Answer
$$L = \frac{{17}}{6}$$
Work Step by Step
$$\eqalign{
& 24xy = {y^4}{\text{ + 48 from }}y = 2{\text{ to }}y = 4 \cr
& {\text{Solving for }}x \cr
& x = \frac{{{y^4} + 48}}{{24y}} \cr
& x = \frac{{{y^3}}}{{24}} + \frac{2}{y} \cr
& {\text{Calculate }}\frac{{dx}}{{dy}} \cr
& \frac{{dx}}{{dy}} = \frac{d}{{dy}}\left[ {\frac{{{y^3}}}{{24}} + \frac{2}{y}} \right] \cr
& \frac{{dy}}{{dx}} = \frac{3}{{24}}{y^2} - \frac{2}{{{y^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{8}{y^2} - 2{y^{ - 2}} \cr
& {\text{Use }}L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} dy} \cr
& L = \int_2^4 {\sqrt {1 + {{\left( {\frac{1}{8}{y^2} - 2{y^{ - 2}}} \right)}^2}} dy} \cr
& L = \int_2^4 {\sqrt {1 + {{\left( {\frac{1}{8}{y^2}} \right)}^2} - 2\left( {\frac{1}{8}{y^2}} \right)\left( {2{y^{ - 2}}} \right) + {{\left( {2{y^{ - 2}}} \right)}^2}} dy} \cr
& L = \int_2^4 {\sqrt {{{\left( {\frac{1}{8}{y^2}} \right)}^2} + \frac{1}{2} + {{\left( {2{y^{ - 2}}} \right)}^2}} dy} \cr
& L = \int_2^4 {\sqrt {{{\left( {\frac{1}{8}{y^2} + 2{y^{ - 2}}} \right)}^2}} dy} \cr
& L = \int_2^4 {\left( {\frac{1}{8}{y^2} + 2{y^{ - 2}}} \right)dy} \cr
& {\text{Integrating}} \cr
& L = \left[ {\frac{1}{{24}}{y^3} - \frac{2}{y}} \right]_2^4 \cr
& L = \left[ {\frac{1}{{24}}{{\left( 4 \right)}^3} - \frac{2}{4}} \right] - \left[ {\frac{1}{{24}}{{\left( 2 \right)}^3} - \frac{2}{2}} \right] \cr
& L = \frac{{13}}{6} + \frac{2}{3} \cr
& L = \frac{{17}}{6} \cr} $$
