Application of Inequality in Definite Integration - Practice Questions & MCQ

Property 1

If f(x) ≤ g(x) ≤ h(x) for all x in [a, b], then

${\int }_a^{b}f(x)dx\le {\int }_a^{b}g(x)dx\le {\int }_a^{b}h(x)dx$

As from the figure, Area of $ABDC\le$ Area of $ABFE\le$ Area of $ABHG$

Property 2

If m is the least value (global minimum) and M is the greatest value (global maximum) of the function f(x) on the interval [a, b], then

$\mathbf{m}(\mathbf{b}-\mathbf{a})\le {\int }_{\mathbf{a}}^{\mathbf{b}}\mathbf{f}(\mathbf{x})\mathrm{dx}\le \mathbf{M}(\mathbf{b}-\mathbf{a})$

Proof:

It is given that m ≤  f(x) ≤  M for all x in [a, b]

$\begin{array}{ll}\therefore & {\int }_a^b\mathrm{mdx}\le {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{x})\mathrm{dx}\le {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{M}\mathrm{dx}\\ \Rightarrow & \mathrm{m}(\mathrm{b}-\mathrm{a})\le {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{x})\mathrm{dx}\le \mathrm{M}(\mathrm{b}-\mathrm{a})\end{array}$

Property 3

$|{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{x})\mathrm{dx}|\le {\int }_{\mathrm{a}}^{\mathrm{b}}|\mathbf{f}(\mathrm{x})|\mathrm{dx}$

We know that, $-|\mathrm{f}(\mathrm{x})|\le \mathrm{f}(\mathrm{x})\le |\mathrm{f}(\mathrm{x})|,\mathrm{\forall }\mathrm{x}\in [\mathrm{a},\mathrm{b}]$
$\begin{array}{rl}& \therefore -{\int }_a^b|\mathrm{f}(\mathrm{x})|\mathrm{dx}\le {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{x})\mathrm{dx}\le {\int }_{\mathrm{a}}^{\mathrm{b}}|\mathrm{f}(\mathrm{x})|\mathrm{dx}\\ & \Rightarrow \\ & |{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{x})\mathrm{dx}|\le {\int }_{\mathrm{a}}^{\mathrm{b}}|\mathrm{f}(\mathrm{x})|\mathrm{dx}\end{array}$

If $f^2(x)$ and $g^2(x)$ are integrable function on the interval $[a,b]$, then
$|{\int }_a^{b}f(x)g(x)dx|\le \sqrt{({\int }_a^b{f}^2(x)dx)({\int }_a^b{g}^2(x)dx)}.$

Proof:

Let $\mathrm{F}(\mathrm{x})=\{\mathrm{f}(\mathrm{x})-\lambda \mathrm{g}(\mathrm{x}){\}}^2$, where $\lambda$ is a real number.
Since, $\{f(x)-\lambda g(x){\}}^2\ge 0$
So,
$\begin{array}{rl}& {\int }_a^b\{f(x)-\lambda g(x){\}}^{2}dx\ge 0\\ \Rightarrow & {\int }_a^b\{f^2(x)-2\lambda f(x)g(x)+{\lambda }^2{g}^2(x)\}dx\ge 0\\ \Rightarrow & {\lambda }^2{\int }_a^b\{g^2(x)\}dx-2\lambda {\int }_a^b\{f(x)g(x)\}dx+{\int }_a^b\{f^2(x)\}dx\ge 0\end{array}$

Discriminant is non positive, i.e. $B^2-4AC\le 0$
$\begin{array}{rl}& \Rightarrow 4{\lambda }^2{\{{\int }_a^b\{f(x)g(x)\}dx\}}^2\le 4\left\{{\lambda }^2{\int }_a^b\{g^2(x)\}dx\right\}\left\{{\int }_a^b\{f^2(x)\}dx\right\}\\ & \Rightarrow |{\int }_a^{b}f(x)g(x)dx|\le \sqrt{({\int }_a^b{f}^2(x)dx)({\int }_a^b{g}^2(x)dx)}\end{array}$