Question:
If [ ] denotes the greatest integer function, then the integral \(\int_0^{\pi}[\cos x]\,dx\) is equal to:
If [ ] denotes the greatest integer function, then the integral \(\int_0^{\pi}[\cos x]\,dx\) is equal to:
Show Hint
Find where \([\cos x]\) equals 0 and where it equals \(-1\) on \([0,\pi]\) and integrate piecewise.
Updated On: Nov 11, 2026
- \(\frac{\pi}{2}\)
- \(0\)
- \(-1\)
- \(-\frac{\pi}{2}\)
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The Correct Option is D
Solution and Explanation
On \(0<x\le\frac{\pi}{2}\): \(0\le\cos x<1\), so \([\cos x]=0\).
On \(\frac{\pi}{2}<x\le\pi\): \(-1\le\cos x<0\), so \([\cos x]=-1\).
(The single point \(x=0\) where \(\cos x=1\) does not affect the integral.)
\(\int_0^{\pi}[\cos x]dx=\int_0^{\pi/2}0\,dx+\int_{\pi/2}^{\pi}(-1)dx=-\left(\pi-\frac{\pi}{2}\right)=-\frac{\pi}{2}\).
Answer: (4) \(-\frac{\pi}{2}\)
On \(\frac{\pi}{2}<x\le\pi\): \(-1\le\cos x<0\), so \([\cos x]=-1\).
(The single point \(x=0\) where \(\cos x=1\) does not affect the integral.)
\(\int_0^{\pi}[\cos x]dx=\int_0^{\pi/2}0\,dx+\int_{\pi/2}^{\pi}(-1)dx=-\left(\pi-\frac{\pi}{2}\right)=-\frac{\pi}{2}\).
Answer: (4) \(-\frac{\pi}{2}\)
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