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

May 25, 2018

$L = \frac{1}{6} \left(7 \sqrt{197} - \sqrt{5}\right) + \frac{1}{12} \ln \left(\frac{14 + \sqrt{197}}{2 \sqrt{5}}\right)$ units.

#### Explanation:

f(x)=(3x−2)^2

f'(x)=6x−4

Arc length is given by:

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

Apply the substitution $6 x - 4 = u$:

$L = \frac{1}{6} {\int}_{2}^{14} \sqrt{1 + {u}^{2}} \mathrm{du}$

Apply the substitution $u = \tan \theta$:

$L = \frac{1}{6} \int {\sec}^{3} \theta d \theta$

This is a known integral:

$L = \frac{1}{12} \left[\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right]$

Reverse the last substitution:

$L = \frac{1}{12} {\left[u \sqrt{1 + {u}^{2}} + \ln | u + \sqrt{1 + {u}^{2}} |\right]}_{2}^{14}$

Hence

$L = \frac{1}{6} \left(7 \sqrt{197} - \sqrt{5}\right) + \frac{1}{12} \ln \left(\frac{14 + \sqrt{197}}{2 \sqrt{5}}\right)$