Question

If \([x]\) denotes the greatest integer less than or equal to \(x\), then the value of the integral \(\int_{-\pi/2}^{\pi/2}[[x]-\sin x]\,dx\) is equal to :

• A. \(-\pi\)
• B. \(\pi\)
• C. 0
• D. 1

Hint:

Since \([x]\) is an integer, \([[x]-\sin x]=[x]+[-\sin x]\). Use \([f]+[-f]=-1\) for non-integer \(f\).

Answer 1

The Correct Option is A

\([[x]-\sin x]=[x]+[-\sin x]\).
\(\int_{-\pi/2}^{\pi/2}[x]\,dx\): on \(\left(-\frac\pi2,-1\right)\) the value is \(-2\) (length \(\frac\pi2-1\)); on \([-1,0)\) it is \(-1\) (length 1); on \([0,1)\) it is 0; on \(\left[1,\frac\pi2\right)\) it is 1 (length \(\frac\pi2-1\)). Sum \(=-2\left(\frac\pi2-1\right)-1+\left(\frac\pi2-1\right)=-\frac\pi2\).
For \(g(x)=[-\sin x]\): \(g(x)+g(-x)=[-\sin x]+[\sin x]=-1\) except at points where \(\sin x\) is an integer. So \(\int_{-\pi/2}^{\pi/2}g\,dx=\frac12\int_{-\pi/2}^{\pi/2}(g(x)+g(-x))dx=-\frac\pi2\).
Total \(=-\frac\pi2-\frac\pi2=-\pi\).
Answer: (1) \(-\pi\)