Application of Even- Odd Properties in Definite Integration - Practice Questions & MCQ

Property 7

$\int_{-a}^a f(x) d x=\left\{\begin{array}{cc}0, & \text { if } f \text { is an odd function } \\ \text { i.e. } f(-x)=-f(x) \\ 2 \int_0^a f(x) d x, & \text { if } f \text { is an even function } \\ \text { i.e. } f(-x)=f(x)\end{array}\right.$

Proof: 

$\begin{aligned} \int_{-\mathrm{a}}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x} & =\underbrace{\int_{-a}^0 f(x) d x}_{x=-t}+\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx} \\ & =\int_a^0 f(-t)(-d t)+\int_0^a f(x) d x \\ & =\int_0^a f(-x)(d x)+\int_0^a f(x) d x \\ & =\left\{\begin{array}{cc}-\int_0^a f(x) d x+\int_0^a f(x) d x, & \text { if } \mathrm{f}(\mathrm{x}) \text { is odd } \\ \int_0^a f(x) d x+\int_0^a f(x) d x, & \text { if } \mathrm{f}(\mathrm{x}) \text { is even }\end{array}\right. \\ & =\left\{\begin{array}{cc}0, & \text { if } f \text { is an odd function } \\ 2 \int_0^a f(x) d x, & \text { if } f \text { is an even function }\end{array}\right.\end{aligned}$

Proof using Graph

The graph of the odd function is symmetric about the origin, as shown in the above figure

So, if $\int_0^a f(x) d x=\alpha$ then, $\int_{-a}^0 f(x) d x=-\alpha$
$
\therefore \quad \int_{-a}^a f(x) d x=0
$

The graph of the even function is symmetric about the y-axis, as shown in the above figure

So, $\quad \int_{-a}^0 f(x) d x=\int_0^a f(x) d x=\alpha$
$
\therefore \quad \int_{-a}^a f(x) d x=2 \alpha=2 \int_0^a f(x) d x
$

Corollary:
$
\int_0^{2 a} f(x) d x=\left\{\begin{array}{cc}
2 \int_0^a f(x) d x, & \text { if } f(2 a-x)=f(x) \\
0, & \text { if } f(2 a-x)=-f(x)
\end{array}\right.
$

Property 9

If f(x) is a periodic function with period T, then the area under f(x) for n periods would be n times the area under f(x) for one period, i.e.

$\int_0^{n T} f(x) d x=n \int_0^T f(x) d x$

Proof:

Graphical Method

f(x) is a periodic function with period T. Consider the following graph of function f(x).

The graph of the function is the same in each of the intervals (0, T), (T, 2T), (2T, 3T) ……..

So,

$\begin{aligned} \int_0^{\mathrm{nT}} \mathrm{f}(\mathrm{x}) \mathrm{d} & =\text { total shaded area } \\ & =\mathrm{n} \times(\text { area in the interval }(0, \mathrm{~T})) \\ & =\mathrm{n} \int_0^{\mathrm{T}} \mathrm{f}(\mathrm{x}) \mathrm{dx}\end{aligned}$

Property 10

$\int_a^{a+n T} f(x) d x=\int_0^{a T} f(x) d x=n \int_0^T f(x) d x$

Proof:

Let, $\quad \mathrm{I}=\int_{\mathrm{a}}^{\mathrm{a}+\mathrm{nT}} \mathrm{f}(\mathrm{x}) \mathrm{dx}$
$
=\int_{\mathrm{a}}^0 \mathrm{f}(\mathrm{x}) \mathrm{dx}+\int_0^{\mathrm{nT}} \mathrm{f}(\mathrm{x}) \mathrm{dx}+\underbrace{\int_{n T}^{a+n T} f(x) d x}_{x=y+n T}
$
$\Rightarrow \mathrm{dx}=\mathrm{dy}$ and when $\mathrm{x}=\mathrm{nT}$ then $\mathrm{y}=0$ and $\mathrm{x}=\mathrm{a}+\mathrm{nT}, \mathrm{y}=\mathrm{a}$
$
\begin{aligned}
& \quad=\int_a^0 f(x) d x+\int_0^{n T} f(x) d x+\int_0^a f(y) d y \\
& \quad=n \int_0^{n T} f(x) d x \\
& {\left[\because \int_0^a f(y) d y=\int_0^a f(x) d x \text { and } \int_0^a f(x) d x=-\int_a^0 f(x) d x\right]}
\end{aligned}
$

Property 11
$
\int_{a+n T}^{b+n T} f(x) d x=\int_a^b f(x) d x
$

Property 12
$
\int_{\mathrm{mT}}^{\mathrm{nT}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=(\mathrm{n}-\mathrm{m}) \int_0^{\mathrm{T}} \mathrm{f}(\mathrm{x}) \mathrm{dx}
$

Where ‘T’ is the period and m and n are Integers.