# Inequalities – Property 1

## Inequalities – Property 1

Property: If at every point x of an interval [a, b], the inequalities $$g\left( x \right)\le f\left( x \right)\le h\left( x \right)$$ are fulfilled, then $$\int\limits_{a}^{b}{g\left( x \right)}.dx\le \int\limits_{a}^{b}{f\left( x \right)}.dx\le \int\limits_{a}^{b}{h\left( x \right)}.dx$$, where a < b.

Proof:

In fig. Area of curvilinear trapezoid aAFb ≤ Area of curvilinear trapezoid aBEb ≤ Area of curvilinear trapezoid aCDb

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

Example: Prove that $$0<\int\limits_{0}^{1}{\frac{{{x}^{7}}.dx}{\sqrt[3]{\left( 1+{{x}^{8}} \right)}}}<\frac{1}{8}$$

Solution:

$$0<\int\limits_{0}^{1}{\frac{{{x}^{7}}.dx}{\sqrt[3]{\left( 1+{{x}^{8}} \right)}}}<{{x}^{7}}$$$$\forall 0<x<l$$

$$\int\limits_{0}^{1}{0.dx}<\int\limits_{0}^{1}{\frac{{{x}^{7}}}{\sqrt[3]{\left( 1+{{x}^{8}} \right)}}.dx}<\int\limits_{0}^{1}{{{x}^{7}}.dx}$$ $$0<\int\limits_{0}^{1}{\frac{{{x}^{7}}.dx}{\sqrt[3]{\left( 1+{{x}^{8}} \right)}}}<{{\left( \frac{{{x}^{8}}}{8} \right)}_{0}}^{1}$$

Hence $$0<\int\limits_{0}^{1}{\frac{{{x}^{7}}.dx}{\sqrt[3]{\left( 1+{{x}^{8}} \right)}}}<\frac{1}{8}$$