Question

$\underset{-\mathrm{\pi }/2}{\overset{\mathrm{\pi }/2}{\int }}\sin \left|x\right|dx$ is equal to

(a) 1
(b) 2
(c) − 1
(d) − 2

Answer 1

(b) 2

$$\begin{gathered} {\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\sin \left|x\right|dx \\ =-{\int }_{-\frac{\mathrm{\pi }}{2}}^0\sin xdx+{\int }_0^{\frac{\mathrm{\pi }}{2}}\sin xdx \\ =-{\left[-\cos x\right]}_{-\frac{\mathrm{\pi }}{2}}^0+{\left[-\cos x\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ =1-0-0+1 \\ =2 \end{gathered}$$