Calculus : Integral Inequality?

Calculus is one of the major topics in Mathematics across the world. . The problems involving the definite integrals require advanced knowledge. In this lecture with the help of the problem defined here we try to get the insights of the definite integrals calculations.

What is meant by inequalities in integral calculus?

Consider an integral inequality like,

$a<f(x)<b\Rightarrow a(q-p)<\underset{p}{\overset{q}{\int }}f(x)\mathrm{dx}<b(q-p)$

To understand the above inequality, let us draw an analogy to simple algebraic inequality such as ,

(x-1) < x < (x+1) for any real value of x.

The proof of above simple inequality is ,

First consider left part of the inequality, (x-1)<x implies , -1< 0. This is always true.

Similarly if the right part of the inequality is taken, x<(x+1) implies , 0<1. This is always true.

Let us expand our this understanding to integral inequality we just considered. From the basic principles of definite integrals,

$a<f(x)<b\Rightarrow a(q-p)<\underset{p}{\overset{q}{\int }}f(x)\mathrm{dx}<b(q-p)$ can also be proved very easily.