Upper and Lower Bounds of Integral

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Theorem

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

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


Then:

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

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

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


Then:

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


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 $\closedint a b$ 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.

$\blacksquare$


Sources