Question

${\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\left[2\sin x\right]dx$ where [.] denotes G.I.F $\le x$

• A. $-\pi$
• B. $0$
• C. $-\frac{\pi }{2}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C $-\frac{\pi }{2}$

${\int }_{\pi /2}^{3\pi /2}\left[2\sin x\right]dx={\int }_{\pi /2}^{5\pi /6}\left[2\sin x\right]dx+{\int }_{5\pi /6}^{\pi }\left[2\sin x\right]dx$ $+{\int }_{\pi }^{7\pi /6}\left[2\sin x\right]dx+{\int }_{7\pi /6}^{3\pi /2}\left[2\sin x\right]dx$

$={\int }_{\pi /2}^{5\pi /6}1dx+{\int }_{5\pi /6}^{\pi }0dx+{\int }_{\pi }^{7\pi /6}(-1)dx+{\int }_{7\pi /6}^{3\pi /2}-2dx$

$=(\frac{5\pi }{6}-\frac{\pi }{2})+0-(\frac{7\pi }{6}-\pi )-2(\frac{3\pi }{2}-\frac{7\pi }{6})$

$=\frac{2\pi }{6}-\frac{\pi }{6}-2\left(\frac{2\pi }{6}\right)=\frac{\pi }{6}-\frac{4\pi }{6}$

$=-\frac{\pi }{2}$