Answer
$L = \int_0^2 {\sqrt {1 + 9{x^4}} dx} $
Work Step by Step
$$\eqalign{
& {\text{Let }}y = {x^3},{\text{ }}0 \leqslant x \leqslant 2 \cr
& {\text{Differentiating}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{x^3}} \right] \cr
& \frac{{dy}}{{dx}} = 3{x^2} \cr
& \cr
& {\text{Use the arc lenght formula }} \cr
& L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} \cr
& {\text{Substituting the limits and the derivative of the function}} \cr
& L = \int_0^2 {\sqrt {1 + {{\left( {3{x^2}} \right)}^2}} dx} \cr
& {\text{Simplifying}} \cr
& L = \int_0^2 {\sqrt {1 + 9{x^4}} dx} \cr} $$
