Answer
$$\frac{{\sqrt 3 }}{2} - \frac{\pi }{{24}}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {\left| {\frac{1}{2} - \sin x} \right|dx} \cr
& {\text{By the definition of the aboslute value}} \cr
& \int_0^{\pi /2} {\left| {\frac{1}{2} - \sin x} \right|dx} = \int_0^{\pi /6} {\left( { - \frac{3}{\pi }x + \frac{1}{2}} \right)dx} + \int_{\pi /6}^{\pi /2} {\sin xdx} \cr
& {\text{Integrating}} \cr
& {\text{ }} = \left[ { - \frac{3}{{2\pi }}{x^2} + \frac{1}{2}x} \right]_0^{\pi /6} - \left[ {\cos x} \right]_{\pi /6}^{\pi /2} \cr
& {\text{ }} = \left[ { - \frac{3}{{2\pi }}{{\left( {\frac{\pi }{6}} \right)}^2} + \frac{1}{2}\left( {\frac{\pi }{6}} \right)} \right] - \left[ {\cos \left( {\frac{\pi }{2}} \right) - \cos \left( {\frac{\pi }{6}} \right)} \right] \cr
& {\text{Simplify}} \cr
& {\text{ }} = - \frac{{3\pi }}{{24}} + \frac{\pi }{{12}} + \frac{{\sqrt 3 }}{2} \cr
& {\text{ }} = \frac{{\sqrt 3 }}{2} - \frac{\pi }{{24}} \cr} $$
