Mathematics - Definite Integration Question with Solution | TestHub

We have,

$I={\int }_{-\pi /2}^{\pi /2}\left(\left[\left[x\right]-\sin x\right]\right)dx \dots \left(\mathrm{i}\right)$

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

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

$\Rightarrow I=-2{\int }_{-\pi /2}^{-1}dx-{\int }_{-1}^{0}dx-1{\int }_0^{1}dx+{\int }_1^{\frac{\mathrm{\pi }}{2}}0·dx$

$\Rightarrow I=-2\left(-1+\frac{\mathrm{\pi }}{2}\right)-1-1$

$\Rightarrow I=-\pi$