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

Jun 27, 2018

$L = \frac{\sqrt{5}}{2} \left(\sqrt{21} - 1\right) + \frac{1}{4} \ln \left(\frac{2 \sqrt{5} + \sqrt{21}}{2 + \sqrt{5}}\right)$

#### Explanation:

Given

$f \left(x\right) = \sqrt{x - 1}$, $x \in \left[2 , 6\right]$

Let $y = f \left(x\right)$, we thus have:

$x = {y}^{2} + 1$, $y \in \left[1 , \sqrt{5}\right]$

Take the derivative with respect to $y$:

$x ' = 2 y$

Arc length is given by:

$L = {\int}_{1}^{\sqrt{5}} \sqrt{1 + 4 {y}^{2}} \mathrm{dy}$

Apply the substitution $2 y = \tan \theta$:

$L = \frac{1}{2} \int {\sec}^{3} \theta d \theta$

This is a known integral. If you do not have it memorized apply integration by parts or look it up in a table of integrals:

$L = \frac{1}{4} \left[\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right]$

Reverse the substitution:

$L = \frac{1}{4} {\left[2 y \sqrt{1 + 4 {y}^{2}} + \ln | 2 y + \sqrt{1 + 4 {y}^{2}} |\right]}_{1}^{\sqrt{5}}$

Insert the limits of integration:

$L = \frac{\sqrt{5}}{2} \left(\sqrt{21} - 1\right) + \frac{1}{4} \ln \left(\frac{2 \sqrt{5} + \sqrt{21}}{2 + \sqrt{5}}\right)$