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

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 $[0, a]$.