Q. $\int_{- \pi/2}^{\pi/2}{\sin|x|dx}$is equal to
A) 0
B) 1
C) 2
D) $\pi$
Explanation
(c)
Let $I = \int_{- \pi/2}^{\pi/2}{\sin|x|dx = 2\int_{0}^{\pi/2}{\sin{x\ dx}}}$
$= 2\left\lbrack - \cos x \right\rbrack_{0}^{\pi/2} = 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.