Answer
$${\text{area}} = 6 + \frac{{25}}{2}{\sin ^{ - 1}}\left( {\frac{4}{5}} \right)$$
Work Step by Step
$$\eqalign{
& {\text{Let }}y = \sqrt {25 - {x^2}} ,\,\,\,\,y = 0,\,\,\,x = 0,\,\,\,x = 4 \cr
& \cr
& {\text{We can represent the enclosed by the curves as}} \cr
& {\text{area}} = \int_0^4 {\sqrt {25 - {x^2}} } dx \cr
& \cr
& {\text{integrate by tables using the formula }} \cr
& \int {\sqrt {{a^2} - {x^2}} } dx = \frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\left( {\frac{x}{a}} \right) + C \cr
& {\text{area}} = \int_0^4 {\sqrt {25 - {x^2}} } dx = \left[ {\frac{x}{2}\sqrt {25 - {x^2}} + \frac{{25}}{2}{{\sin }^{ - 1}}\left( {\frac{x}{5}} \right)} \right]_0^4 \cr
& {\text{area}} = \left[ {\frac{4}{2}\sqrt {25 - {4^2}} + \frac{{25}}{2}{{\sin }^{ - 1}}\left( {\frac{4}{5}} \right)} \right] - \left[ {\frac{0}{2}\sqrt {25 - {0^2}} + \frac{{25}}{2}{{\sin }^{ - 1}}\left( {\frac{0}{5}} \right)} \right] \cr
& {\text{simplify}} \cr
& {\text{area}} = \left[ {2\sqrt 9 + \frac{{25}}{2}{{\sin }^{ - 1}}\left( {\frac{4}{5}} \right)} \right] \cr
& {\text{area}} = 6 + \frac{{25}}{2}{\sin ^{ - 1}}\left( {\frac{4}{5}} \right) \cr} $$
