# Upper and Lower Bounds of Integral

## Theorem

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.

Let $\ds \int_a^b \map f x \rd x$ be the definite integral of $\map f x$ over $\closedint a b$.

Then:

$\ds m \paren {b - a} \le \int_a^b \map f x \rd x \le M \paren {b - a}$

where:

$M$ is the maximum of $f$
$m$ is the minimum of $f$

on $\closedint a b$.

### Corollary

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.

Suppose that $\forall t \in \closedint a b: \size {\map f t} < \kappa$.

Then:

$\ds \forall \xi, x \in \closedint a b: \size {\int_x^\xi \map f t \rd t} < \kappa \size {x - \xi}$

## Proof

This follows directly from the definition of definite integral:

From Continuous Image of Closed Interval is Closed Interval it follows that $m$ and $M$ both exist.

The closed interval $\closedint a b$ is a finite subdivision of itself.

By definition, the upper sum is $M \paren {b - a}$, and the lower sum is $m \paren {b - a}$.

The result follows.

$\blacksquare$