Let
.
A **partition** of the interval is a collection
of points

Let
be a bounded function defined on
and let
be a partition of
.
Let

and

for
.
The **lower sum** of
for
is

and the **upper sum** of
for
is

We notice that: if
,
then

Therefore if
and
are partitions of
with
and
,
then the corresponding lower sums and upper sums satisfy the inequalities

In case partitions
and
of the same interval
satisfy

it is possible to find partitions
so that

with
.
The above observation shows now that the corresponding lower sums and upper
sums satisfy the inequalities

Consider two arbitrary partitions
and
of
.
Then the partition

satisfies

The above observation now shows

Since it is always true that
,
we see that

for any two partitions
,
of
.
Hence

and

A bounded function defined on
is call **integrable** if there is exactly one number
satisfying

for any partitions
and
of
.
This unique number is called the **definite integral** of
on
and is denoted

The above discussion shows that

Theorem. If
is a bounded function defined on
,
then
is integrable on
if and only if for each
there exists a partition
of
such that

Theorem. If is continuous on , then is integrable on

Proof. Let
.
We show that there exists a partition
of
with
.
As was shown earlier the continuous function
defined on the closed and bounded interval
is **uniformly continuous.** Thus we can find
such that

Choose any partition
consisting of

with
for all
.
It follows that

for
.
Therefore

Theorem. Let
be integrable on
.
Define
on
by

If
is continuous at
in
then
is differentiable at
,
and

Proof. Let
.
We show that there is
such that

whenever
.
Since
is continuous at
,
there is a
such that
for

Case (i)
.
Since
is a partition of the interval
,
it follows that

Therefore

Case (ii)
.
Since
is a partition of the interval
,
it follows that

Therefore

Theorem. If
is integrable on
and
for some function
,
then

Proof. We need to show that

for every partition

of

By the **Mean Value Theorem** there is some point
in
such that

If

and

then

that
is

Adding
these equations for
,
we have

Therefore