Answer
$$A = \frac{1}{3}\pi $$
Work Step by Step
$$\eqalign{
& {\text{The area of the region is given by}} \cr
& A = \int_0^1 {\frac{2}{{\sqrt {4 - {x^2}} }}} dx \cr
& A = 2\int_0^1 {\frac{1}{{\sqrt {4 - {x^2}} }}} dx \cr
& {\text{Integrate using basic integration rules}} \cr
& A = 2\left[ {{{\sin }^{ - 1}}\left( {\frac{x}{2}} \right)} \right]_0^1 \cr
& A = 2\left[ {{{\sin }^{ - 1}}\left( {\frac{1}{2}} \right) - {{\sin }^{ - 1}}\left( {\frac{0}{2}} \right)} \right] \cr
& A = 2\left( {\frac{1}{6}\pi - 0} \right) \cr
& A = \frac{1}{3}\pi \cr} $$
