Answer
$A \approx 4$
Work Step by Step
$$\eqalign{
& {\text{Let }}y = \arcsin \left( {\frac{1}{2}x} \right){\text{ and }}y = 2 - {x^2} \cr
& {\text{From the graph we have the intersection points}} \cr
& x \approx - 1.75{\text{ and }}x = 1.17 \cr
& {\text{The area of the region bounded by the curves is given by:}} \cr
& A = \int_{ - 1.75}^{1.17} {\left[ {\left( {2 - {x^2}} \right) - \arcsin \left( {\frac{1}{2}x} \right)} \right]} dx \cr
& \cr
& {\text{* Integrating }}\int {\arcsin \left( {\frac{1}{2}x} \right)} dx \cr
& u = \arcsin \left( {\frac{1}{2}x} \right),{\text{ }}du = \frac{1}{{\sqrt {4 - {x^2}} }}dx \cr
& dv = dx,{\text{ }}v = x \cr
& = x\arcsin \left( {\frac{1}{2}x} \right) - \int {\frac{x}{{\sqrt {4 - {x^2}} }}} dx \cr
& = x\arcsin \left( {\frac{1}{2}x} \right) + \sqrt {4 - {x^2}} + C \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \underbrace {A = \int_{ - 1.75}^{1.17} {\left[ {\left( {2 - {x^2}} \right) - \arcsin \left( {\frac{1}{2}x} \right)} \right]} dx}_ \Downarrow \cr
& A = \left[ {2x - \frac{1}{3}{x^3} - x\arcsin \left( {\frac{1}{2}x} \right) - \sqrt {4 - {x^2}} } \right]_{ - 1.75}^{1.17} \cr
& {\text{Evaluating}} \cr
& A = \left[ {2.34 - \frac{1}{3}{{\left( {1.17} \right)}^3} - \left( {1.17} \right)\arcsin \left( {\frac{1}{2}\left( {1.17} \right)} \right) - \sqrt {4 - {{\left( {1.17} \right)}^2}} } \right] \cr
& - \left[ { - 3.5 - \frac{1}{3}{{\left( { - 1.75} \right)}^3} - \left( {-1.75} \right)\arcsin \left( {\frac{1}{2}\left( {-1.75} \right)} \right) - \sqrt {4 - {{\left( {-1.75} \right)}^2}} } \right] \cr
& {\text{Simplifying}} \cr
& A \approx - 0.547 - \left( {4.546} \right) \cr
& A \approx 4 \cr} $$
