Question

${\int }_0^{\frac{\pi }{2}}(x-[\sin x\left]\right)dx=$ _________.
(where $\left[x\right]=$ greatest integer not greater than $x$).

• A. $\frac{{\pi }^2}{8}-2$
• B. $\frac{{\pi }^2}{4}-1$
• C. $\frac{{\pi }^2}{8}-1$
• D. $\frac{{\pi }^2}{8}$

Answer 1

Answer: (D)

$\frac{{\pi }^2}{8}$

${\int }_0^{\pi /2}(x-[\sin x\left]\right)dx={\int }_0^{\pi /2}xdx-{\int }_0^{\pi /2}\left[\sin x\right]dx$

$={\left(\frac{x^2}{2}\right)}_0^{\pi /2}-{\int }_0^{\pi /2}\left[\sin x\right])dx$

$=\frac{{\pi }^2}{8}$ (for all values of $x\in (0,\frac{\pi }{2})$, $\left[\sin x\right]=0$ and hence, the second integral is 0.)

This is the required answer.