where it is.
And I need to know the top and
I need to know this side.
Those are the four sides
of the figure.
If I don't incorporate them
into the information, I'll
never get anything out.
So I need to know everything.
And I need to know exactly those
things, in order to have
a formula for the area.
Now, when this gets carried out
in practice, as we will do
now in our first example,
it's more
complicated than it looks.
So here's our first example:
Find the area between x = y ^2
and y = x - 2.
This is our first example.
Let me make sure that I chose
the example that I wanted to.
Yeah.
Now, there's a first step in
figuring these things out.
And this is that you must
draw a picture.
If you don't draw a picture
you'll never figure out what
this area is, because you'll
never figure out what's what
between these curves.
The first curve, y = x
^2, is a parabola.
But x is a function of y.
It's pointing this way.
So it's this parabola here.
That's y = x^2.
Whoops, x = y ^2.
The second curve is a line, a
straight line of slope 1,
starting at (x = 2, y = 0).
It goes through this place here,
which is 2 over and has
slope 1, so it does this.
And this shape in here is what
we mean by the area between
the curves.
Now that we see what it is,
we have a better idea of
what our goal is.
If you haven't drawn it,
you have no hope.
Now, I'm going to describe
two ways of getting
at this area here.
And the first one is motivated
by the shape that I just
described right here.
Namely, I'm going to use it
in a straightforward way.
I'm going to chop things up
into these vertical pieces
just as I did right there.
Now, here's the difficulty
with that.
The difficulty is that the
upper curve here has one
formula but the lower curve
shifts from being a part of
the parabola to being a part
of the straight line.
That means that there are two
different formulas for the
lower function.
And the only way to accommodate
that is to
separate this up into
two halves.
Separate it out into
two halves.
I'm going to have to divide
it right here.
So we must break it into two
pieces and find the integral
of one half and the other half.
Question?
STUDENT: [INAUDIBLE]
PROFESSOR: So, you're one
step ahead of me.
We'll also have to be sure to
distinguish between the top
branch and the bottom branch of
the parabola, which we're
about to do.
Now, in order to distinguish
what's going on I actually
have to use multi colors here.
And so we will do that.
First there's the top part,
which is orange.
That's the top part.
I'll call it top.
And then there's the bottom
part, which has two halves.
They are pink, and I
guess this is blue.
Alright, so now let's see
what's happening.
The most important two points
that I have to figure out in
order to get started here.
Well, really I'm going
to have to figure out
three points, I claim.
I'm going to have to figure
out where this point is.
Where this point is, and
where that point is.
If I know where these three
points are, then I have a
chance of knowing where
to start, where
to end, and so forth.
Another question.
STUDENT: [INAUDIBLE]
PROFESSOR: Could you speak up?
STUDENT: [INAUDIBLE]
PROFESSOR: The question
is, why do we need to
split up the area.
And I think in order to answer
that question further, I'm
going to have to go into the
details of the method, and
then you'll see where
it's necessary.
So the first step is that I'm
going to figure out what these
three points are.
This one is kind of easy;
it's the point (0, 0).
This point down here and
this point up here are
intersections of
the two curves.
I can identify them by the
following equation.
I need to see where these
curves intersect.
At what, well, if I plug in x
= y^2, I get y = y^2 - 2.
And then I can solve this
quadratic equation. y
^2 - y - 2 = 0.
So (y - 2)( y + 1) = 0.
And this is telling me that
y = 2 or y = - 1.
So I've found y = - 1.
That means this point down here
has second entry - 1.
Its first entry, its x value,
I can get from this formula
here or the other formula.
I have to square,
this - 1 ^2 = 1.
So that's the formula
for this point.
And the other point has
second entry 2.
And, again, with his formula y
= x ^2, I have to square y to
get x, so this is 4.
Now, I claim I have enough
data to get started.
But maybe I'll identify
one more thing.
I need the top, the bottom left,
and the bottom right.
The top is the formula for this
branch of x = y ^2, which
is in the positive y region.
And that is y = the
square root of x.
The bottom curve, part of the
parabola, so this is the
bottom left, is y = minus
square root x.
That's the other branch
of the square root.
And this is exactly what
you were asking before.
And this is, we have to
distinguish between these two.
And the point is, these formulas
really are different.
They're not the same.
Now, the last bit is the bottom
right chunk here, which
is this pink part.
Bottom right.
And that one is the formula
for the line.
And that's y = x - 2.
Now I'm ready to
find the area.
It's going to be
in two chunks.
This is the left part,
plus the right part.
And the left part, and I want to
set it up as an integral, I
want there to be a dx and here
I want to set up an integral
and I want it to be dx.
I need to figure out what
the range of x is.
So, first I'm going to - well,
let's leave ourselves a little
more room than that.
Just to be safe.
OK, here's the right.
So here we have our dx.
Now, I need to figure out
the starting place
and the ending place.
So the starting place is
the leftmost place.
The leftmost place
is over here.
And x = 0 there.
So we're going to travel from
this vertical line to the
green line.
Over here.
And that's from 0 to 1.
And the difference between the
orange curve and the blue
curve is what I call top and
bottom left, over there.
So that is square root of x
minus minus square root of x.
Again, this is what I call
top, and this was bottom.
But only the left.
I claim that's giving me the
left half of this, the left
section of this diagram.
Now I'm going to do the right
section of the diagram.
I start at 1.
The lower limit is 1.
And I go all the way
to this point here.
Which is the last bit.
And that's going to be x = 4.
The upper limit here is 4.
And now I have to take the
difference between the top and
the bottom again.
The top is square root
of x all over again.
But the bottom has changed.
The bottom is now the
quantity (x - 2).
Please don't forget
your parenthesis.
There's going to be minus
signs and cancellations.
Now, this is almost the
end of the problem.
The rest of it is routine.
We would just have to evaluate
these integrals.
And, fortunately, I'm going
to spare you that.
We're not going to
bother to do it.
That's the easy part.
We're not going to do it.
But I'm going to show you that
there's a much quicker way
with this integral.
And with this area calculation.
Right now.
The quicker way is what
you see when you see
how long this is.
And you see that there's another
device that you can
use that looks similar in
principle to this, but
reverses the roles of x and y.
And the other device, which
I'll draw over here,
schematically.
No, maybe I'll draw it on
this blackboard here.
So, Method 2, if you like,
this was Method 1, and we
should call it the hard way.
Method 2, which is better
in this case, is to use
horizontal slices.
Let me draw the picture,
at least schematically.
Here's our picture that
we had before.
And now instead of slicing it
vertically, I'm going to slice
it horizontally.
Like this.
Now, the dimensions have
different names.
But the principle is similar.
The width, we now call dy.
Because it's the change in y.
And this distance here, from the
left end to the right end,
we have to figure out what the
formulas for those things are.
So on the left, maybe I'll draw
them color coded again.
So here's a left.
And, whoops, orange
is right, I guess.
So here we go.
So we have the left, which
is this green.
Is x = y ^2.
And the right, which is
orange, is y = x - 2.
And now in order to use this,
it's going to turn out that we
want to write x as, we want
to reverse roles.
So we want to write this as
x is a function of y.
So we'll use it in this form.
And now I want to set up
the integral for you.
This time, the area
is equal to an
integral in the dy variable.
And its starting place
is down here.
And it's ending place
is up there.
This is the lowest value
of y, and this is the
top value of y.
And we've already computed
those things.
The lowest level of y is - 1.
So this is y = - 1.
And this top value is y = 2.
So this goes from - 1 to 2.
And now the difference is this
distance here, the distance
between the rightmost point
and the leftmost point.
Those are the two dimensions.
So again, it's a rectangle but
its horizontal is long and its
vertical is very short.
And what are they?
It's the difference between
the right and the left.
The right-hand is (y + 2), and
the right-hand is y ^2.
So this is the formula.
STUDENT: [INAUDIBLE]
PROFESSOR: What was
the question?
Why is it right minus left?
That's very important.
Why is it right minus left?
And that's actually the point
that I was about to make.
Which is this.
That y + 2, which is the right,
is bigger than y ^2,
which is the left.
So that means that y +
2 - y ^2 is positive.
If you do it backwards, you'll
always get a negative number
and you'll always get
the wrong answer.
So this is the right-hand end
minus the left-hand end gives
you a positive number.
And it's not obvious, actually,
where you are.
There's another double-check,
by the way.
When you look at this quantity,
you see that the
ends pinch.
And that's exactly the
crossover points.
That is, when y = - 1,
y + 2 - y ^2 = 0.
And when y = 2, y
+ 2 - y ^2 = 0.
And that's not an accident,
that's exactly the geometry of
the shape that we picked
out there.
So this is the technique.
Now, this is a much more
routine integral.
I'm not going to carry it out,
I'll just do one last step.
Which is that this is (y ^2
/ 2) + 2y - (y ^3 / 3),
evaluated at - 1 and 2.
Which, if you work
it out, is 9/2.
So we're done for today.
And tomorrow we'll do more
volumes, more things including
three dimensions.
