Question

The value of ${\int }_{\pi }^{2\pi }\left[2\sin x\right]dx$ is equal to (where $[\cdot ]$ is the G.I.F)

• A. $-\pi$
• B. $-2\pi$
• C. $-\frac{5\pi }{3}$
• D. $\frac{5\pi }{3}$

Answer 1

The correct option is A $-\pi$

For $xϵ$ $(\pi ,\frac{7\pi }{6}]$ and $[\frac{11\pi }{6},2\pi )$

$\left[2\sin x\right]=-1$.

For $xϵ(\frac{7\pi }{6},\frac{11\pi }{6}),$ $\left[2\sin x\right]=-2$.

Therefore the given integral becomes,

${\int }_{\pi }^{7\pi /6}(-1)dx$ $+{\int }_{7\pi /6}^{11\pi /6}(-2)dx$ $+{\int }_{11\pi /6}^{2\pi }(-1)dx$

$=(-1{)}_{\pi }^{7\pi /6}+(-2{)}_{7\pi /6}^{11\pi /6}+(-1{)}_{11\pi /6}^{2\pi }$

$=\frac{-5\pi }{3}$

Therefore, the correct answer is C.