Question:
If $[x]$ is the greatest integer $\leq x$, then
$\pi^{2} \int_{0}^{2}\left(\sin \frac{\pi x}{2}\right)(x-[x])^{[x]} d x$ is equal to:
Correct Option: , 2
Solution:
$\pi^{2}\left[\int_{0}^{1} \sin \frac{\pi x}{2} d x+\int_{1}^{2} \sin \frac{\pi x}{2}(x-1) d x\right]$
$=\pi^{2}\left[-\frac{2}{\pi}\left(\cos \frac{\pi x}{2}\right)+\left((x-1)\left(-\frac{2}{\pi} \cos \frac{\pi x}{2}\right)\right)_{1}^{2}-\int_{1}^{2}-\frac{2}{\pi} \cos \frac{\pi x}{2} d x\right]$
$=\pi^{2}\left[0+\frac{2}{\pi}+\frac{2}{\pi}+\frac{2}{\pi} \cdot \frac{2}{\pi}\left(\sin \frac{\pi x}{2}\right)_{1}^{2}\right]$
$=4 \pi-4=4(\pi-1)$