# What is the arc length of f(x)= x ^ 3 / 6 + 1 / (2x)  on x in [1,3]?

Nov 11, 2015

$\frac{14}{3} \approx 4.667$

#### Explanation:

If $f : x \mapsto {x}^{3} / 6 + \frac{1}{2 x}$, the length is
$L = {\int}_{1}^{3} \sqrt{1 + f ' {\left(x\right)}^{2}} \mathrm{dx}$.

Because $f ' \left(x\right) = {x}^{2} / 2 - \frac{1}{2 {x}^{2}} = \frac{{x}^{4} - 1}{2 {x}^{2}}$, you have
$f ' {\left(x\right)}^{2} + 1 = \frac{{x}^{8} - 2 {x}^{4} + 1}{4 {x}^{4}} + 1 = \frac{{x}^{8} + 2 {x}^{4} + 1}{4 {x}^{4}} = {\left(\frac{{x}^{4} + 1}{2 {x}^{2}}\right)}^{2}$,
so, you can write
$L = {\int}_{1}^{3} \frac{{x}^{4} + 1}{2 {x}^{2}} \mathrm{dx} = {\int}_{1}^{3} \left({x}^{2} / 2 + \frac{1}{2 {x}^{2}}\right) \mathrm{dx}$
$L = {\left[{x}^{3} / 6 - \frac{1}{2 x}\right]}_{1}^{3} = \frac{14}{3}$