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