Answer
$$L = \sqrt {65} + \frac{1}{8}\ln \left( {8 + \sqrt {65} } \right)$$
Work Step by Step
$$\eqalign{
& y = 2{x^2},{\text{ 0}} \leqslant x \leqslant {\text{2}} \cr
& {\text{Calculate the arc length of the curve using the formula}}{\text{.}} \cr
& L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} \cr
& {\text{,then}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {2{x^2}} \right] = 4x \cr
& L = \int_0^2 {\sqrt {1 + {{\left( {4x} \right)}^2}} dx} \cr
& L = \frac{1}{4}\int_0^2 {\sqrt {1 + {{\left( {4x} \right)}^2}} \left( 4 \right)dx} \cr
& {\text{Integrate using the formula }} \cr
& \int {\sqrt {{a^2} + {u^2}} du = \frac{u}{2}\sqrt {{a^2} + {u^2}} + \frac{{{a^2}}}{2}\ln \left( {u + \sqrt {{u^2} + {a^2}} } \right) + C} \cr
& L = \frac{1}{4}\left[ {\frac{{4x}}{2}\sqrt {1 + {{\left( {4x} \right)}^2}} + \frac{{{1^2}}}{2}\ln \left( {4x + \sqrt {1 + {{\left( {4x} \right)}^2}} } \right)} \right]_0^2 \cr
& L = \left[ {\frac{2}{2}\sqrt {1 + {{\left( 8 \right)}^2}} + \frac{1}{8}\ln \left( {8 + \sqrt {1 + {{\left( 8 \right)}^2}} } \right)} \right] - \left[ {0 + \frac{1}{8}\ln \left( 1 \right)} \right] \cr
& L = \sqrt {65} + \frac{1}{8}\ln \left( {8 + \sqrt {65} } \right) \cr} $$
