# Inequalities – Property – II

## Inequalities – Property – II

Property – II: If m is the value(global minimum) and M is the greatest value(global maximum) of the function f(x) on the interval [a, b] (estimation of an integral), then $$m\left( b-a \right)\le \int\limits_{a}^{b}{f\left( x \right)}\ dx\le M\left( b-a \right)$$.

Proof:

It is given that $$m\le f\left( x \right)\le M$$. Then

$$\int\limits_{a}^{b}{m.dx}\le \int\limits_{a}^{b}{f\left( x \right).dx}\le \int\limits_{a}^{b}{M.dx}$$,

$$m\left( b-a \right)\le \int\limits_{a}^{b}{f\left( x \right)\ dx\le \ }M\left( b-a \right)$$

It is clear from fig

Area of $$AB{{B}^{‘}}A\le \int\limits_{a}^{b}{f\left( x \right)}\ dx\le Area\ \ of\ \ ABB”A”$$

i.e., $$m\left( b-a \right)\le \int\limits_{a}^{b}{f\left( x \right)}\ dx\le M\left( b-a \right)$$.