Lower Sum, Upper Sum

Let $a<b$. A partition of the interval is a collection $P$ of points
MATH

Let $f$ be a bounded function defined on $[a,b]$ and let $P$ be a partition of $[a,b]$. Let
MATH
and
MATH
for MATH. The lower sum of $f$ for $P$ is
MATH
and the upper sum of $f$ for $P$ is
MATH
We notice that: if $x_{i-1}<z<x_i$, then
MATH

MATH
Therefore if $P$ and $Q$ are partitions of $[a,b]$ with $P\subset Q$ and $|Q\backslash P|=1$, then the corresponding lower sums and upper sums satisfy the inequalities
MATH
In case partitions $P$ and $Q$ of the same interval $[a,b]$ satisfy
MATH
it is possible to find partitions $R_i$ so that
MATH
with MATH. The above observation shows now that the corresponding lower sums and upper sums satisfy the inequalities
MATH

Consider two arbitrary partitions $P$ and $Q$ of $[a,b]$. Then the partition
MATH
satisfies
MATH

The above observation now shows
MATH
Since it is always true that $L(f,R)\leq U(f,R)$, we see that
MATH
for any two partitions $P$, $Q$ of $[a,b]$. Hence
MATH
and
MATH

Integral of a Bounded Function

A bounded function defined on $[a,b]$ is call integrable if there is exactly one number $I$ satisfying
MATH
for any partitions $P$ and $Q$ of $[a,b]$. This unique number is called the definite integral of $f$ on $[a,b]$ and is denoted
MATH
The above discussion shows that
MATH

Characterization of an Integrable Function

Theorem. If $f$ is a bounded function defined on $[a,b]$, then $f$ is integrable on $[a,b]$ if and only if for each $\epsilon >0$ there exists a partition $P$ of $[a,b]$ such that
MATH

Every Continuous Function is Integrable

Theorem. If $f$ is continuous on $[a,b]$, then $f$ is integrable on $[a,b].$

Proof. Let $\epsilon >0$. We show that there exists a partition $P$ of $[a,b] $ with MATH. As was shown earlier the continuous function $f$ defined on the closed and bounded interval $[a,b]$ is uniformly continuous. Thus we can find $\delta >0$ such that
MATH
Choose any partition $P$ consisting of
MATH
with MATH for all $i$. It follows that
MATH
for MATH. Therefore
MATH

MATH

The First Fundamental Theorem of Calculus

Theorem. Let $f$ be integrable on $[a,b]$. Define $F$ on $[a,b]$ by
MATH

If $f$ is continuous at $c$ in $[a,b]$ then $F$ is differentiable at $c$, and
MATH

Proof. Let $\epsilon >0$. We show that there is $\delta >0$ such that
MATH
whenever $0<|x-c|<\delta $. Since $f$ is continuous at $c$, there is a $\delta >0$ such that MATH for $|x-c|<\delta .$

Case (i) $x<c$. Since $x<c$ is a partition of the interval $[x,c]$, it follows that
MATH

Therefore
MATH

Case (ii) $c<x$. Since $c<x$ is a partition of the interval $[c,x]$, it follows that
MATH

Therefore
MATH

The Second Fundamental Theorem of Calculus

Theorem. If $f$ is integrable on $[a,b]$ and $f=g^{\prime }$ for some function $g$, then
MATH

Proof. We need to show that
MATH
for every partition $P$
MATH
of $[a,b].$

By the Mean Value Theorem there is some point $z_{i}$ in $[x_{i-1},x_{i}]$ such that
MATH
If
MATH
and
MATH
then
MATH
that is
MATH
Adding these equations for MATH, we have
MATH

MATH

MATH
Therefore
MATH

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