The value of $\int\limits_{-\pi / 2}^{\pi / 2} \sin |x| d x$ is equal to :

2 -2 1 0

2

$I=\int\limits_{-\pi / 2}^0-\sin x d x+\int\limits_0^{\pi / 2} \sin x d x$ $|\cos x|_{-\pi / 2}^0-|\cos x|_0^{\pi / 2}=(1-0)-(0-1)=2$ Hence (1) is the correct answer.