Show that \[\int\limits_{ - a}^a {f(x)dx = \left\{ {\begin{array}{*{20}{c}}
  {2\int\limits_0^a {f(x)dx,\,if\,f(x){\text{ is an even function}}} } \\
  {0,\,if\,f(x){\text{ is a odd function}}}
\end{array}} \right.} \]

Hint: To solve this problem we should know about the following term:
Condition for checking whether its function is odd: The integrand of an odd function over a symmetric interval is zero. This is because the region below the x-axis is symmetric to the region above the x-axis.
Condition for checking whether its function is Even: if integration for a function over a symmetrical interval will equal to twice of integrand for the same function over half of interval. Then this will be an even function.
First check if the interval is symmetric or not. Then we will do integration of a given function of a given function as per given interval then by observing the result we can define whether the function is odd or even.

Complete step by step answer:
Given
Let $I = \int\limits_{ - a}^a {f\{ x\} dx} $
$I = \int\limits_{ - a}^0 {f\{ x\} dx} + \int\limits_0^a {f\{ x\} dx} $
Now if $f\{ x\} $ is even, $f\{ x\} = f\{ - x\} $
$I = \int\limits_{ - a}^0 {f\{ x\} dx} + \int\limits_0^a {f\{ x\} dx} $
$\Rightarrow I = \int\limits_{ - ( - a)}^{ - 0} {f\{ - x\} \{ - dx\} } + \int\limits_0^a {f\{ x\} dx} $
(Substituting$x\,by\, - x$)
$\Rightarrow I = \int\limits_0^a {f\{ - x\} dx} + \int\limits_0^a {f\{ x\} dx} $
But $f\{ x\} = f\{ - x\} $
$\Rightarrow I = \int\limits_0^a {f\{ x\} dx} + \int\limits_0^a {f\{ x\} dx} $
\[\Rightarrow I = 2\int\limits_0^a {f\{ x\} dx} \] if $f\{ x\} $ is an even.
Now
If $f\{ x\} $ is an odd, $f\{ x\} = - f\{ - x\} $
$I = \int\limits_{ - a}^0 {f\{ x\} dx} + \int\limits_0^a {f\{ x\} dx} $
$\Rightarrow I = \int\limits_{ - ( - a)}^{ - 0} {f\{ - x\} \{ - dx\} } + \int\limits_0^a {f\{ x\} dx} $
Substituting $x\,by\, - x$
$ I = \int\limits_0^a {f\{ - x\} dx} + \int\limits_0^a {f\{ x\} dx} $
But $f\{ x\} = - f\{ - x\} $
$\Rightarrow I = - \int\limits_0^a {f\{ x\} dx} + \int\limits_0^a {f\{ x\} dx} $
$\therefore I = 0\,if\,f\{ x\} \,$is odd.
Hence proved.

Note:Differentiation is a method of finding the derivative of a function. Differentiation is a process, in Maths, where we find the instantaneous rate of change in function based on one of its variables. Theorem If $f\{ x\} $ is a continuous function on the closed, bounded interval [a,b], then f is integrable on [a,b]. ... If $f\{ x\} $ is continuous on the closed, bounded interval [a,b] then f is uniformly continuous on [a,b].