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

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

Explanation

Solution

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 ∫2x−3dx\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 ∫2x−3dx\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 ∫baf(x)dx\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(x)f\left( x \right) is discontinuous between [a,b], then that definite integral ∫baf(x)dx\int_{b}^{a}{f\left( x \right)dx} is said to be improper integral.
So, in the integration ∫−112x−3dx\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 ∫−112x−3dx\int_{-1}^{1}{2{{x}^{-3}}dx}.
Using the formula of integration: ∫xndx=xn+1n+1\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}, we can write
∫−112x−3dx=[x−3+1−3+1]−11\int_{-1}^{1}{2{{x}^{-3}}dx}=\left[ \dfrac{{{x}^{-3+1}}}{-3+1} \right]_{-1}^{1}
The above can also be written as
∫−112x−3dx=[x−2−2]−11\int_{-1}^{1}{2{{x}^{-3}}dx}=\left[ \dfrac{{{x}^{-2}}}{-2} \right]_{-1}^{1}
∫−112x−3dx=[1−2x2]−11\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
⇒∫−112x−3dx=[1−2×12−1−2×(−1)2]\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
⇒∫−112x−3dx=[1−2−1−2]\Rightarrow \int_{-1}^{1}{2{{x}^{-3}}dx}=\left[ \dfrac{1}{-2}-\dfrac{1}{-2} \right]
The above can also be written as
⇒∫−112x−3dx=0\Rightarrow \int_{-1}^{1}{2{{x}^{-3}}dx}=0
Now, we can say that ∫−112x−3dx\int_{-1}^{1}{2{{x}^{-3}}dx} is a proper integral. And, we have evaluated it. The value of ∫−112x−3dx\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:
∫xndx=xn+1n+1\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}
We should know about proper and improper integrals.