Q136 / 200 · Manipal (MET) 2010 · Mathematics
The value of the integral $\int_{\pi / 2}^{3 \pi / 2}[\sin x] d x$, where [.] denotes the greatest integer function, is
- A$\frac{\pi}{2}$
- Correct answerB$-\frac{\pi}{2}$
- C0
- D$\pi$
Solution
$\int_{\pi / 2}^{3 \pi / 2}[\sin x] d x=\int_{\pi / 2}^\pi[\sin x] d x+\int_\pi^{3 \pi / 2}[\sin x] d x$
$=\int_{\pi / 2}^\pi 0 d x+\int_\pi^{3 \pi / 2}(-1) d x=-[x]_x^{3 \pi / 2}$
$=-\left(\frac{3 \pi}{2}-\pi\right)=-\frac{\pi}{2}$
Hence, the correct answer is option: 'B'.