Answer
$$L = 2\sqrt 3 - \frac{4}{3}$$
Work Step by Step
$$\eqalign{
& y = {x^{1/2}} - \frac{{{x^{3/2}}}}{3}{\text{ on }}\left[ {1,3} \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{ }} \cr
& {\text{The length of the curve 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) = {x^{1/2}} - \frac{{{x^{3/2}}}}{3}{\text{ and }}\left[ {1,3} \right] \to a = 1{\text{ and }}b = 3.{\text{ then}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^{1/2}} - \frac{{{x^{3/2}}}}{3}} \right] \cr
& f'\left( x \right) = \frac{1}{2}{x^{ - 1/2}} - \frac{1}{3}\left( {\frac{3}{2}{x^{1/2}}} \right) \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_1^3 {\sqrt {1 + {{\left( {\frac{1}{2}{x^{ - 1/2}} - \frac{1}{2}{x^{1/2}}} \right)}^2}} } dx \cr
& {\text{simplifying}} \cr
& L = \int_1^3 {\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_1^3 {\sqrt {1 + {{\left( {\frac{1}{2}{x^{ - 1/2}}} \right)}^2} - \frac{1}{2} + {{\left( {\frac{1}{2}{x^{1/2}}} \right)}^2}} } dx \cr
& L = \int_1^3 {\sqrt {{{\left( {\frac{1}{2}{x^{ - 1/2}}} \right)}^2} + \frac{1}{2} + {{\left( {\frac{1}{2}{x^{1/2}}} \right)}^2}} } dx \cr
& {\text{factor}} \cr
& L = \int_1^3 {\sqrt {{{\left( {\frac{1}{2}{x^{ - 1/2}} + \frac{1}{2}{x^{1/2}}} \right)}^2}} } dx \cr
& L = \int_1^3 {\left( {\frac{1}{2}{x^{ - 1/2}} + \frac{1}{2}{x^{1/2}}} \right)} dx \cr
& L = \frac{1}{2}\int_1^3 {\left( {{x^{ - 1/2}} + {x^{1/2}}} \right)} dx \cr
& {\text{integrate}} \cr
& L = \frac{1}{2}\left( {2{x^{1/2}} + \frac{2}{3}{x^{3/2}}} \right)_1^3 \cr
& {\text{evaluate the limits}} \cr
& L = \frac{1}{2}\left( {2{{\left( 3 \right)}^{1/2}} + \frac{2}{3}{{\left( 3 \right)}^{3/2}}} \right) - \frac{1}{2}\left( {2{{\left( 1 \right)}^{1/2}} + \frac{2}{3}{{\left( 1 \right)}^{3/2}}} \right) \cr
& L = \frac{1}{2}\left( {2\sqrt 3 + 2\sqrt 3 } \right) - \frac{1}{2}\left( {\frac{8}{3}} \right) \cr
& L = 2\sqrt 3 - \frac{4}{3} \cr} $$
