# What is the arc length of f(x)=x^2-2x+35 on x in [1,7]?

Mar 7, 2018

The arc length is $3 \sqrt{145} + \frac{1}{4} \ln \left(12 + \sqrt{145}\right)$ units.

#### Explanation:

$y = {x}^{2} - 2 x + 35$

$y ' = 2 x - 2$

Arc length is given by:

$L = {\int}_{1}^{7} \sqrt{1 + {\left(2 x - 2\right)}^{2}} \mathrm{dx}$

Apply the substitution $2 x - 2 = \tan \theta$:

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

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:

$L = \frac{1}{4} {\left[\left(2 x - 2\right) \sqrt{1 + {\left(2 x - 2\right)}^{2}} + \ln | \left(2 x - 2\right) + \sqrt{1 + {\left(2 x - 2\right)}^{2}} |\right]}_{1}^{7}$

Insert the limits of integration:

$L = 3 \sqrt{145} + \frac{1}{4} \ln \left(12 + \sqrt{145}\right)$