Question

If $\left[y\right]$ = the greatest integer less than or equal to $y$ then ${\int }_{\pi /2}^{3\pi /2}\left[2\sin x\right]$dx is

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

Answer 1

Answer: (C)

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

$I={\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\left[2sinx\right]dx$ ...(1)
Using property ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$I={\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\left[2sin(\frac{3\pi }{2}+\frac{\pi }{2}-x)\right]dx={\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\left[2sin(-x)\right]dx={\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}[-2sin\left(x\right)]dx$ ...(2)
Adding (1) and (2) and using $\left[x\right]+[-x]=-1$
We get
$2I={\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}(-1)dx=-{\left[x\right]}_{\frac{\pi }{2}}^{\frac{3\pi }{2}}=-\pi I=-\frac{\pi }{2}$