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PROFESSOR: Today we're going
to continue with integration.
And we get to do the-- probably
the most important thing
of this entire course.
Which is appropriately named.
It's called the fundamental
theorem of calculus.
And we'll be abbreviating
it FTC and occasionally I'll
put in a 1 here, because there
will be two versions of it.
But this is the
one that you'll be
using the most in this class.
The fundamental theorem of
calculus says the following.
It says that if F' =
f, so F'(x) = f(x),
there's a capital
F and a little f,
then the integral from a to b
of f(x) is equal to F(b) - F(a).
That's it.
That's the whole theorem.
And you may recognize it.
Before, we had the
notation that F
was the antiderivative,
that is, capital F
was the integral of f(x).
We wrote it this way.
This is this
indefinite integral.
And now we're putting
in definite values.
And we have a connection
between the two
uses of the integral sign.
But with the definite
values, we get real numbers
out instead of a function.
Or a function up to a constant.
So this is it.
This is the formula.
And it's usually also written
with another notation.
So I want to introduce that
notation to you as well.
So there's a new notation here.
Which you'll find
very convenient.
Because we don't always
have to give a letter f
to the functions involved.
So it's an abbreviation.
For right now there'll be
a lot of f's, but anyway.
So here's the abbreviation.
Whenever I have a difference
between a function at two
values, I also can
write this as F(x)
with an a down here
and a b up there.
So that's the
notation that we use.
And you can also, for
emphasis, and this sometimes
turns out to be important, when
there's more than one variable
floating around in the problem.
To specify that
the variable is x.
So this is the same
thing as x = a.
And x = b.
It indicates where
you want to plug in,
what you want to plug in.
And now you take the top
value minus the bottom value.
So F(b) - F(a).
So this is just a notation, and
in that notation, of course,
the theorem can be written
with this set of symbols here.
Equally well.
So let's just give a
couple of examples.
The first example
is the one that we
did last time very laboriously.
If you take the function F(x),
which happens to be x^3 / 3,
then if you differentiate
it, you get, well,
the the factor of 3 cancels.
So you get x^2,
that's the derivative.
And so by the
fundamental theorem,
so this implies by the
fundamental theorem,
that the integral from say, a to
b of x^3 over - sorry, x^2 dx,
that's the derivative here.
This is the function we're
going to use as f(x) here -
is equal to this function
here, F(b) - F(a), that's here.
This function here.
So that's F(b) - F(a), and
that's equal to b^3 / 3 -
a^3 / 3.
Now, in this new
notation, we usually
don't have all of these letters.
All we write is the following.
We write the
integral from a to b,
and I'm going to
do the case 0 to b,
because that was the one that
we actually did last time.
So I'm going to set a = 0 here.
And then, the problem we
were faced last time as this.
And as I said we did
it very laboriously.
But now you can see that we
can do it in ten seconds,
let's say.
Well, the antiderivative
of this is x^3 / 3.
I'm going to evaluate it
at 0 and at b and subtract.
So that's going to
be b^3 / 3 - 0^3 / 3.
Which of course is b^3 / 3.
And that's the end,
that's the answer.
So this is a lot
faster than yesterday.
I hope you'll agree.
And we can dispense with
those elaborate computations.
Although there's a conceptual
reason, a very important one,
for understanding the
procedure that we went through.
Because eventually you're
going to be using integrals
and these quick ways
of doing things,
to solve problems like finding
the volumes of pyramids.
In other words, we're going
to reverse the process.
And so we need to understand
the connection between the two.
I'm going to give a
couple more examples.
And then we'll go on.
So the second
example would be one
that would be quite difficult
to do by this Riemann sum
technique that we
described yesterday.
Although it is possible.
It uses much higher
mathematics to do it.
And that is the area under one
hump of the sine curve, sin x.
Let me just draw
a picture of that.
The curve goes like this, and
we're talking about this area
here.
It starts out at
0, it goes to pi.
That's one hump.
And so the answer is, it's the
integral from 0 to pi of sin
x dx.
And so I need to take the
antiderivative of that.
And that's -cos x.
That's the thing whose
derivative is sin x.
Evaluating it at 0 and pi.
Now, let's do this
one carefully.
Because this is where I see
a lot of arithmetic mistakes.
Even though this is the
easy part of the problem.
It's hard to pay attention
and plug in the right numbers.
And so, let's just pay
very close attention.
I'm plugging in pi.
That's -cos pi.
That's the first term.
And then I'm
subtracting the value
at the bottom, which is -cos 0.
There are already five
opportunities for you
to make a transcription
error or an arithmetic
mistake in what I just did.
And I've seen all five of them.
So the next one is
that this is -(-1).
Minus negative 1, if you like.
And then this is minus,
and here's another -1.
So altogether we have 2.
So that's it.
That's the area.
This area, which is hard
to guess, this is area 2.
The third example
is maybe superfluous
but I'm going to say it anyway.
We can take the integral,
say, from 0 to 1, of x^100.
Any power, now, is
within our power.
So let's do it.
So here we have the
antiderivative is x^101 / 101,
evaluated at 0 and 1.
And that is just 1 / 101.
That's that.
So that's the
fundamental theorem.
Now this, as I say,
harnesses a lot
of what we've already learned,
all about antiderivatives.
Now, I want to give you an
intuitive interpretation.
So let's try that.
We'll talk about a proof
of the fundamental theorem
a little bit later.
It's not actually that hard.
But we'll give an intuitive
reason, interpretation,
if you like.
Of the fundamental theorem.
So this is going to
be one which is not
related to area, but rather
to time and distance.
So we'll consider x(t) is
your position at time t.
And then x'(t), which is dx/dt,
is going to be what we know
as your speed.
And then what the theorem is
telling us is the following.
It's telling us the integral
from a to b of v(t) dt -
so, reading the relationship
- is equal to x (b) - x(a).
And so this is some
kind of cumulative sum
of your velocities.
So let's interpret the
right-hand side first.
This is the distance traveled.
And it's also what you
would read on your odometer.
Right, from the beginning
to the end of the trip.
That's what you would
read on your odometer.
Whereas this is what you would
read on your speedometer.
So this is the interpretation.
Now, I want to just
go one step further
into this
interpretation, to make
the connection with the Riemann
sums that we had yesterday.
Because those are very
complicated to understand.
And I want you to
understand them viscerally
on several different levels.
Because that's how you'll
understand integration better.
The first thing that
I want to imagine,
so we're going to do a
thought experiment now,
which is that you are
extremely obsessive.
And you're driving
your car from time a
to time b, place Q
to place R, whatever.
And you check your
speedometer every second.
OK, so you've read your
speedometer in the i-th second,
and you've read that
you're going at this speed.
Now, how far do you
go in that second?
Well, the answer is
you go this speed
times the time interval,
which in this case
we're imagining as 1 second.
All right?
So this is how far you went.
But this is the time interval.
And this is the
distance traveled
in that-- second number
i, in the i-th second.
The distance traveled in
the i-th second, that's
a total distance you traveled.
Now, what happens if you
go the whole distance?
Well, you travel the sum
of all these distances.
So it's some massive sum, where
n is some ridiculous number
of seconds.
3600 seconds or
something like that.
Whatever it is.
And that's going to turn out
to be very similar to what you
would read on your odometer.
Because during that second,
you didn't change velocity
very much.
So the approximation
that the speed at one
time that you spotted it is
very similar to the speed
during the whole second.
It doesn't change that much.
So this is a pretty
good approximation
to how far you traveled.
And so the sum is a very
realistic approximation
to the entire integral.
Which is denoted this way.
Which, by the
fundamental theorem,
is exactly how far you traveled.
So this is x(b) - x(a) Exactly.
The other one is approximate.
OK, again this is
called a Riemann sum.
All right, so that's the intro
to the fundamental theorem.
And now what I need to do
is extend it just a bit.
And the way I'm going to
extend it is the following.
I'm going to do it on
this example first.
And then we'll do
it more formally.
So here's this example
where we went someplace.
But now I just want to draw
you an additional picture here.
Imagine I start here
and I go over to there
and then I come back.
And maybe even I
do a round trip.
I come back to the same place.
Well, if I come back
to the same place,
then the position is unchanged
from the beginning to the end.
In other words, the
difference is 0.
And the velocity, technically
rather than the speed.
It's the speed to the right
and the speed to the left
maybe are the same,
but one of them
is going in the positive
direction and one of them
is going in the
negative direction,
and they cancel each other.
So if you have this
kind of situation,
we want that to be reflected.
We like that
interpretation and we
want to preserve it even
when-- in the case when
the function v is negative.
And so I'm going to now extend
our notion of integration.
So we'll extend integration
to the case f negative.
Or positive.
In other words, it
could be any sign.
Actually, there's no change.
The formulas are all the same.
We just-- If this v is
going to be positive,
we write in a positive number.
If it's going to be negative,
we write in a negative number.
And we just leave it alone.
And the real-- So here's--
Let me carry out an example
and show you how it works.
I'll carry out the example
on this blackboard up here.
Of the sine function.
But we're going
to try two humps.
We're going to try the
first hump and the one that
goes underneath.
There.
So our example here is
going to be the integral
from 0 to 2pi of sin x dx.
And now, because the fundamental
theorem is so important, and so
useful, and so
convenient, we just
assume that it be true
in this case as well.
So we insist that this is going
to be -cos x, evaluated at 0
and 2pi, with the difference.
Now, when we carry
out that difference,
what we get here is
-cos 2pi - (-cos 0).
Which is -1 - (-1), which is 0.
And the interpretation
of this is the following.
Here's our double hump,
here's pi and here's 2pi.
And all that's happening is that
the geometric interpretation
that we had before of
the area under the curve
has to be taken with a grain
of salt. In other words,
I lied to you before when I said
that the definite integral was
the area under the curve.
It's not.
The definite
integral is the area
under the curve when
it's above the curve,
and it counts negatively
when it's below the curve.
So yesterday, my geometric
interpretation was incomplete.
And really just a plain lie.
So the true geometric
interpretation
of the definite integral
is plus the area
above the axis,
above the x-axis,
minus the area below the x-axis.
As in the picture.
I'm just writing
it down in words,
but you should think
of it visually also.
So that's the setup here.
And now we have the complete
definition of integrals.
And I need to list for you
a bunch of their properties
and how we deal with integrals.
So are there any
questions before we go on?
Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: Right.
So the question was,
wouldn't the absolute value
of the velocity
function be involved?
The answer is yes.
That is, that's one
question that you could ask.
One question you
could ask is what's
the total distance traveled.
And in that case,
you would keep track
of the absolute value of
the velocity as you said,
whether it's
positive or negative.
And then you would get the
total length of this curve here.
That's, however, not what the
definite integral measures.
It measures the net
distance traveled.
So it's another thing.
In other words, we can do that.
We now have the
tools to do both.
We could also-- So if you
like, the total distance
is equal to the
integral of this.
From a to b.
But the net distance is the
one without the absolute value
signs.
So that's correct.
Other questions?
All right.
So now, let's talk about
properties of integrals.
So the properties of integrals
that I want to mention to you
are these.
The first one doesn't
bear too much comment.
If you take the cumulative
integral of a sum,
you're just trying to get the
sum of the separate integrals
here.
And I won't say much about that.
That's because sums come
out, the because the integral
is a sum.
Incidentally, you know
this strange symbol here,
there's actually a reason
for it historically.
If you go back to
old books, you'll
see that it actually looks
a little bit more like an S.
This capital sigma is a sum.
S for sum, because
everybody in those days
knew Latin and Greek.
And this one is also
an S, but gradually it
was such an important S
that they made a bigger.
And then they stretched it out
and made it a little thinner,
because it didn't fit into
one typesetting space.
And so just for typesetting
reasons it got stretched.
And got a little bit skinny.
Anyway, so it's really an
S. And in fact, in French
they call it sum.
Even though we call it integral.
So it's a sum.
So it's consistent
with sums in this way.
And similarly, similarly we can
factor constants out of sums.
So if you have an integral like
this, the constant factors out.
But definitely don't try to
get a function out of this.
That won't happen.
OK, in other words, c
has to be a constant.
Doesn't depend on x.
The third property.
What do I want to call
the third property here?
I have sort of a preliminary
property, yes, here.
Which is the following.
And I'll draw a picture of it.
I suppose you have three
points along a line.
So then I'm going to
draw a picture of that.
And I'm going to use the
interpretation above the curve,
even though that's
not the whole thing.
So here's a, here's
b and here's c.
And you can see that
the area of this piece,
of the first two pieces
here, when added together,
gives you the area of the whole.
And that's the rule that
I'd like to tell you.
So if you integrate
from a to b, and you
add to that the
integral from b to c,
you'll get the
integral from a to c.
This is going to be just
a little preliminary,
because the rule is a
little better than this.
But I will explain
that in a minute.
The fourth rule is
a very simple one.
Which is that the integral
from a to a of f(x) dx
is equal to 0.
Now, that you can see very
obviously because there's
no area.
No horizontal movement there.
The rectangle is
infinitely thin,
and there's nothing there.
So this is the case.
You can also interpret
it a F(a) - F(a).
So that's also consistent
with our interpretation.
In terms of the fundamental
theorem of calculus.
And it's perfectly reasonable
that this is the case.
Now, the fifth property
is a definition.
It's not really a property.
But it's very important.
The integral from a to b of f(x)
dx equal to minus the integral
from b to a, of f( x) dx.
Now, really, the right-hand side
here is an undefined quantity
so far.
We never said you
could ever do this
where the a is less than the b.
Because this is
working backwards here.
But we just have a convention
that that's the definition.
Whenever we write
down this number,
it's the same as minus
what that number is.
And the reason for
all of these is again
that we want them to be
consistent with the fundamental
theorem of calculus.
Which is the thing that
makes all of this work.
So if you notice the left-hand
side here is F(b) - F(a),
capital F, the
antiderivative of little f.
On the other hand, the
other side is minus,
and if we just ignore that,
we say these are letters,
if we were a machine, we didn't
know which one was bigger than
which, we just plugged them in,
we would get here F(a) - F(b),
over here.
And to make these two
things equal, what we want
is to put that minus sign in.
Now it's consistent.
So again, these
rules are set up so
that everything is consistent.
And now I want to
improve on rule 3 here.
And point out to you - so
let me just go back to rule 3
for a second - that now that
we can evaluate integrals
regardless of the order, we
don't have to have a < b,
b < c in order to make
sense out of this.
We actually have the possibility
of considering integrals
where the a's and the
b's and the c's are
in any order you want.
And in fact, with
this definition,
with this definition 5, 3 works
no matter what the numbers are.
So this is much more convenient.
We don't, this is not necessary.
Not necessary.
It just works
using convention 5.
OK, with 5.
Again, before I go
on, let me emphasize:
we really want to respect
the sign of this velocity.
We really want the net
change in the position.
And we don't want this
absolute value here.
Because otherwise, all of our
formulas are going to mess up.
We won't always
be able to check.
Sometimes you have
letters rather than
actual numbers here,
and you won't know
whether a is bigger than b.
So you'll want to know that
these formulas work and are
consistent in all situations.
OK, I'm going to
trade these again.
In order to preserve the
ordering 1 through 5.
And now I have a sixth property
that I want to talk about.
This one is called estimation.
And it says the following.
If f(x) <= g(x), then the
integral from a to b of f(x) dx
is less than or equal to the
integral from a to b of g(x)
dx.
Now, this one says that if I'm
going more slowly than you,
then you go farther than I do.
OK.
That's all it's saying.
For this one, you'd
better have a < b.
You need it.
Because we flip the signs when
we flip the order of a and b.
So this one, it's essential
that the lower limit be smaller
than the upper limit.
But let me just emphasize,
because we're dealing
with the generalities of this.
Actually if one of
these is negative
and the other one is
negative, then it also works.
This one ends up being, if
f is more negative than g,
then this added up thing is
more negative than that one.
Again, under the assumption
that a is less than b.
So as I wrote it it's
in full generality.
Let's illustrate this one.
And then we have one more
property to learn after that.
So let me give you an
example of estimation.
The example is the same as
one that I already gave you.
But this time, because we
have the tool of integration,
we can just follow our
noses and it works.
I start with the
inequality, so I'm
trying to illustrate
estimation, so I
want to start with
an inequality which
is what the hypothesis is here.
And I'm going to
integrate the inequality
to get this conclusion.
And see what conclusion it is.
The inequality that I want
to take is that e^x >= 1,
for x >= 0.
That's going to be
our starting place.
And now I'm going
to integrate it.
That is, I'm going
to use estimation
to see what that gives.
Well, I'm going to
integrate, say, from 0 to b.
I can't integrate below 0
because it's only true above 0.
This is e^x dx greater than or
equal to the integral from 0
to b of 1 dx.
Alright, let's work out
what each of these is.
The first one, e^x dx, is,
the antiderivative is e^x,
evaluated at 0 and b.
So that's e^b - e^0.
Which is e^b - 1.
The other one,
you're supposed to be
able to get by
the rectangle law.
This is one rectangle
of base b and height 1.
So the answer is b.
Or you can do it by
antiderivatives, but it's b.
That means that our inequality
says if I just combine these
two things together,
that e^b - 1 >= b.
And that's the same
thing as e^b >= 1 + b.
Again, this only
works for b >= 0.
Notice that if b were
negative, this would be a well
defined quantity.
But this estimation
would be false.
We need that the b > 0 in
order for this to make sense.
So this was used.
And that's a good thing, because
this inequality is suspect.
Actually, it turns out to
be true when b is negative.
But we certainly
didn't prove it.
I'm going to just
repeat this process.
So let's repeat it.
Starting from the
inequality, the conclusion,
which is sitting right here.
But I'll write it in a form
e^x >= 1 + x, for x >= 0.
And now, if I
integrate this one,
I get the integral from 0 to b,
e^x dx is greater than or equal
to the integral from
0 to b, (1 + x) dx,
and I remind you that we've
already calculated this one.
This is e^b - 1.
And the other one is
not hard to calculate.
The antiderivative
is x + x^2 / 2.
We're evaluating
that at 0 and b.
So that comes out
to be b + b^2 / 2.
And so our conclusion is that
the left side, which is e^b -
1 >= b + b^2 / 2.
And this is for b >= 0.
And that's the same thing
as e^b >= 1 + b + b^2 / 2.
This one actually is
false for b negative,
so that's something
that you have
to be careful with
the b positive's here.
So you can keep on
going with this,
and you didn't have to think.
And you'll produce a very
interesting polynomial,
which is a good
approximation to e^b.
So that's it for the
basic properties.
Now there's one tricky property
that I need to tell you about.
It's not that tricky,
but it's a little tricky.
And this is change of variables.
Change of variables
in integration,
we've actually already done.
We called that, the last
time we talked about it,
we called it substitution.
And the idea here,
if you may remember,
was that if you're faced
with an integral like this,
you can change it to, if you put
in u = u(x) and you have a du,
which is equal to
u'(x) du-- dx, sorry.
Then you can change the
integral as follows.
This is the same as
g(u(x)) u'(x) dx.
This was the general
procedure for substitution.
What's new today is that we're
going to put in the limits.
If you have a limit here,
u_1, and a limit here, u_2,
you want to know what
the relationship is
between the limits here and
the limits when you change
variables to the new variables.
And it's the simplest
possible thing.
Namely the two limits over here
are in the same relationship
as u(x) is to this
symbol u here.
In other words, u_1 =
u(x_1), and u_2 = u(x_2).
That's what works.
Now there's only
one danger here,
there's one subtlety
which is, this only works
if u' does not change sign.
I've been worrying a little
bit about going backwards
and forwards, and
I allowed myself
to reverse and do all
kinds of stuff, right,
with these integrals.
So we're sort of free to do it.
Well, this is one case where
you want to avoid it, OK?
Just don't do it.
It is possible, actually,
to make sense out of it,
but it's also possible to get
yourself infinitely confused.
So just make sure
that-- Now, it's
OK if u' is always negative,
or always going one way,
so OK if u' is always
positive, you're always
going the other way,
but if you mix them
up you'll get yourself mixed up.
Let me give you an example.
The example will be maybe
close to what we did last time.
When we first did
substitution, I mean.
So the integral from 1 to 2,
this time I'll put in definite
limits, of x^2 plus-- sorry,
maybe I call this x^3. x^3 + 2,
let's say, I don't know,
to the 5th power, x^2 dx.
So this is an example
of an integral
that we would have tried to
handle by substitution before.
And the substitution we would
have used is u = x^3 + 2.
And that's exactly what
we're going to do here.
But we're just going to also
take into account the limits.
The first step, as in any
substitution or change
of variables, is this.
And so we can fill
in the things that we
would have done previously.
Which is that this is the
integral and this is u^5.
And then because this is
3x^2, we see that this is 3.
Sorry, let's write
it the other way.
1/3 du = x^2 dx.
So that's what I'm going to
plug in for this factor here.
So here's 1/3 du,
which replaces that.
But now there's
the extra feature.
The extra feature is the limits.
So here, really in
disguise, because, and now
this is incredibly important.
This is one of the reasons why
we use this notation dx and du.
We want to remind
ourselves which variable
is involved in the integration.
And especially if you're the
one naming the variables,
you may get mixed
up in this respect.
So you must know which variable
is varying between 1 and 2.
And the answer is, it's
x is the one that's
varying between 1 and 2.
So in disguise, even
though I didn't write it,
it was contained in
this little symbol here.
This reminded us which variable.
You'll find this amazingly
important when you
get to multivariable calculus.
When there are many
variables floating around.
So this is an incredibly
important distinction to make.
So now, over here
we have a limit.
But of course it's supposed
to be with respect to u, now.
So we need to calculate what
those corresponding limits are.
And indeed it's just, I plug in
here u_1 is going to be equal
to what I plug in for x = 1,
that's going to be 1^3 + 2,
which is 3.
And then u_2 is 2^3 + 2,
which is equal to 10, right?
8 + 2 = 10.
So this is the integral
from 3 to 10, of u^5 1/3 du.
And now I can
finish the problem.
This is 1/18 u^6, from 3 to 10.
And this is where the
most common mistake occurs
in substitutions of this type.
Which is that if
you ignore this,
and you plug in
these 1 and 2 here,
you think, oh I should just
be putting it at 1 and 2.
But actually, it
should be, the u-value
that we're interested in,
and the lower limit is u = 3
and u = 10 is the upper limit.
So those are suppressed here.
But those are the
ones that we want.
And so, here we go.
It's 1/18 times some ridiculous
number which I won't calculate.
10^6 - - 3^6.
Yes, question.
STUDENT: [INAUDIBLE]
PROFESSOR: So, if
you want to do things
with where you're worrying
about the sign change,
the right strategy is,
what you suggested works.
And in fact I'm going to
do an example right now
on this subject.
But, the right strategy is
to break it up into pieces.
Where u' has one sign
or the other, OK?
Let me show you an example.
Where things go wrong.
And I'll tell you how
to handle it, roughly.
So here's our warning.
Suppose you're integrating
from -1 to 1, x^2 dx.
Here's an example.
And you have the temptation
to plug in u = x^2.
Now, of course, we know
how to integrate this.
But let's just pretend we
were stubborn and wanted
to use substitution.
Then we have du = 2x dx.
And now if I try to
make the correspondence,
notice that the limits
are u_1 = (-1)^2,
that's the bottom limit.
And u_2 is the upper limit.
That's 1^2, that's
also equal to 1.
Both limits are 1.
So this is going from 1 to 1.
And no matter what it is,
we know it's going to be 0.
But we know this is not 0.
This is the integral
of a positive quantity.
And the area under a curve is
going to be a positive area.
So this is a positive quantity.
It can't be 0.
If you actually plug it in,
it looks equally strange.
You put in here this u and then,
so that would be for the u^2.
And then to plug in for dx,
you would write dx = 1/(2x) du.
And then you might
write that as this.
And so what I should put in
here is this quantity here.
Which is a perfectly
OK integral.
And it has a value, I
mean, it's what it is.
It's 0.
So of course this is not true.
And the reason is that
u was equal to x^2,
and u'(x) was equal to 2x, which
was positive for x positive,
and negative for x negative.
And this was the sign change
which causes us trouble.
If we break it off into its
two halves, then it'll be OK
and you'll be able to use this.
Now, there was a mistake.
And this was essentially
what you were saying.
That is, it's possible to see
this happening as you're doing
it if you're very careful.
There's a mistake
in this process,
and the mistake is
in the transition.
This is a mistake here.
Maybe I haven't used
any red yet today,
so I get to use some red here.
Oh boy.
This is not true, here.
This step here.
So why isn't it true?
It's not true for
the standard reason.
Which is that really, x = plus
or minus square root of u.
And if you stick to
one side or the other,
you'll have a coherent
formula for it.
One of them will be the plus and
one of them will be the minus
and it will work out when you
separate it into its pieces.
So you could do that.
But this is a can of worms.
So I avoid this.
And just do it in a place where
the inverse is well defined.
And where the
function either moves
steadily up or steadily down.
