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

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

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

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.