Mathematics - Definite Integration Question with Solution | TestHub

$\mathrm{f}\left(\mathrm{x}+1\right)=\left\{\begin{array}{cc}\left[\mathrm{x}+1\right],& \mathrm{x}+1\le 2\\ 0,& \mathrm{x}+1>2\end{array}\right.=\left\{\begin{array}{cc}\left[\mathrm{x}\right]+1,& \mathrm{x}\le 1\\ 0,& \mathrm{x}>1\end{array}\right.$
$=\left\{\begin{array}{cc}0,& -1\le \mathrm{x}<0\\ 1,& 0\le \mathrm{x}<1\\ 0,& 1<\mathrm{x}<2\end{array}\right.$
$\mathrm{f}\left({\mathrm{x}}^2\right)=\left\{\begin{array}{cc}\left[{\mathrm{x}}^2\right].& {\mathrm{x}}^2\le 2\\ 0,& {\mathrm{x}}^2>2\end{array}\right.$

$=\left\{\begin{array}{cc}\left[{\mathrm{x}}^2\right].& \left|\mathrm{x}\right|\le \sqrt{2}\\ 0,& \left|\mathrm{x}\right|>\sqrt{2}\end{array}\right.$
$=\{\begin{array}{cc}\text{0,}& -1\le \text{x}<0\\ \text{0,}& 0\le \text{x}<1\\ \text{1,}& 1\le \text{x}<\sqrt{2}\\ \text{0,}& \sqrt{2}<\text{x}<2\end{array}$
$\mathrm{l}=\underset{-1}{\overset{0}{\int }}0\mathrm{d}\mathrm{x}+\underset{0}{\overset{1}{\int }}0\mathrm{d}\mathrm{x}+\underset{1}{\overset{\sqrt{2}}{\int }}\frac{\mathrm{x}\mathrm{d}\mathrm{x}}{2+0}+\underset{\sqrt{2}}{\overset{2}{\int }}0\mathrm{d}\mathrm{x}$
$=\frac{1}{4}{\left[{\mathrm{x}}^2\right]}_1^{\sqrt{2}}=\frac{1}{4}\left[2-1\right]=\frac{1}{4}$
$4\mathrm{l}=1$
$4\mathrm{l}-1=0$