Question

If $\left[x\right]$ is the greatest integer $\le x$, then ${\pi }^2\underset{0}{\overset{2}{\int }}\left(\sin \frac{\pi x}{2}\right){(x-[x\left]\right)}^{\left[x\right]}dx$ is equal to

• A. $4(\pi +1)$
• B. $2(\pi -1)$
• C. $4(\pi -1)$
• D. $2(\pi +1)$

Answer 1

The correct option is C $4(\pi -1)$

${\pi }^2\underset{0}{\overset{2}{\int }}\left(\sin \frac{\pi x}{2}\right)(x-[x]{)}^{\left[x\right]}dx$
$={\pi }^2\left[\underset{0}{\overset{1}{\int }}\sin \frac{\pi x}{2}dx+\underset{1}{\overset{2}{\int }}(x-1)\sin \frac{\pi x}{2}dx\right]$

$={\pi }^2[{(-\frac{2}{\pi }\cos \frac{\pi x}{2})}_0^1-\frac{2}{\pi }{\left(\right(x-1\left)\cos \frac{\pi x}{2}\right)}_1^2+\underset{1}{\overset{2}{\int }}\frac{2}{\pi }\cos \frac{\pi x}{2}dx]$
$={\pi }^2[\frac{2}{\pi }+\frac{2}{\pi }+\frac{4}{{\pi }^2}{\left(\sin \frac{\pi x}{2}\right)}_1^2]$
$={\pi }^2[\frac{4}{\pi }-\frac{4}{{\pi }^2}]=4(\pi -1)$