Let f: ℝ → ℝ be a function defined by f(x) = \ cases [x], & x ≤ 2...

f\left(x + 1\right) = \left\{\begin{cases}\left[x + 1\right], & x + 1 \le 2 \\ 0, & x + 1 > 2\end{cases}\right = \left\{\begin{cases}\left[x\right] + 1, & x \le 1 \\ 0, & x > 1\end{cases}\right)<br />= \left\{$\begin{cases}0, & - 1 \le x < 0 \\ 1, & 0 \le x < 1 \\ 0, & 1 < x < 2\end{cases}$\right)
f\left({x}^{2}\right) = \left\{\begin{cases}\left[{x}^{2}\right] . & {x}^{2} \le 2 \\ 0, & {x}^{2} > 2\end{cases}\right)<br /><br />= \left\{$\begin{cases}\left[{x}^{2}\right] . & | x | \le \sqrt{2} \\ 0, & | x | > \sqrt{2}\end{cases}$\right)
= {\begin{cases}\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{cases}$$<br />$l = \mathop{∫}_{- 1}^{0}0dx + \mathop{∫}_{0}^{1}0dx + \mathop{∫}_{1}^{\sqrt{2}}\dfrac{xdx}{2 + 0} + \mathop{∫}_{\sqrt{2}}^{2}0dx$<br />$= \dfrac{1}{4}{\left[{x}^{2}\right]}_{1}^{\sqrt{2}} = \dfrac{1}{4}\left[2 - 1\right] = \dfrac{1}{4}$<br />$4l = 1$<br />$4l - 1 = 0$