Question

How do you evaluate $\int{\dfrac{1}{{{x}^{3}}}}dx$ from 3 to $\infty $ ?

Answer 1

Hint: In this question, we have to find the value of definite integral. Thus, we will use the integration and the basic mathematical rules to get the solution. First, we will apply the exponent formula $\dfrac{1}{{{x}^{n}}}={{x}^{-n}}$ in the problem. After that, we will apply the integration formula $\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}$ in the integral. Then, we will again apply the integration formula $\dfrac{1}{{{x}^{n}}}={{x}^{-n}}$ in the equation. After that, we will again apply the integral formula $\int\limits_{a}^{b}{f(x)dx=f(b)-f(a)}$ . In the end, we will make the necessary mathematical calculations, to get the required solution for the problem.

Complete step by step solution:
According to the question, we have to find the value of definite integral.
Thus, we will use the integration formula and the basic mathematical rules to get the solution.
The integral to be solved is $\int{\dfrac{1}{{{x}^{3}}}}dx$ for $[3,\infty ]$ -------------- (1)
First, we will apply the exponent formula $\dfrac{1}{{{x}^{n}}}={{x}^{-n}}$ in equation (1), we get
$\Rightarrow \int\limits_{3}^{\infty }{{{x}^{-3}}dx}$
Now, we will apply the integration formula $\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}$ in the above equation, we get
$\Rightarrow \left[ \dfrac{{{x}^{-3+1}}}{-3+1} \right]_{3}^{\infty }$
On further simplification, we get
$\Rightarrow \left[ \dfrac{{{x}^{-2}}}{-2} \right]_{3}^{\infty }$
Now, we will again apply the exponent formula $\dfrac{1}{{{x}^{n}}}={{x}^{-n}}$ in the above equation, we get
$\Rightarrow \left[ \dfrac{-1}{2{{x}^{2}}} \right]_{3}^{\infty }$
In the last, we will apply the definite integral formula $\int\limits_{a}^{b}{f(x)dx=f(b)-f(a)}$ in the above equation, we get
$\Rightarrow \left( \dfrac{-1}{2{{\left( \infty \right)}^{2}}} \right)-\left( \dfrac{-1}{2{{\left( 3 \right)}^{2}}} \right)$
Therefore, we get
$\Rightarrow \left( \dfrac{-1}{2\times \infty } \right)-\left( \dfrac{-1}{18} \right)$
Now, we know that $\dfrac{1}{\infty }=0$ , thus we get
$\Rightarrow 0+\left( \dfrac{1}{18} \right)$
Thus, we get
$\Rightarrow \dfrac{1}{18}$

Therefore, for the definite integral $\int{\dfrac{1}{{{x}^{3}}}}dx$ from 3 to $\infty $ , its value is equal to $\dfrac{1}{18}$

Note: While solving this problem, do the step-by-step calculations properly to avoid the mathematical error. Do mention the formulas you are using to get an accurate answer. Always remember that $\dfrac{1}{0}=\infty $ , therefore $\dfrac{1}{\infty }=0$