# How do you find the lengths of the curve y=x^3/12+1/x for 1<=x<=3?

May 3, 2018

$= \frac{17}{6}$

#### Explanation:

$s = {\int}_{1}^{3} \setminus \sqrt{1 + {\left(y '\right)}^{2}} \setminus \setminus \mathrm{dx}$

$s = {\int}_{1}^{3} \setminus \sqrt{1 + {\left({x}^{2} / 4 - \frac{1}{x} ^ 2\right)}^{2}} \setminus \setminus \mathrm{dx}$

$= {\int}_{1}^{3} \setminus \sqrt{1 + {x}^{4} / 16 - \frac{1}{2} + \frac{1}{x} ^ 4} \setminus \setminus \mathrm{dx}$

$= {\int}_{1}^{3} \setminus \sqrt{{x}^{4} / 16 + \frac{1}{2} + \frac{1}{x} ^ 4} \setminus \setminus \mathrm{dx}$

$= {\int}_{1}^{3} \setminus \sqrt{{\left({x}^{2} / 4 + \frac{1}{x} ^ 2\right)}^{2}} \setminus \setminus \mathrm{dx}$

$= {\left(\setminus {x}^{3} / \left(12\right) - \frac{1}{x}\right)}_{1}^{3}$

$= \left(\frac{27}{12} - \frac{1}{3}\right) - \left(\frac{1}{12} - 1\right) = \frac{17}{6}$