Upper and Lower Bounds of Integral

Theorem
Let $f$ be a real function which is continuous on the closed interval $\left[{a \,.\,.\, b}\right]$.

Let $\displaystyle \int_a^b f \left({x}\right) \rd x$ be the definite integral of $f \left({x}\right)$ over $\left[{a \,.\,.\, b}\right]$.

Then:
 * $\displaystyle m \left({b - a}\right) \le \int_a^b f \left({x}\right) \rd x \le M \left({b - a}\right)$

where:
 * $M$ is the maximum of $f$
 * $m$ is the minimum of $f$

on $\left[{a \,.\,.\, b}\right]$.

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 $\left[{a \,.\,.\, b}\right]$ is a finite subdivision of itself.

By definition, the upper sum is $M \left({b - a}\right)$, and the lower sum is $m \left({b - a}\right)$.

The result follows.