Question

What is ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}|\sin x|dx$ equal to ?

• A. 2
• B. 1
• C. $\pi$
• D. 0

Answer 1

The correct option is A 2

$\sin x$ is positive for $xϵ[0,\frac{\pi }{2}]$ and negative for $xϵ[\frac{-\pi }{2},0]$

Then, we have the integral,

${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left|\sin x\right|dx=-{\int }_{-\frac{\pi }{2}}^0\sin x.dx+{\int }_0^{\frac{\pi }{2}}\sin x.dx$

$=-|-\cos x{|}_{-\frac{\pi }{2}}^0-|\cos x{|}_0^{\frac{\pi }{2}}$

$=-(-\cos 0+0)-(0-(\cos 0\left)\right)=2$

Hence, A is correct.