Answer
$$s = \frac{8}{{15}}\left( {6\sqrt 3 + 1} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{4}{5}{x^{5/4}},{\text{ }}\left[ {0,4} \right] \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{4}{5}{x^{5/4}}} \right] \cr
& f'\left( x \right) = {x^{1/4}} \cr
& {\text{Use the arc length formula}} \cr
& s = \int_a^b {\sqrt {1 + {{\left[ {f'\left( x \right)} \right]}^2}} dx} \cr
& s = \int_0^4 {\sqrt {1 + {{\left[ {{x^{1/4}}} \right]}^2}} dx} \cr
& s = \int_0^4 {\sqrt {1 + \sqrt x } dx} \cr
& {\text{Let }}u = 1 + \sqrt x ,{\text{ }}\sqrt x = u - 1,{\text{ }}du = \frac{1}{{2\sqrt x }}dx \cr
& {\text{New limits of integration}} \cr
& x = 0 \to u = 1 \cr
& x = 4 \to u = 3 \cr
& {\text{Substituting}} \cr
& s = \int_0^4 {\sqrt {1 + \sqrt x } dx} = \int_1^3 {\sqrt u \left( {2\sqrt x } \right)du} \cr
& s = \int_1^3 {\sqrt u \left( {u - 1} \right)du} \cr
& s = \int_1^3 {\left( {{u^{3/2}} - {u^{1/2}}} \right)du} \cr
& {\text{Integrating}} \cr
& s = \left[ {\frac{2}{5}{u^{5/2}} - \frac{2}{3}{u^{3/2}}} \right]_1^3 \cr
& s = \left[ {\frac{2}{5}{{\left( 3 \right)}^{5/2}} - \frac{2}{3}{{\left( 3 \right)}^{3/2}}} \right] - \left[ {\frac{2}{5}{{\left( 1 \right)}^{5/2}} - \frac{2}{3}{{\left( 1 \right)}^{3/2}}} \right] \cr
& s = \frac{{18}}{5}\sqrt 3 - \frac{6}{3}\sqrt 3 + \frac{4}{{15}} \cr
& s = \frac{8}{5}\sqrt 3 + \frac{4}{{15}} \cr
& s = \frac{8}{{15}}\left( {6\sqrt 3 + 1} \right) \cr} $$
