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

$\int_a^b f(x) d x \leq \int_a^b g(x) d x \leq \int_a^b h(x) d x$

As from the figure, Area of $A B D C \leq$ Area of $A B F E \leq$ Area of $A B H G$

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

$\mathbf{m}(\mathbf{b}-\mathbf{a}) \leq \int_{\mathbf{a}}^{\mathbf{b}} \mathbf{f}(\mathbf{x}) \mathrm{dx} \leq \mathbf{M}(\mathbf{b}-\mathbf{a})$

Proof:

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

$\begin{array}{ll}\therefore & \int_a^b \mathrm{mdx} \leq \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx} \leq \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{M} \mathrm{dx} \\ \Rightarrow & \mathrm{m}(\mathrm{b}-\mathrm{a}) \leq \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx} \leq \mathrm{M}(\mathrm{b}-\mathrm{a})\end{array}$

Property 3

$
\left|\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}\right| \leq \int_{\mathrm{a}}^{\mathrm{b}}|\mathbf{f}(\mathrm{x})| \mathrm{dx}
$

We know that, $\quad-|\mathrm{f}(\mathrm{x})| \leq \mathrm{f}(\mathrm{x}) \leq|\mathrm{f}(\mathrm{x})|, \forall \mathrm{x} \in[\mathrm{a}, \mathrm{b}]$
$
\begin{aligned}
& \therefore \quad-\int_a^b|\mathrm{f}(\mathrm{x})| \mathrm{dx} \leq \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx} \leq \int_{\mathrm{a}}^{\mathrm{b}}|\mathrm{f}(\mathrm{x})| \mathrm{dx} \\
& \Rightarrow \\
& \quad\left|\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}\right| \leq \int_{\mathrm{a}}^{\mathrm{b}}|\mathrm{f}(\mathrm{x})| \mathrm{dx}
\end{aligned}
$

If $f^2(x)$ and $g^2(x)$ are integrable function on the interval $[a, b]$, then
$
\left|\int_a^b f(x) g(x) d x\right| \leq \sqrt{\left(\int_a^b f^2(x) d x\right)\left(\int_a^b g^2(x) d x\right)} .
$

Proof:

Let $\mathrm{F}(\mathrm{x})=\{\mathrm{f}(\mathrm{x})-\lambda \mathrm{g}(\mathrm{x})\}^2$, where $\lambda$ is a real number.
Since, $\{f(x)-\lambda g(x)\}^2 \geq 0$
So,
$
\begin{aligned}
& \int_a^b\{f(x)-\lambda g(x)\}^2 d x \geq 0 \\
\Rightarrow \quad & \int_a^b\left\{f^2(x)-2 \lambda f(x) g(x)+\lambda^2 g^2(x)\right\} d x \geq 0 \\
\Rightarrow \quad & \lambda^2 \int_a^b\left\{g^2(x)\right\} d x-2 \lambda \int_a^b\{f(x) g(x)\} d x+\int_a^b\left\{f^2(x)\right\} d x \geq 0
\end{aligned}
$

Discriminant is non positive, i.e. $\quad B^2-4 A C \leq 0$
$
\begin{aligned}
& \Rightarrow \quad 4 \lambda^2\left\{\int_a^b\{f(x) g(x)\} d x\right\}^2 \leq 4\left\{\lambda^2 \int_a^b\left\{g^2(x)\right\} d x\right\}\left\{\int_a^b\left\{f^2(x)\right\} d x\right\} \\
& \Rightarrow\left|\int_a^b f(x) g(x) d x\right| \leq \sqrt{\left(\int_a^b f^2(x) d x\right)\left(\int_a^b g^2(x) d x\right)}
\end{aligned}
$