Given: $$f: \mathrm{R} \rightarrow \mathrm{R} f(x)=\left\{\begin{aligned} {[x], } & x \leq 2 \\ o, & x>2\end{aligned}\right.$$

$$\text { And } \quad I=\int_\limits{-1}^2 \frac{x f\left(x^2\right)}{2+f(x+1)} d x$$

So, $$f\left(x^2\right)=\left\{\begin{array}{cl} {\left[x^2\right],} & x^2 \leq 2, \quad x \in[-\sqrt{2}, \sqrt{2} \\ 0, & x^2>2, \quad x \in(-\infty,-\sqrt{2}) \cup(\sqrt{2}, \infty) \end{array}\right.$$

$$\text { And } f(x+1)=\left\{\begin{array}{cll} {[x+1],} & x+1 \leq 2, & x \leq 1 \\ 0, & x+1>2, & x>1 \end{array}\right.$$

$$\text { So, } \mathrm{I}=\int_\limits{-1}^0 \frac{x f\left[x^2\right]}{2+f(x+1)} d x+\int_\limits0^1 \frac{x f\left(x^2\right)}{2+f(x+1)} d x +\int_\limits1^{\sqrt{2}} \frac{x f\left(x^2\right)}{2+f(x+1)} d x+\int_\limits2^2 \frac{x f\left(x^2\right)}{2+f(x+1)} d x$$

$$\Rightarrow \mathrm{I}=\int_\limits{-1}^0 \frac{x\left[x^2\right]}{2+[x+1]} d x+\int_\limits0^1 \frac{x\left[x^2\right]}{2+[x+1]} d x+\int_\limits1^{\sqrt{2}} \frac{x\left[x^2\right]}{2+0} d x +\int_\limits{\sqrt{2}}^2 \frac{x \cdot 0}{2+0} d x$$

$$\Rightarrow \mathrm{I}=\int_\limits{-1}^0 \frac{x\left[x^2\right]}{2+[x+1]} d x+\int_\limits0^1 \frac{x\left[x^2\right]}{2+[x+1]} d x+\int_\limits1^{\sqrt{2}} \frac{x\left[x^2\right]}{2} d x$$

Using the property of greatest integer function, For $$x \in(-1,0),[x+1]=0$$ and $$\left[x^2\right]=0$$ and, For $$x \in(0,1),[x+1]=1$$ and $$\left[x^2\right]=0$$ and, For $$x \in(1, \sqrt{2}),\left[x^2\right]=1$$

$$\begin{aligned} & \Rightarrow \quad \mathrm{I}=\int_{-1}^0 \frac{x .0}{2+0} d x+\int_0^1 \frac{x .0}{2+1} d x+\int_\limits1^{\sqrt{2}} \frac{x .1}{2} d x \\ & \Rightarrow \quad \mathrm{I}=\int_1^{\sqrt{2}} \frac{x}{2} d x \\ & \Rightarrow \quad \mathrm{I}=\frac{1}{2}\left[\frac{x^2}{2}\right]_1^{\sqrt{2}} \\ & \Rightarrow \quad \mathrm{I}=\frac{1}{2}\left[\frac{2}{2}-\frac{1}{2}\right] \\ & \Rightarrow \quad \mathrm{I}=\frac{1}{4} \\ & \Rightarrow \quad 4 \mathrm{I}=1 \\ & \Rightarrow 4 \mathrm{I}-1=0 \\ \end{aligned}$$

Hint:

(i) Find $$f\left(x^2\right)$$ and $$f(x+1)$$ using composite function.

(ii) Split the given integral using the property of the greatest integer function.

(iii) Find the value of the definite integral using, if $$\int g(x) d x=\mathrm{G}(x) \Rightarrow \int_a^b g(x) d x=[\mathrm{G}(b)-\mathrm{G}(a)]$$