If x is the greatest integer ≤ x , then π 2 ∫ 0 2 sin π x 2 x - x [ x ] d x is equal…

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

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

$I={\pi }^2{\int }_0^1\sin \left(\frac{\pi x}{2}\right)x^{0}dx+{\pi }^2{\int }_1^2\sin \left(\frac{\pi x}{2}\right){\left(x-1\right)}^{1}dx$

$I={\pi }^2{\int }_0^1\sin \left(\frac{\pi x}{2}\right)dx+{\pi }^2{\int }_1^2\sin \left(\frac{\pi x}{2}\right)\left(x-1\right)dx$

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

$I={\pi }^2{\left[-\frac{2}{\pi }\cos \left(\frac{\pi x}{2}\right)\right]}_0^1+{\pi }^2{\left[\left(x\right)\frac{2}{\pi }\right.\overline{)\left(-\cos \left(\frac{\pi x}{2}\right)\right)}}_1^2\left.-{\int }_1^2\frac{2}{\pi }\left(-\cos \frac{\pi \mathrm{x}}{2}\right)dx\right]+{\mathrm{\pi }}^2\times \frac{2}{\mathrm{\pi }}{\left[\cos \left(\frac{\mathrm{\pi x}}{2}\right)\right]}_1^2$

$I={\pi }^2{\left[-\frac{2}{\pi }\cos \left(\frac{\pi x}{2}\right)\right]}_0^1+{\pi }^2{\left[\left(x\right)\frac{2}{\pi }\right.\overline{)\left(-\cos \left(\frac{\pi x}{2}\right)\right)}}_1^2\left.+{\left(\frac{2}{\mathrm{\pi }}\right)}^2{\left[\sin \left(\frac{\mathrm{\pi x}}{2}\right)\right]}_1^2\right]+{\mathrm{\pi }}^2\times \frac{2}{\mathrm{\pi }}{\left[\cos \left(\frac{\mathrm{\pi x}}{2}\right)\right]}_1^2$

$I=-{\pi }^2\left[\frac{2}{\pi }\cos \left(\frac{\pi }{2}\right)-\frac{2}{\pi }\cos \left(0\right)\right]+{\pi }^2\left[2\times \frac{2}{\pi }\left(-\cos \pi \right)+1\times \frac{2}{\pi }\times \overline{)\left(\cos \frac{\mathrm{\pi }}{2}\right)}\right]\left.+4\left[\sin \left(\mathrm{\pi }\right)-\sin \frac{\mathrm{\pi }}{2}\right]\right]+{\mathrm{\pi }}^2\times \frac{2}{\mathrm{\pi }}\left[\cos \left(\mathrm{\pi }\right)-\cos \frac{\mathrm{\pi }}{2}\right]$

$I=2\pi +4\pi +4(0-1)-2\mathrm{\pi }$

$I=4\pi -4$

$I=4\left(\pi -1\right)$