Answer
$ \frac{59}{24}$
Work Step by Step
$y = \frac{x^{3}}{3} + \frac{1}{4x}$
then $y' = x^{2} - \frac{1}{4x^{2}}$
$1+(y')^{2} = 1 + (x^{4} - \frac{1}{2} + \frac{1}{16x^{4}})$
$ = x^{4} + \frac{1}{2} + \frac{1}{16x^{4}}$
$ = (x^{2} + \frac{1}{4x^2})^{2}$
$L = \int^{2}_{1} \sqrt{1+(y')^{2}} dx$
$ = \int ^{2}_{1} (x^{2} + \frac{1}{4x^{2}}) dx$
$ = [\frac{1}{3} x^{3} - \frac{1}{4x}]^{2}_{1}$
$ = \frac{59}{24}$
