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Question: How do you evaluate the integral \(\int {\dfrac{1}{{{x^3}}}dx} \) from \(3\) to \(\infty \)?...

How do you evaluate the integral ∫1x3dx\int {\dfrac{1}{{{x^3}}}dx} from 33 to ∞\infty ?

Explanation

Solution

Here given an integral of a function of which we have to find the integration of the function, here function is called the integrand. There are two types of integrals, which are definite integrals and indefinite integrals. Definite integrals are those integrals for which there is a particular limit on the integrals. Whereas for the indefinite integrals there are no such limits on the integral.

Complete step-by-step solution:
Consider the given integral, as shown below:
⇒∫1x3dx\Rightarrow \int {\dfrac{1}{{{x^3}}}dx}
Now the above integral is called an indefinite integral, as it has no lower limits and upper limits on the integral.
Now applying the lower limit and upper limits on the integral of the given integration, so now by applications of the limits on the integral, the integral becomes a definite integral.
⇒∫3∞1x3dx=∫3∞x−3dx\Rightarrow \int\limits_3^\infty {\dfrac{1}{{{x^3}}}dx} = \int\limits_3^\infty {{x^{ - 3}}dx}
Now applying the basic formula of integration which is ∫xndx=xn+1n+1+c\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c} , in the integral as shown:
⇒x−3+1−3+1∣3∞\Rightarrow \left. {\dfrac{{{x^{ - 3 + 1}}}}{{ - 3 + 1}}} \right|_3^\infty
Now simplifying the solved integral so as to substitute the limits.
⇒x−2−2∣3∞=−12x2∣3∞\Rightarrow \left. {\dfrac{{{x^{ - 2}}}}{{ - 2}}} \right|_3^\infty = \left. { - \dfrac{1}{{2{x^2}}}} \right|_3^\infty
Now substituting the lower and the upper limits of the integral.
⇒−[12(∞)2−12(3)2]\Rightarrow - \left[ {\dfrac{1}{{2{{\left( \infty \right)}^2}}} - \dfrac{1}{{2{{\left( 3 \right)}^2}}}} \right]
We know that the reciprocal of infinite is zero.
⇒−[0−12(9)]\Rightarrow - \left[ {0 - \dfrac{1}{{2(9)}}} \right]
⇒118\Rightarrow \dfrac{1}{{18}}
∴∫3∞1x3dx=118\therefore \int\limits_3^\infty {\dfrac{1}{{{x^3}}}dx} = \dfrac{1}{{18}}

The value of the integral is ∫3∞1x3dx=118\int\limits_3^\infty {\dfrac{1}{{{x^3}}}dx} = \dfrac{1}{{18}}

Note: Note that the definite integrals are generally used to find the areas under a given curve or to find the area between the curves, or to find the area of a function, given the limits of the definite integral. The limits are given in order to find the area of the function within that particular limit, i.e, to integrate the function from the upper limit to the lower limit.