Answer
$$L = 4\sqrt {17} + \ln \left( {4 + \sqrt {17} } \right)$$
Work Step by Step
$$\eqalign{
& {\text{Let }}y = \frac{{{x^2}}}{4}{\text{ on the interval }}\left[ {0,8} \right] \cr
& {\text{Find }}\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{{x^2}}}{4}} \right] = \frac{x}{2} \cr
& {\text{Calculate the arc length }}L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} d} x \cr
& L = \int_0^8 {\sqrt {1 + {{\left( {\frac{x}{2}} \right)}^2}} d} x \cr
& L = \int_0^8 {\sqrt {\frac{{4 + {x^2}}}{4}} d} x \cr
& L = \frac{1}{2}\int_0^8 {\sqrt {4 + {x^2}} d} x \cr
& {\text{Use the table of integrals formula 76: }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\int {\sqrt {{a^2} + {x^2}} dx} = \frac{x}{2}\sqrt {{a^2} + {x^2}} + \frac{{{a^2}}}{2}\ln \left( {x + \sqrt {{a^2} + {x^2}} } \right) + C \cr
& {\text{Therefore,}} \cr
& L = \frac{1}{2}\int_0^8 {\sqrt {4 + {x^2}} d} x = \frac{1}{2}\left[ {\frac{x}{2}\sqrt {4 + {x^2}} + \frac{4}{2}\ln \left( {x + \sqrt {4 + {x^2}} } \right)} \right]_0^8 \cr
& {\text{Simplify}} \cr
& L = \frac{1}{2}\left[ {\frac{8}{2}\sqrt {4 + {8^2}} + \frac{4}{2}\ln \left( {8 + \sqrt {4 + {8^2}} } \right)} \right] - \frac{1}{2}\left[ {0 + \frac{4}{2}\ln \left( {\sqrt {4 + {0^2}} } \right)} \right] \cr
& L = \frac{1}{2}\left[ {4\sqrt {68} + 2\ln \left( {8 + \sqrt {68} } \right)} \right] - \frac{1}{2}\left[ {2\ln \left( 2 \right)} \right] \cr
& L = 2\sqrt {68} + \ln \left( {8 + \sqrt {68} } \right) - \ln 2 \cr
& L = 4\sqrt {17} + \ln \left( {\frac{{8 + 2\sqrt {17} }}{2}} \right) \cr
& L = 4\sqrt {17} + \ln \left( {4 + \sqrt {17} } \right) \cr} $$
