The Cauchy Mean Value Theorem is the basic tool needed to prove a theorem
which facilitates evaluation of limits of the form
when
Theorem. Suppose that
and and suppose also that
exists. Then
exists, and
PROOF. The hypothesis that exists contains two implicit assumptions:
(1) there is an interval such that and exist for all in except, perhaps, for ,
(2) in this interval with, once again, the possible exception of
On the other hand,
and
are not even assumed to be defined at
.
If we define
(changing the previous values of
and
,
if necessary), then
and
are continuous at
.
If
,
then the Mean Value Theorem and the Cauchy Mean Value Theorem apply to
and
on the interval
(and a similar statement holds for
).
First applying the Mean Value Theorem to
,
we see that
,
for if
there would be some
in
with
,
contradicting (2). Now applying the Cauchy Mean Value Theorem to
and
,
we see that there is a number
.
in
such that
or
Now
approaches
as
approaches
,
because
,
is in
;
since
exists, it follows that
Definition. The function is uniformly continuous on an interval if for every there is some such that, for all and in
Lemma. Let and let be continuous on the interval Let , and suppose that statements (i) and (ii) hold:
(i) if and are in and , then ,
(ii) if and are in and , then
Then there is a
0 such that
Proof. Since
is continuous at
,
there is a
0 such that
It follows that
(iii) if and then
Choose to be the minimum of and . We claim that this works. In fact, suppose that and are any two points in with . If and are both in then by (i); and if and are both in then by (ii). The only other possibility is that
or
In either case, since , we also have and So by (iii).
THEOREM If is continuous on , then is uniformly continuous on
PROOF: For let's say that is good on if there is some such that, for all and in if , then . Then we're trying to prove that is -good on for all
Consider any particular . Let
Then
(since
a is in
),
and
is bounded above (by
),
so
has a least upper bound
.
We really should write
,
since
and
might depend on
.
But we won't since we intend to prove that
,
no matter what
is.
Suppose that we had . Since is continuous at , there is some such that, if , then . Consequently, if and , then So is surely -good on the interval . On the other hand, since is the least upper bound of , it is also clear that is -good on . Then the above Lemma implies that is -good on , so is in A, contradicting the fact that is an upper bound.
To complete the proof we just have to show that is actually in . The argument for this is practically the same: Since is continuous at , there is some such that, if , then So is -good on . But is also -good on , so the Lemma implies that is -good on .
Definition. If
,
define
It follows from the Fundamental Theorem of Calculus that
is differentiable and
For each fixed
,
define
.
Then
.
Therefore there is a constant
so that
Setting
,
we see that
Therefore
Corollary. If
is a natural number and
,
then
Corollary. If
,
then
Since
for all
,
is an increasing function on
.
It is unbounded from above:
and it is unbounded from below:
Therefore the inverse function
of
exists. The exponential function
is defined as
.
Thus
is the unique positive number such that
holds for all real number
Since
holds for all
,
it follows that, writing
and
Therefore
is differentiable at any
with
Write
,
Then
,
.
Then
so
Definition:
Thus
is the unique number satisfying
As
and
it follows that
Since
,
it follows that
holds for all integers
and
holds for all integers
with
,
i.e.,
holds for all rational numbers.
Definition. For any number ,
Note that if
is rational, then
Definition. If
,
then for any real number
In case , this definition is consistent with the earlier definition.
(1) If , then for all
(2) If , then , for all
Theorem. If
is differentiable and
then there is a number
such that
Proof. Consider
Then
Therefore
must be constant
Hereditary: If is integrable on , then is integrable on for any
Order Preserving: If
and
are integrable on
and
,
then
Corollary: If
is integrable on
,
then
is integrable on
and
Proof:
Linearity: If
and
are integrable on
,
then
Cauchy-Schwarz Inequality
Theorem. If
and
are integrable on
,
then
unless there exists a constant such that
Proof. Write
The algebraic properties of integrals show that
(i) unless
(ii)
(iii)
(iii')
We need to show that
Case (1) . The result holds since both sides are zero.
Case
(2).
Then
Therefore
If the equality holds then
consequently
Triangular Inequality
If
and
are integrable on
,
then
Proof.
Recall that the first fundamental theorem of calculus asserts that if is integrable on and if is continuous at then the function defined by is differentiable at and . Note that the function is not assumed to be continuous throughout . The following result, however, does hold true.
Theorem. If
is integrable on
and
is defined by
then
is continuous on
Proof. Fix . Since is integrable on it is, by definition, bounded on . Choose so that for all . Let be given. Set . We now show that for
Case (1).
.
Then
Since
it follows that
Therefore,
Case (2).
.
Then
Since
it follows that
Therefore,
Thus
in both cases.
Theorem. Suppose that
is continuous on
and that
is integrable and nonnegative on
.
Then there exists
in
such that
Corollary. Suppose that
is continuous on
.
Then there exists
in
such that
Proof. Since
is continuous on
,
there exists
in
such that
and there exists
in
such that
It follows from the inequality
that
If
then the result clearly holds. If
is not identically
then
and so
Now the function
given by
is positive for
and negative for
.
It follows from the Intermediate Value Theorem that there exists
(between
and
with
Let
.
Let
be a partition of the interval
consisting of
For a bounded function
defined on
and for each
fix a point
in the interval
.
Then
Any sum of the form
is called a Riemann sum of
for
Theorem. Suppose that
is continuous on
.
Then for every
there is some
such that, if
be a partition of the interval
consisting of
with
for all
,
then
for any Riemann sum formed by choosing
in
Proof. Given
,
choose
so that for all
in
This is possible since
is uniformly continuous on
.
Now for any partition
satisfies the stated condition we have
But we have also
and
The result follows from the above three inequalities.
Let a curve be given parametrically by
For any partition
of
the total length of the polygonal line segments connecting
where
and
for all
,
is given by
We now show that for each given
there exists a
such that if all
in the partition
are
,
then
(1) For the given
,
there exists
so that
whenever
is a partition
with
for all
.
(2) Since the function is uniformly continuous for all , it follows that there exists such that
(3) Since the function
is continuous on
,
it is uniformly continuous there. Thus there exists
such that
Let
.
Consider a partition
consisting of
with
for all
.
The Mean Value Theorem shows that there exist
and
in
such that
Since
,
it follows that
and so
The desired result follows since
If
and
are continuous on
,
the length of the parametric curve
is defined to be the definite integral
If
is continuous on
,
then the length of its graph is the same as the length of the parametric curve
Therefore it is given by
If the curve is given by the polar equation
It is just a particular instance of the parametric equations
Since
and
it follows that
so the length of its graph is given by
Example. Find
Set
.
Then
and
.
Thus
Since
,
,
we have
Example.
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