# How do you evaluate the integral int 1/x^3dx from 3 to oo?

Jan 24, 2018

Using limits to evaluate the improper integral, the answer is $\frac{1}{18}$

#### Explanation:

This is an improper limit, so we need to use a limit to solve it.

First, let's go ahead and find the antiderivative of $\frac{1}{x} ^ 3$ so we can use it later to solve the integral.

$\int \frac{1}{x} ^ 3 \mathrm{dx} = \int {x}^{-} 3 \mathrm{dx} = \frac{{x}^{-} 2}{-} 2 = - \frac{1}{2 {x}^{2}}$

Now, since we can't technically plug in $\infty$ to an expression like this, we need to use a limit to evaluate the integral, like this:

${\int}_{3}^{\infty} \frac{1}{x} ^ 3 \mathrm{dx} = {\lim}_{b \to \infty} {\int}_{3}^{b} \frac{1}{x} ^ 3 \mathrm{dx}$

Now, we can evaluate the integral.

${\lim}_{b \to \infty} {\int}_{3}^{b} \frac{1}{x} ^ 3 \mathrm{dx}$

$= {\lim}_{b \to \infty} {\left[- \frac{1}{2 {x}^{2}}\right]}_{3}^{b}$

$= {\lim}_{b \to \infty} \left(- \frac{1}{2 {b}^{2}}\right) - \left(- \frac{1}{2 {\left(3\right)}^{2}}\right)$

$= {\lim}_{b \to \infty} \left(- \frac{1}{2 {b}^{2}}\right) + \frac{1}{18}$

Since the numerator of the first fraction is 1, and the denominator is approaching infinity, the fraction will approach 0. Therefore:

${\lim}_{b \to \infty} \left(- \frac{1}{2 {b}^{2}}\right) + \frac{1}{18}$

$= 0 + \frac{1}{18}$

$= \frac{1}{18}$