Application of Even-Odd Properties in Definite Integrals

Application of Even-Odd Properties in Definite Integration

Definite integration calculates the area under a curve between two specific points on the x-axis.

Let f be a function of x defined on the closed interval [a, b]. F be another function such that $\frac{d}{d x}(F(x))=f(x)$ for all x in the domain of f, then $\int_a^b f(x) d x=[F(x)+c]_a^b=F(b)-F(a)$is called the definite integral of the function f(x) over the interval [a, b], where a is called the lower limit of the integral and b is called the upper limit of the integral.

Definite integrals have properties that relate to the limits of integration.

Property 1

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

Proof:

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

Proof using Graph

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

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

$
\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

$\begin{array}{ll}\text { So, } & \int_{-a}^0 \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\alpha \\ \therefore & \int_{-a}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=2 \alpha=2 \int_0^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}\end{array}$

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 2

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.

$\int_0^{\mathrm{nT}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\mathrm{n} \int_0^{\mathrm{T}} \mathrm{f}(\mathrm{x}) \mathrm{dx}$

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 interval (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 3

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

Proof:

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

Property 4

$
\int_{\mathrm{a}+\mathrm{nT}}^{\mathrm{b}+\mathrm{nT}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}
$

Property 5

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

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

Solved Example Based on Application of Even-Odd Properties in Definite Integration

Example 1: $\int_{-3 \pi / 2}^{-\pi / 2}\left[(x+\pi)^3+\cos ^2(x+3 \pi)\right] d x$ is equal to

1) $\frac{\pi^4}{32}$
2) $\frac{\pi^4}{32}+\frac{\pi}{2}$
3) $\frac{1}{2}$
4) $\frac{\pi}{2}-1$

Solution

As learnt in concept

Properties of definite integration -

If $f(x)$ is an EVEN function of x: then integral of the function from - a to a is the same as twice the integral of the same function from o to a.

$\int_{-a}^a f(x) d x=2\left\{\int_o^a f(x) d x\right\}$

wherein

Check even function $f(-x)=f(x)$ and symmetrical about y axis.
$
I=\int_{\frac{-3 \pi}{2}}^{-\frac{\pi}{2}}\left[(x+\pi)^3+\cos ^2(x+3 \pi)\right] d x
$
Put $x+\pi=t$

$
\begin{aligned}
& I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left[t^3+\cos ^2 t\right] d t \\
& I=2 \int_0^{\frac{\pi}{2}} \cos ^2 t d \\
& =\int_0^{\frac{\pi}{2}}[1+\operatorname{Cos} 2 t] d t \\
& {[t]_0^{\frac{\pi}{2}}+\left[\frac{\operatorname{Sin} 2 t}{2}\right]_0^{\frac{\pi}{2}}} \\
& \frac{\pi}{2}+0
\end{aligned}
$

Example 2: $\int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x$ is

1) $\pi^2 / 4$
2) $\pi^2$
3) 0
4) $\pi / 2$

Solution

As we learnt in

Properties of definite integration -

If $f(x)$ is an EVEN function of x: then integral of the function from - a to a is the same as twice the integral of the same function from o to a.

$\int_{-a}^a f(x) d x=2\left\{\int_o^a f(x) d x\right\}$

- wherein

Check even function $f(-x)=f(x)$ and symmetrical about y axis.

And,

Properties of Definite Integration -

If $f(x)$ is an odd function of $x$ then integral of the function from -a to a is ZERO

$\int_{-a}^a f(x) d x=0$
- wherein

Check

Odd function $f(-x)=-f(x)$

And,

Properties of Definite Integration -

$\int_0^{2 a} f(x) d x=\int_0^a[f(x)+f(-x)] d x$

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

- $I=\int_{-\pi}^\pi \frac{2 x(1+\sin x) d x}{1+\cos ^2 x}$

$I=\int_0^\pi\left(\frac{2 x(1+\sin x)-2 x+2 x \sin x}{\left(1+\cos ^2 x\right)}\right) d x$

$\begin{aligned} & I=\int_0^\pi \frac{4 x \sin x}{\left(1+\cos ^2 x\right)} d x \\ & I=\int_0^{\frac{\pi}{2}} \frac{4 \sin x}{1+\cos ^2 x}(x+\pi-x) d x \\ & I=4 \pi \int_0^{\frac{\pi}{2}} \frac{\sin x d x}{1+\cos ^2 x} \\ & I=4 \pi\left[\tan ^{-1}(\cos x)\right]^\pi \\ & I=4 \pi\left[0+\frac{\pi}{4}\right]=\pi^2\end{aligned}$

Example 3: The value of the integral $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^4 x\left[1+\log \left[\frac{2+\sin x}{2-\sin x}\right]\right] d x$ is :

1) 0
2) $\frac{3}{4}$
3) $\frac{3}{8} \pi$
4) $\frac{3}{16} \pi$

Solution

Properties of definite integration -

If $f(x)$ is an EVEN function of x: then integral of the function from - a to a is the same as twice the integral of the same function from o to a.

$\int_{-a}^a f(x) d x=2\left\{\int_0^a f(x) d x\right\}$

- wherein

Check even function $f(-x)=f(x)$ and symmetrical about y axis.

Check

Odd function $f(-x)=-f(x)$

$\int_{-\pi}^{\frac{1}{2}} \sin ^4 x\left(1+\log \left(\frac{2+\sin x}{2-\sin x}\right)\right) d x$

$=2 \int_0^{\frac{\pi}{2}} \sin ^4 x d x+2 \int_0^{\frac{x}{2}} \sin ^4 x \log \frac{2+\sin x}{2-\sin x} d x$

Since value of $\int_0^{\frac{5}{2}} \sin ^4 x \log \frac{2+\sin x}{2-\sin x} d x=0$ (Odd function)

So,

$\begin{aligned} & \sin ^4 x=\frac{1}{4}(1-\cos 2 x)^2 \\ & \sin ^4 x=\frac{1}{4}\left(1+\cos ^2 2 x-2 \cos 2 x\right) \\ & \sin ^4 x=\frac{1}{4}\left(1+\frac{1+\cos 4 x}{2}-2 \cos 2 x\right)\end{aligned}$

$\begin{aligned} & I=2 \cdot \int_0^{\frac{\pi}{2}} \sin ^4(x) d x \\ & I=2 \cdot \int_0^{\frac{\pi}{2}} \frac{1}{4}\left(\frac{3+\cos (4 x)-8 \cos (2 x)}{2}\right) d x \\ & =\frac{3 \pi}{8}\end{aligned}$

Example 4: $\int_{-\Pi / 2}^{\Pi / 2} \frac{\cos x}{1+e^x} d x$ is equal to?

1) $1$

2) $2 $

3) $-1$

4) $-2$

Solution

Properties of definite integration -

If $f(x)$ is an EVEN function of x: then integral of the function from - a to a is the same as twice the integral of the same function from o to a.

$\int_{-a}^a f(x) d x=2\left\{\int_0^a f(x) d x\right\}$

- wherein

Check even function $f(-x)=f(x)$ and symmetrical about y axis.

$\begin{aligned} & I=\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^x} d x=\int_0^{\pi / 2}\left(\frac{\cos x}{1+e^x}+\frac{\cos (-x)}{1+e^x}\right) d x=\int_0^{\pi / 2} \cos x d x= \\ & {[\sin x]_0^{\pi / 2}=1}\end{aligned}$

Example 5: $\int_{-1}^1 \frac{\log \left(x+\sqrt{1+x^2}\right)}{x+\log \left(x+\sqrt{1+x^2}\right)}(f(x)-f(-x)) d x$ is euqal to?

1) $0$

2) $2 \int_0^1 \frac{\log \left(x+\sqrt{1+x^2}\right)}{x+\log \left(x+\sqrt{1+x^2}\right)}(f(x)-f(-x)) d x$

3) $2 f(x)$

4) none of these

Solution

As we learnt

Properties of definite integration -

If $f(x)$ is an EVEN function of x: then integral of the function from - a to a is the same as twice the integral of the same function from o to a.

$\int_{-a}^a f(x) d x=2\left\{\int_o^a f(x) d x\right\}$

- wherein

Check even function $f(-x)=f(x)$ and symmetrical about y axis.

$\operatorname{Let} \phi(x)=\frac{\log \left(x+\sqrt{1+x^2}\right)}{x+\log \left(x+\sqrt{1+x^2}\right)} ; g(x)=(f(x)-f(-x))$

f(x) is an even function $\int_{-1}^1(\phi(x) \cdot g(x)) d x=0$

g(x) is an odd function.

List of Topics Related to Application of Even-Odd Properties in Definite Integrals

Explore key topics related to the application of even and odd function properties in definite integrals, alongside integral applications, particular functions, indefinite integrals, integration by parts, and inequalities. Understanding these concepts helps simplify evaluation and leverage symmetry in integral problems effectively.

Application of Integrals

Integral of Particular Functions

Indefinite Integrals

Integration by parts

Application of Inequality in Definite Integration

NCERT Resources

Explore comprehensive NCERT resources for Class 12 Maths Chapter 7 on Integrals, including detailed notes, complete solutions, and exemplar problem sets. These expertly curated materials support thorough understanding and effective exam preparation.

NCERT Class 12 Maths Notes for Chapter 7 - Integrals

NCERT Class 12 Maths Solutions for Chapter 7 - Integrals

NCERT Class 12 Maths Exemplar Solutions for Chapter 7 - Integrals

Practice Questions based on Application of Even-Odd Properties in Definite Integrals

Strengthen your grasp of even-odd function properties in definite integrals with these practice questions. These problems enhance your ability to apply symmetry concepts for simplifying and solving integrals confidently.

Application Of Even Odd Properties In Definite Integration- Practice Question MCQ

We have provided below the practice questions related to different concepts of integration to improve your understanding: