Question

The value of ${\int }_{\pi /2}^{3\pi /2}\left[2\sin x\right]dx$ is equal to?

($\left[x\right]$ denotes the greatest integer function)

• A. $-\pi$
• B. $0$
• C. $-\pi /2$
• D. $\pi /2$

Answer 1

The correct option is C $-\pi /2$

Given : ${\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\left[2\sin x\right]dx$

$I={\int }_{\frac{\pi }{2}}^{\frac{5\pi }{6}}dx+{\int }_{\frac{5\pi }{6}}^{\pi }0dx+{\int }_{\pi }^{\frac{7\pi }{6}}(-1)dx+{\int }_{\frac{7\pi }{6}}^{\frac{3\pi }{2}}(-1)dxI={\left[x\right]}_{\frac{\pi }{2}}^{\frac{5\pi }{6}}-{\left[x\right]}_{\pi }^{\frac{7\pi }{6}}-2{\left[x\right]}_{\frac{7\pi }{6}}^{\frac{3\pi }{2}}I=(\frac{5\pi }{6}-\frac{3\pi }{2})-(\frac{7\pi }{6}-\pi )-2(\frac{3\pi }{2}-\frac{7\pi }{6})I=\frac{\pi }{3}-\frac{\pi }{6}-\frac{2\pi }{3}=-\frac{\pi }{3}-\frac{\pi }{6}=-\frac{\pi }{2}$

Hence the correct answer is $-\frac{\pi }{2}$