Answer
$$L = \frac{1}{{4a}}\left( {20a\sqrt {400{a^2} + 1} + \ln \left( {20a + \sqrt {400{a^2} + 1} } \right)} \right)$$
Work Step by Step
$$\eqalign{
& y = a{x^2},{\text{ from }}x = 0{\text{ to }}x = 10 \cr
& {\text{Find }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {a{x^2}} \right] \cr
& \frac{{dy}}{{dx}} = 2ax \cr
& {\text{The arc length is }} \cr
& L = \int_0^{10} {\sqrt {1 + {{\left( {2ax} \right)}^2}} dx} \cr
& L = \frac{1}{{2a}}\int_0^{10} {\sqrt {1 + {{\left( {2ax} \right)}^2}} \left( {2a} \right)dx} \cr
& {\text{Integrate by tables}} \cr
& \int {\sqrt {{u^2} + 1} du = } \frac{1}{2}\left( {u\sqrt {{u^2} + 1} + \ln \left| {u + \sqrt {{u^2} + 1} } \right|} \right) + C \cr
& L = \frac{1}{{4a}}\left[ {2ax\sqrt {{{\left( {2ax} \right)}^2} + 1} + \ln \left| {2ax + \sqrt {{{\left( {2ax} \right)}^2} + 1} } \right|} \right]_0^{10} \cr
& {\text{Evaluating}} \cr
& L = \frac{1}{{4a}}\left[ {20a\sqrt {{{\left( {20a} \right)}^2} + 1} + \ln \left| {20a + \sqrt {{{\left( {20a} \right)}^2} + 1} } \right|} \right] \cr
& L = \frac{1}{{4a}}\left( {20a\sqrt {400{a^2} + 1} + \ln \left( {20a + \sqrt {400{a^2} + 1} } \right)} \right) \cr} $$
