Answer
$$A = \frac{{22}}{3}$$
Work Step by Step
$$\eqalign{
& {\text{From the graph we can note that the area is given by}} \cr
& A = \int_0^4 {\left( {\sqrt x - \left( { - \frac{1}{4}x} \right)} \right)} dx \cr
& A = \int_0^4 {\left( {{x^{1/2}} + \frac{1}{4}x} \right)} dx \cr
& {\text{Integrate}} \cr
& A = \left[ {\frac{2}{3}{x^{3/2}} + \frac{1}{8}{x^2}} \right]_0^4 \cr
& {\text{Evaluate }} \cr
& A = \left[ {\frac{2}{3}{{\left( 4 \right)}^{3/2}} + \frac{1}{8}{{\left( 4 \right)}^2}} \right] - \left[ {\frac{2}{3}{{\left( 0 \right)}^{3/2}} + \frac{1}{8}{{\left( 0 \right)}^2}} \right] \cr
& {\text{Simplify}} \cr
& A = \frac{{22}}{3} + 0 \cr
& A = \frac{{22}}{3} \cr} $$