
If $\int\limits_{-1}^{1}{f(x)dx=0}$, then
A. $f(x)=f(-x)$
B. $f(-x)=-f(x)$
C. $f(x)=2f(x)$
D. None of these
Answer
162.3k+ views
Hint: In this question, we are to find the type of the function i.e., an even function or an odd function. Here the interval of the integral is in the form of $[-a, a]$. So, we can split the integral into two parts among the intervals $[-a,0]$ and $[0, a]$.
Formula Used:Definite integral:
Consider a function $f(x)$ is defined on $[a,b]$. If the integral of this function, $\int{f(x)dx=F(x)}$, then $F(b)-F(a)$ is called the definite integral of the function $f(x)$ over $[a,b]$.
I.e.,$\int\limits_{a}^{b}{f(x)dx}=\left. F(x) \right|_{a}^{b}=F(b)-F(a)$
Here $a$ (lower limit) and $b$(upper limit).
\[\int\limits_{a}^{b}{f(x)dx}=\int\limits_{a}^{b}{f(t)dt}\]
Some of the properties of the definite integrals are:
1) Interchanging the limits: \[\int\limits_{a}^{b}{f(x)dx=-}\int\limits_{b}^{a}{f(x)dx}\]
2) If $a3) $\int\limits_{0}^{a}{f(x)dx}=\int\limits_{0}^{a}{f(a-x)dx}$
4) $\int\limits_{-a}^{a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx}$ if $f(x)$ is an even function
$\int\limits_{-a}^{a}{f(x)dx}=0$ if $f(x)$ is an odd function
5) $\begin{align}
& \int\limits_{0}^{2a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx};\text{if }f(2a-x)=f(x) \\
& \text{ }=0\text{ if }f(2a-x)=-f(x) \\
\end{align}$
6) $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\sin x)}{f(\sin x)+f(\cos x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\cos x)}{f(\sin x)+f(\cos x)}dx=\dfrac{\pi }{4}}}$
7) $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\tan x)}{f(\tan x)+f(\cot x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\cot x)}{f(\tan x)+f(\cot x)}dx=\dfrac{\pi }{4}}}$
Complete step by step solution:Given integral is
$I=\int\limits_{-1}^{1}{f(x)dx=0}$
Since the interval is among $[-1,1]$ we can split the integral into two parts with the intervals $[-1,0]$ and $[0,1]$. I.e.,
$\begin{align}
& \int\limits_{-1}^{1}{f(x)dx=0} \\
& \Rightarrow \int\limits_{-1}^{0}{f(x)dx+\int\limits_{0}^{1}{f(x)dx=0}} \\
& \Rightarrow \int\limits_{-1}^{0}{f(x)dx}=-\int\limits_{0}^{1}{f(x)dx} \\
& \Rightarrow \int\limits_{0}^{1}{f(-x)dx=}\int\limits_{0}^{1}{\left[ -f(x) \right]dx} \\
\end{align}$
Therefore, we can conclude that
$f(-x)=-f(x)$
So, the function in the given integral is an odd function.
Option ‘B’ is correct
Note: Here we need to remember that the given integral should have the interval in the form of $[-a, a]$. So, we can evaluate the integral by parts with the intervals $[-a,0]$ and $[0, a]$.
Formula Used:Definite integral:
Consider a function $f(x)$ is defined on $[a,b]$. If the integral of this function, $\int{f(x)dx=F(x)}$, then $F(b)-F(a)$ is called the definite integral of the function $f(x)$ over $[a,b]$.
I.e.,$\int\limits_{a}^{b}{f(x)dx}=\left. F(x) \right|_{a}^{b}=F(b)-F(a)$
Here $a$ (lower limit) and $b$(upper limit).
\[\int\limits_{a}^{b}{f(x)dx}=\int\limits_{a}^{b}{f(t)dt}\]
Some of the properties of the definite integrals are:
1) Interchanging the limits: \[\int\limits_{a}^{b}{f(x)dx=-}\int\limits_{b}^{a}{f(x)dx}\]
2) If $a
4) $\int\limits_{-a}^{a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx}$ if $f(x)$ is an even function
$\int\limits_{-a}^{a}{f(x)dx}=0$ if $f(x)$ is an odd function
5) $\begin{align}
& \int\limits_{0}^{2a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx};\text{if }f(2a-x)=f(x) \\
& \text{ }=0\text{ if }f(2a-x)=-f(x) \\
\end{align}$
6) $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\sin x)}{f(\sin x)+f(\cos x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\cos x)}{f(\sin x)+f(\cos x)}dx=\dfrac{\pi }{4}}}$
7) $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\tan x)}{f(\tan x)+f(\cot x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\cot x)}{f(\tan x)+f(\cot x)}dx=\dfrac{\pi }{4}}}$
Complete step by step solution:Given integral is
$I=\int\limits_{-1}^{1}{f(x)dx=0}$
Since the interval is among $[-1,1]$ we can split the integral into two parts with the intervals $[-1,0]$ and $[0,1]$. I.e.,
$\begin{align}
& \int\limits_{-1}^{1}{f(x)dx=0} \\
& \Rightarrow \int\limits_{-1}^{0}{f(x)dx+\int\limits_{0}^{1}{f(x)dx=0}} \\
& \Rightarrow \int\limits_{-1}^{0}{f(x)dx}=-\int\limits_{0}^{1}{f(x)dx} \\
& \Rightarrow \int\limits_{0}^{1}{f(-x)dx=}\int\limits_{0}^{1}{\left[ -f(x) \right]dx} \\
\end{align}$
Therefore, we can conclude that
$f(-x)=-f(x)$
So, the function in the given integral is an odd function.
Option ‘B’ is correct
Note: Here we need to remember that the given integral should have the interval in the form of $[-a, a]$. So, we can evaluate the integral by parts with the intervals $[-a,0]$ and $[0, a]$.
Recently Updated Pages
If tan 1y tan 1x + tan 1left frac2x1 x2 right where x frac1sqrt 3 Then the value of y is

Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main Eligibility Criteria 2025

NIT Delhi Cut-Off 2025 - Check Expected and Previous Year Cut-Offs

JEE Main Seat Allotment 2025: How to Check, Documents Required and Fees Structure

JEE Mains 2025 Cut-Off GFIT: Check All Rounds Cutoff Ranks

NIT Durgapur JEE Main Cut-Off 2025 - Check Expected & Previous Year Cut-Offs

JEE Main 2024 Cut-off for NIT Surathkal

Other Pages
JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

JEE Advanced 2025 Notes

List of Fastest Century in IPL History

Verb Forms Guide: V1, V2, V3, V4, V5 Explained
