Answer
$s \approx 7.6337$
Work Step by Step
$$\eqalign{
& {\text{Let the function }}y = {x^{2/3}}{\text{ on the interval }}\left[ {1,8} \right] \cr
& {\text{The arc length of the function }}y{\text{ on the interval }}\left[ {a,b} \right]{\text{ is}} \cr
& s = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} \cr
& {\text{Differentiating}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{x^{2/3}}} \right] \cr
& \frac{{dy}}{{dx}} = \frac{2}{3}{x^{ - 1/3}} \cr
& {\text{Substituting }}\frac{{dy}}{{dx}}{\text{ into the arc length formula}} \cr
& s = \int_1^8 {\sqrt {1 + {{\left( {\frac{2}{3}{x^{ - 1/3}}} \right)}^2}} dx} \cr
& s = \int_1^8 {\sqrt {1 + \frac{4}{9}{x^{ - 2/3}}} dx} \cr
& {\text{Integrate using a graphing utility or a computer}}{\text{, we obtain}} \cr
& s \approx 7.6337 \cr} $$
