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

How do you evaluate the definite integral x3dx\int{{{x}^{3}}dx} from [1,1]\left[ -1,1 \right]

Explanation

Solution

The definite integral of f(x)f(x) from aa and bb abf(x)dx=[F(x)]ab\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 x3dx\int{{{x}^{3}}dx}
The limit given to us is [1,1]\left[ -1,1 \right].
Let I=11x3dxI=\int\limits_{-1}^{1}{{{x}^{3}}dx}
On integration x3{{x}^{3}} becomes x44\dfrac{{{x}^{4}}}{4}.
On integration dxdx becomes 11.
I=11x3dxI=\int\limits_{-1}^{1}{{{x}^{3}}dx}
=[x44]11=\left[ \dfrac{{{x}^{4}}}{4} \right]_{-1}^{1}
Here, 11 will become the upper limit and 1-1 will be a lower limit.
I=[x44]11I=\left[ \dfrac{{{x}^{4}}}{4} \right]_{-1}^{1}
=[144(1)44]=\left[ \dfrac{{{1}^{4}}}{4}-\dfrac{{{\left( -1 \right)}^{4}}}{4} \right]
=1414=0=\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'a' that is at bottom of the integral sign is called the lower limit of integral and the number b'b' at the top of integral sign is called the upper limit of integral.
(4) aa and bb were given as an integral need to be smaller than upper limit.