# What is the arc length of f(x)=2/x^4-1/x^6 on x in [3,6]?

Jun 23, 2018

$\approx 3.000205746$

#### Explanation:

$f \left(x\right) = 2 \cdot {x}^{- 4} - {x}^{- 6}$
so we get

$f ' \left(x\right) = 8 {x}^{- 5} + 6 {x}^{- 7}$

and we have to solve

${\int}_{3}^{6} \sqrt{1 + {\left(8 {x}^{- 5} + 6 {x}^{- 7}\right)}^{2}} \mathrm{dx}$
this is $\approx 3.0002057846$

Jun 25, 2018

$L = 3 + \frac{1}{3} {\sum}_{n = 1}^{\infty} {\sum}_{m = 0}^{2 n} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}2 n \\ m\end{matrix}\right) \frac{1}{1 + 10 n + 2 m} {\left(\frac{8}{243}\right)}^{2 n} {\left(\frac{- 1}{12}\right)}^{m} \left(1 - \frac{1}{2} ^ \left(1 + 10 n + 2 m\right)\right)$ units.

#### Explanation:

$f \left(x\right) = \frac{2}{x} ^ 4 - \frac{1}{x} ^ 6$

$f ' \left(x\right) = - \frac{8}{x} ^ 5 \left(1 - \frac{3}{4 {x}^{2}}\right)$

Arc length is given by:

$L = {\int}_{3}^{6} \sqrt{1 + \frac{64}{x} ^ 10 {\left(1 - \frac{3}{4 {x}^{2}}\right)}^{2}} \mathrm{dx}$

For $x \in \left[3 , 6\right]$, $\frac{64}{x} ^ 10 {\left(1 - \frac{3}{4 {x}^{2}}\right)}^{2} < 1$. Take the series expansion of the square root:

$L = {\int}_{3}^{6} {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(\frac{64}{x} ^ 10 {\left(1 - \frac{3}{4 {x}^{2}}\right)}^{2}\right)}^{n} \mathrm{dx}$

Isolate the $n = 0$ term and simplify:

$L = {\int}_{3}^{6} \mathrm{dx} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {2}^{6 n} {\int}_{3}^{6} \frac{1}{x} ^ \left(10 n\right) {\left(1 - \frac{3}{4 {x}^{2}}\right)}^{2 n} \mathrm{dx}$

Apply binomial expansion:

$L = 3 + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {2}^{6 n} {\int}_{3}^{6} \frac{1}{x} ^ \left(10 n\right) \left\{{\sum}_{m = 0}^{2 n} \left(\begin{matrix}2 n \\ m\end{matrix}\right) {\left(- \frac{3}{4 {x}^{2}}\right)}^{m}\right\} \mathrm{dx}$

Rearrange:

$L = 3 + {\sum}_{n = 1}^{\infty} {\sum}_{m = 0}^{2 n} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}2 n \\ m\end{matrix}\right) {2}^{6 n} {\left(\frac{- 3}{4}\right)}^{m} {\int}_{3}^{6} \frac{1}{x} ^ \left(10 n + 2 m\right) \mathrm{dx}$

Integrate directly:

$L = 3 + {\sum}_{n = 1}^{\infty} {\sum}_{m = 0}^{2 n} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}2 n \\ m\end{matrix}\right) {2}^{6 n} / \left(1 + 10 n + 2 m\right) {\left(\frac{- 3}{4}\right)}^{m} {\left[\frac{- 1}{x} ^ \left(1 + 10 n + 2 m\right)\right]}_{3}^{6}$

Insert the limits of integration:

$L = 3 + {\sum}_{n = 1}^{\infty} {\sum}_{m = 0}^{2 n} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}2 n \\ m\end{matrix}\right) {2}^{6 n} / \left(1 + 10 n + 2 m\right) {\left(\frac{- 3}{4}\right)}^{m} \left(\frac{1}{3} ^ \left(1 + 10 n + 2 m\right) - \frac{1}{6} ^ \left(1 + 10 n + 2 m\right)\right)$

Simplify:

$L = 3 + \frac{1}{3} {\sum}_{n = 1}^{\infty} {\sum}_{m = 0}^{2 n} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}2 n \\ m\end{matrix}\right) \frac{1}{1 + 10 n + 2 m} {\left(\frac{8}{243}\right)}^{2 n} {\left(\frac{- 1}{12}\right)}^{m} \left(1 - \frac{1}{2} ^ \left(1 + 10 n + 2 m\right)\right)$