Answer
$$\frac{1}{2}$$
Work Step by Step
$$\eqalign{
& \int_{ - \pi }^{2\pi /3} { - \sin x} dx \cr
& = - \int_{ - \pi }^{2\pi /3} {\sin x} dx \cr
& {\text{integrate by using the basic integration rule }}\int {\sin x} dx = - \cos x + C\,\,\,\left( {{\text{page 692}}} \right) \cr
& then \cr
& = - \left( { - \cos x} \right)_{ - \pi }^{2\pi /3} \cr
& = \left( {\cos x} \right)_{ - \pi }^{2\pi /3} \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \cos \left( {\frac{{2\pi }}{3}} \right) - \cos \left( { - \pi } \right) \cr
& {\text{simplifying}} \cr
& = - \frac{1}{2} - \left( { - 1} \right) \cr
& = \frac{1}{2} \cr} $$