Answer
$$L = \frac{{62}}{3}$$
Work Step by Step
$$\eqalign{
& y = \frac{{{x^{3/2}}}}{3} - {x^{1/2}}{\text{ on }}\left[ {4,16} \right] \cr
& {\text{Definition of Arc Length for }}y = f\left( x \right): \cr
& {\text{Let }}f{\text{ have a continuous first derivative on the interval }}\left[ {a,b} \right]{\text{ The length of the curve }} \cr
& {\text{from }}\left( {a,f\left( a \right)} \right){\text{ to }}\left( {b,f\left( b \right)} \right){\text{ is }}L = \int_a^b {\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr
& {\text{Notice that }}y = f\left( x \right) = \frac{{{x^{3/2}}}}{3} - {x^{1/2}}{\text{ and }}\left[ {4,16} \right] \to a = 4{\text{ and }}b = 16.{\text{ then}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^{3/2}}}}{3} - {x^{1/2}}} \right] \cr
& f'\left( x \right) = \frac{3}{2}\left( {\frac{{{x^{1/2}}}}{3}} \right) - \frac{1}{2}{x^{ - 1/2}} \cr
& f'\left( x \right) = \frac{1}{2}{x^{1/2}} - \frac{1}{2}{x^{ - 1/2}} \cr
& {\text{Using the arc length formula}}{\text{, we have}} \cr
& L = \int_4^{16} {\sqrt {1 + {{\left( {\frac{1}{2}{x^{1/2}} - \frac{1}{2}{x^{ - 1/2}}} \right)}^2}} } dx \cr
& {\text{simplifying}} \cr
& L = \int_4^{16} {\sqrt {1 + {{\left( {\frac{1}{2}{x^{1/2}}} \right)}^2} - 2\left( {\frac{1}{2}{x^{1/2}}} \right)\left( {\frac{1}{2}{x^{ - 1/2}}} \right) + {{\left( {\frac{1}{2}{x^{ - 1/2}}} \right)}^2}} } dx \cr
& L = \int_4^{16} {\sqrt {1 + \frac{1}{4}x - \frac{1}{2} + \frac{1}{4}{x^{ - 1}}} } dx \cr
& L = \int_4^{16} {\sqrt {\frac{1}{4}x + \frac{1}{2} + \frac{1}{4}{x^{ - 1}}} } dx \cr
& {\text{factoring }} \cr
& L = \int_4^{16} {\sqrt {{{\left( {\frac{1}{2}{x^{1/2}} + \frac{1}{2}{x^{ - 1/2}}} \right)}^2}} } dx \cr
& L = \int_4^{16} {\left( {\frac{1}{2}{x^{1/2}} + \frac{1}{2}{x^{ - 1/2}}} \right)} dx \cr
& L = \frac{1}{2}\int_4^{16} {\left( {{x^{1/2}} + {x^{ - 1/2}}} \right)} dx \cr
& {\text{integrate and evaluate}} \cr
& L = \frac{1}{2}\left( {\frac{2}{3}{x^{3/2}} + 2{x^{1/2}}} \right)_4^{16} \cr
& L = \frac{1}{2}\left( {\frac{2}{3}{{\left( {16} \right)}^{3/2}} + 2{{\left( {16} \right)}^{1/2}}} \right) - \frac{1}{2}\left( {\frac{2}{3}{{\left( 4 \right)}^{3/2}} + 2{{\left( 4 \right)}^{1/2}}} \right) \cr
& L = \frac{1}{2}\left( {\frac{{128}}{3} + 8} \right) - \frac{1}{2}\left( {\frac{{16}}{3} + 4} \right) \cr
& L = \frac{{76}}{3} - \frac{{14}}{3} \cr
& L = \frac{{62}}{3} \cr} $$
