Answer
$$L = \frac{{80\sqrt {10} - 13\sqrt {13} }}{{27}}$$
Work Step by Step
$$\eqalign{
& y = {x^{2/3}}{\text{ from }}x = 1{\text{ to }}x = 8 \cr
& {\text{Calculate }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{x^{2/3}}} \right] \cr
& \frac{{dy}}{{dx}} = \frac{2}{3}{x^{ - 1/3}} \cr
& {\text{Use }}L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} \cr
& L = \int_1^8 {\sqrt {1 + {{\left( {\frac{2}{3}{x^{ - 1/3}}} \right)}^2}} dx} \cr
& L = \int_1^8 {\sqrt {1 + \frac{4}{9}{x^{ - 2/3}}} dx} \cr
& L = \int_1^8 {\sqrt {1 + \frac{4}{{9{x^{2/3}}}}} dx} \cr
& L = \int_1^8 {\frac{1}{{3{x^{1/3}}}}\sqrt {9{x^{2/3}} + 4} dx} \cr
& L = \frac{1}{{18}}\int_1^8 {\sqrt {9{x^{2/3}} + 4} \left( {6{x^{1/3}}} \right)dx} \cr
& {\text{Integrating}} \cr
& L = \frac{1}{{18}}\left[ {\frac{{2{{\left( {9{x^{2/3}} + 4} \right)}^{3/2}}}}{3}} \right]_1^8 \cr
& L = \frac{1}{{27}}\left[ {{{\left( {9{{\left( 8 \right)}^{2/3}} + 4} \right)}^{3/2}} - {{\left( {9{{\left( 1 \right)}^{2/3}} + 4} \right)}^{3/2}}} \right] \cr
& L = \frac{1}{{27}}\left[ {{{\left( {40} \right)}^{3/2}} - {{\left( {13} \right)}^{3/2}}} \right] \cr
& L = \frac{{80\sqrt {10} - 13\sqrt {13} }}{{27}} \cr} $$
