# What is the arc length of f(x)= sqrt(5x+1)  on x in [0,2]?

May 28, 2018

$L = \frac{1}{10} \left(\sqrt{759} - \sqrt{29}\right) + \frac{5}{4} \ln \left(\frac{2 \sqrt{11} + \sqrt{69}}{2 + \sqrt{29}}\right)$ units.

#### Explanation:

$f \left(x\right) = \sqrt{5 x + 1}$

$f ' \left(x\right) = \frac{5}{2 \sqrt{5 x + 1}}$

Arc length is given by:

$L = {\int}_{0}^{2} \sqrt{1 + \frac{25}{4 \left(5 x + 1\right)}} \mathrm{dx}$

Apply the substitution $5 x + 1 = u$:

$L = \frac{1}{5} {\int}_{1}^{11} \sqrt{1 + \frac{25}{4 u}} \mathrm{du}$

Rearrange:

$L = \frac{1}{5} {\int}_{1}^{11} \frac{\sqrt{4 u + 25}}{2 \sqrt{u}} \mathrm{du}$

Apply the substitution $\sqrt{u} = v$:

$L = \frac{1}{5} {\int}_{1}^{\sqrt{11}} \sqrt{4 {v}^{2} + 25} \mathrm{dv}$

Apply the substitution $2 v = 5 \tan \theta$:

$L = \frac{5}{2} \int {\sec}^{3} \theta d \theta$

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

$L = \frac{5}{4} \left[\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right]$

Reverse the last substitution:

$L = {\left[\frac{1}{10} v \sqrt{4 {v}^{2} + 25} + \frac{5}{4} \ln | 2 v + \sqrt{4 {v}^{2} + 25} |\right]}_{1}^{\sqrt{11}}$

Insert the limits of integration:

$L = \frac{1}{10} \left(\sqrt{759} - \sqrt{29}\right) + \frac{5}{4} \ln \left(\frac{2 \sqrt{11} + \sqrt{69}}{2 + \sqrt{29}}\right)$