# What is the arclength of f(x)=x-sqrt(x+3) on x in [1,3]?

Jun 21, 2018

$L = \frac{4 \sqrt{6} - 1}{8 \sqrt{2}} \sqrt{{\left(4 \sqrt{6} - 1\right)}^{2} + 1} - \frac{35}{8} + \frac{1}{8 \sqrt{2}} \ln \left(\frac{4 \sqrt{6} - 1 + \sqrt{{\left(4 \sqrt{6} - 1\right)}^{2} + 1}}{7 + 5 \sqrt{2}}\right)$ units.

#### Explanation:

$f \left(x\right) = x - \sqrt{x + 3}$

$f ' \left(x\right) = 1 - \frac{1}{2 \sqrt{x + 3}}$

Arclength is given by:

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

Apply the substitution $2 \sqrt{x + 3} = u$:

$L = {\int}_{4}^{2 \sqrt{6}} \sqrt{1 + {\left(1 - \frac{1}{u}\right)}^{2}} \left(\frac{1}{2} u \mathrm{du}\right)$

Simplify:

$L = \frac{1}{2} {\int}_{4}^{2 \sqrt{6}} \sqrt{2 {u}^{2} - 2 u + 1} \mathrm{du}$

Complete the square in the square root:

$L = \frac{1}{2 \sqrt{2}} {\int}_{4}^{2 \sqrt{6}} \sqrt{{\left(2 u - 1\right)}^{2} + 1} \mathrm{du}$

Apply the substitution $2 u - 1 = \tan \theta$:

$L = \frac{1}{4 \sqrt{2}} \int {\sec}^{3} \theta d \theta$

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:

$L = \frac{1}{8 \sqrt{2}} \left[\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right]$

Reverse the substitution:

$L = \frac{1}{8 \sqrt{2}} {\left[\left(2 u - 1\right) \sqrt{{\left(2 u - 1\right)}^{2} + 1} + \ln | \left(2 u - 1\right) + \sqrt{{\left(2 u - 1\right)}^{2} + 1} |\right]}_{4}^{2 \sqrt{6}}$

Insert the limits of integration:

$L = \frac{4 \sqrt{6} - 1}{8 \sqrt{2}} \sqrt{{\left(4 \sqrt{6} - 1\right)}^{2} + 1} - \frac{35}{8} + \frac{1}{8 \sqrt{2}} \ln \left(\frac{4 \sqrt{6} - 1 + \sqrt{{\left(4 \sqrt{6} - 1\right)}^{2} + 1}}{7 + 5 \sqrt{2}}\right)$