How do you evaluate the definite integral \[\int{{{x}^{3}}dx}\] from \[\left[ -1,1 \right]\]

Hint: The definite integral of $f(x)$ from $a$ and $b$ $\int\limits_{a}^{b}{f(x)}dx=[F(x) ]_{a}^{b}$
In this question to find out the value of given definite integral , firstly we will integrate the function after that we will find out the value of function at upper limit and lower limit then to conclude the result we will subtract the value of function at upper limit by value of function at lower limit.

Complete step by step answer:
We have to integrate $\int{{{x}^{3}}dx}$
The limit given to us is $\left[ -1,1 \right]$.
Let $I=\int\limits_{-1}^{1}{{{x}^{3}}dx}$
On integration ${{x}^{3}}$ becomes $\dfrac{{{x}^{4}}}{4}$.
On integration $dx$ becomes $1$.
$I=\int\limits_{-1}^{1}{{{x}^{3}}dx}$
$=\left[ \dfrac{{{x}^{4}}}{4} \right]_{-1}^{1}$
Here, $1$ will become the upper limit and $-1$ will be a lower limit.
$I=\left[ \dfrac{{{x}^{4}}}{4} \right]_{-1}^{1}$
$=\left[ \dfrac{{{1}^{4}}}{4}-\dfrac{{{\left( -1 \right)}^{4}}}{4} \right]$
$=\dfrac{1}{4}-\dfrac{1}{4}=0$

Note:
(1) The definite integral is defined to be exactly the limit and summation to find the not area.
(2) Also note that the notation for the definite integral is very similar to the notation for an infinite integral.
(3) The number $'a'$ that is at bottom of the integral sign is called the lower limit of integral and the number $'b'$ at the top of integral sign is called the upper limit of integral.
(4) $a$ and $b$ were given as an integral need to be smaller than upper limit.