Question

If $f\left(x\right)={\int }_0^x\sin \left[2x\right]dx$

where [x] = greatest integer less than or equal to $x$

then $f\left(\pi /2\right)$ is ?

• A. $\frac{1}{2}\{\sin 1+(\pi -2)\sin 2\}$
• B. $\frac{1}{2}\{\sin 1+\sin 2+(\pi -3)\sin 3\}$
• C. $0$
• D. $\sin 1+(\frac{\pi }{2}-2)\sin 2$

Answer 1

Answer: (B)

$\frac{1}{2}\{\sin 1+\sin 2+(\pi -3)\sin 3\}$

In the interval $[0,\frac{1}{2})$, $\left[2x\right]=0$

In the interval $[\frac{1}{2},1)$, $\left[2x\right]=1$

In the interval $[1,\frac{3}{2})$, $\left[2x\right]=2$

In the interval $[\frac{3}{2},\frac{\pi }{2})$, $\left[2x\right]=3$

Hence,

${\int }_0^{\pi /2}\sin \left[2x\right]dx$

$={\int }_0^{1/2}\left(\sin 0\right)dx+{\int }_{1/2}^1\sin 1dx+{\int }_1^{3/2}\sin 2dx+{\int }_{3/2}^{\pi /2}\sin 3dx$

$=0+\sin 1[x{]}_{1/2}^1+\sin 2[x{]}_1^{3/2}+\sin 3[x{]}_{3/2}^{\pi /2}$

$=\frac{1}{2}\sin 1+\frac{1}{2}\sin 2+\frac{\pi -3}{2}\sin 3$

$=\frac{1}{2}[\sin 1+\sin 2+(\pi -3\left)\sin 3\right]$

Hence, answer is option-(B).