Question

If $I=\underset{-20\pi }{\overset{20\pi }{\int }}\left|\sin x\right|\left[\sin x\right]dx$, where $[\cdot ]$ denotes the greatest integer function, then the value of $I$ is

• A. $-40$
• B. $40$
• C. $20$
• D. $-20$

Answer 1

The correct option is A $-40$

$I=\underset{-20\pi }{\overset{20\pi }{\int }}\left|\sin x\right|\left[\sin x\right]dx\cdots \left(i\right)$
We know
$\underset{-a}{\overset{a}{\int }}f\left(x\right)=\underset{-a}{\overset{a}{\int }}f(-x)$
$\Rightarrow I=\underset{-20\pi }{\overset{20\pi }{\int }}\left|\sin x\right|[-\sin x]dx\cdots \left(ii\right)$

Adding $\left(i\right)$ and $\left(ii\right)$, we get
$\Rightarrow 2I=\underset{-20\pi }{\overset{20\pi }{\int }}\left|\sin x\right|(\left[\sin x\right]+[-\sin x])dx\Rightarrow I=\underset{0}{\overset{20\pi }{\int }}\left|\sin x\right|(\left[\sin x\right]+[-\sin x])dx\Rightarrow I=-40\underset{0}{\overset{\pi /2}{\int }}|\sin x|dx[\because [x]+[-x]=-1,x\ne Z,|\sin x|\text{has period}\pi \text{and}\sin (\pi -x)=\sin x]\Rightarrow I=-40\underset{0}{\overset{\pi /2}{\int }}\sin xdx\Rightarrow I=-40[-\cos x{]}_0^{\pi /2}\therefore I=-40$