Let $f: R \rightarrow R$ be a function defined by $f(x)=\left\{\begin{array}{r}{[x], x \leq 2} \\ 0, x>2\end{array}\right.$, where $[x]$ is the greatest integer less than or equal to $x$. If $I=\int\limits_{-1}^2 \frac{x f\left(x^2\right)}{2+f(x+1)} d x$, then the value, is

$\frac{1}{4}$

We have, $f(x)=\left\{\begin{array}{r}{[x], x \leq 2} \\ 0, x>2\end{array}\right.$

∴  $I=\int\limits_{-1}^2 \frac{x f\left(x^2\right)}{2+f(x+1)} d x$

$\Rightarrow I=\int\limits_{-1}^0 \frac{x \times 0}{2+0} d x+\int\limits_0^1 \frac{x \times 0}{2+1} d x+\int\limits_1^{\sqrt{2}} \frac{x \times 1}{2+0} d x+\int\limits_{\sqrt{2}}^2 \frac{x \times 0}{2+f(x+1)} d x$

$\Rightarrow I=\frac{1}{2}\left[\frac{x^2}{2}\right]_1^{\sqrt{2}}=\frac{1}{2}\left(1-\frac{1}{2}\right)=\frac{1}{4}$