# What is the arclength of f(x)=3x^2-x+4 on x in [2,3]?

Mar 17, 2018

$L = \frac{1}{12} \left(17 \sqrt{290} - 11 \sqrt{122}\right) + \frac{1}{12} \ln \left(\frac{17 + \sqrt{290}}{11 + \sqrt{122}}\right)$ units.

#### Explanation:

f(x)=3x^2−x+4

f'(x)=6x−1

Arclength is given by:

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

Apply the substitution $6 x - 1 = \tan \theta$:

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

This is a known integral:

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

Reverse the substitution:

$L = \frac{1}{12} {\left[\left(6 x - 1\right) \sqrt{1 + {\left(6 x - 1\right)}^{2}} + \ln | \left(6 x - 1\right) + \sqrt{1 + {\left(6 x - 1\right)}^{2}} |\right]}_{2}^{3}$

Hence

$L = \frac{1}{12} \left(17 \sqrt{290} - 11 \sqrt{122}\right) + \frac{1}{12} \ln \left(\frac{17 + \sqrt{290}}{11 + \sqrt{122}}\right)$