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 $f$ have a maximum of $M$ and a minimum of $m$ on $\left[{a \,.\,.\, b}\right]$.

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

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

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

Suppose that $\forall t \in \left[{a \,.\,.\, b}\right]: \left|{f \left({t}\right)}\right| < \kappa$.

Then:
 * $\displaystyle \forall \xi, x \in \left[{a \,.\,.\,b}\right]: \left|{\int_x^\xi f \left({t}\right)\ \mathrm dt}\right| < \kappa \left|{x - \xi}\right|$

Proof
This follows directly from the definition of definite integral.


 * From the Continuity Property it follows that $m$ and $M$ both exist.


 * The closed interval $\left[{a \,.\,.\, b}\right]$ is a 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.

Proof of Corollary
Follows directly from the main proof.