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

Jun 28, 2018

$L = 1 + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{4 n - 1} \left(1 - \frac{1}{2} ^ \left(4 n - 1\right)\right)$ units.

#### Explanation:

$f \left(x\right) = \frac{1}{x}$

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

Arc length is given by:

$L = {\int}_{1}^{2} \sqrt{1 + \frac{1}{x} ^ 4} \mathrm{dx}$

For $x \in \left[1 , 2\right]$, $\frac{1}{x} ^ 4 < 1$. Take the series expansion of the square root:

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

Simplify:

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

Integrate directly:

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

Isolate the $n = 0$ term and simplify:

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