How do you test the improper integral \[\int{2{{x}^{-3}}dx}\] from [-1,1] and evaluate if possible?

Hint: This question is from the topic of calculus. In this question, we have to do the improper integration of the given integration from [-1,1]. In solving this question, we will first find out the integration of the \[\int{2{{x}^{-3}}dx}\] using the formulas of integration. After doing the integration, we will put the limits as [-1,1] in that function and solve the further question. After that, we will get our answer.

Complete step by step solution:
Let us solve this question.
In this question, we have asked to do the integration for \[\int{2{{x}^{-3}}dx}\]. We have given the limits as [-1,1]. We will use the limits in the integration.
Let us first understand about the improper integrals.
The definite integral \[\int_{b}^{a}{f\left( x \right)dx}\] is said to be an improper integral if the limits ‘a’ or ‘b’ or both are infinite. And, also if \[f\left( x \right)\] is discontinuous between [a,b], then that definite integral \[\int_{b}^{a}{f\left( x \right)dx}\] is said to be improper integral.
So, in the integration \[\int_{-1}^{1}{2{{x}^{-3}}dx}\], we can say that it will not be an improper integral. It is a proper integral.
Now, let us evaluate \[\int_{-1}^{1}{2{{x}^{-3}}dx}\].
Using the formula of integration: \[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}\], we can write
\[\int_{-1}^{1}{2{{x}^{-3}}dx}=\left[ \dfrac{{{x}^{-3+1}}}{-3+1} \right]_{-1}^{1}\]
The above can also be written as
\[\int_{-1}^{1}{2{{x}^{-3}}dx}=\left[ \dfrac{{{x}^{-2}}}{-2} \right]_{-1}^{1}\]
\[\int_{-1}^{1}{2{{x}^{-3}}dx}=\left[ \dfrac{1}{-2{{x}^{2}}} \right]_{-1}^{1}\]
Now, we will put the limits and evaluate the value.
So, we can write as
\[\Rightarrow \int_{-1}^{1}{2{{x}^{-3}}dx}=\left[ \dfrac{1}{-2\times {{1}^{2}}}-\dfrac{1}{-2\times {{\left( -1 \right)}^{2}}} \right]\]
The above can also be written as
\[\Rightarrow \int_{-1}^{1}{2{{x}^{-3}}dx}=\left[ \dfrac{1}{-2}-\dfrac{1}{-2} \right]\]
The above can also be written as
\[\Rightarrow \int_{-1}^{1}{2{{x}^{-3}}dx}=0\]
Now, we can say that \[\int_{-1}^{1}{2{{x}^{-3}}dx}\] is a proper integral. And, we have evaluated it. The value of \[\int_{-1}^{1}{2{{x}^{-3}}dx}\] is 0.

Note: We should have a better knowledge in the topic of calculus to solve this type of question easily. We should know how to do the integration with limits. We should remember the following formula:
\[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}\]
We should know about proper and improper integrals.