Question

$\int_{-\pi/2}^{\pi/2} |\sin x| dx$ is

• A. $2$
• B. $0$
• C. $\frac{\pi}{2}$
• D. $1$

Hint:

Even function: $\int_{-a}^a f(x)dx = 2\int_0^a f(x)dx$.

Answer 1

The Correct Option is A

Step 1: $|\sin x|$ is even. So $\int_{-\pi/2}^{\pi/2} |\sin x| dx = 2\int_0^{\pi/2} \sin x dx = 2[-\cos x]_0^{\pi/2} = 2(0+1)=2$.}