Question

If $\left[x\right]$ denotes the greatest integer less than or equal to $x,$ then the value of the integral $\underset{-\pi /2}{\overset{\pi /2}{\int }}\left[\right[x]-\sin x]dx$ is equal to:

• A. $-\pi$
• B. $0$
• C. $\pi$
• D. $1$

Answer 1

The correct option is A $-\pi$

$I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\left[\right[x]-\sin x]dx$
$=\underset{-\pi /2}{\overset{\pi /2}{\int }}\left(\left[x\right]+[-\sin x]\right)dx$
$=\underset{0}{\overset{\pi /2}{\int }}\left(\right[x]+[-x]+(\left[\sin x\right]+[-\sin x]\left)\right)dx$
$=\underset{0}{\overset{\pi /2}{\int }}((-1)+(-1))dx$
$=\underset{0}{\overset{\pi /2}{\int }}(-2)dx=-\pi$