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(A) Given $f(x) = \begin{cases} [x], & x \leq 2 \ 0, & x > 2 \end{cases}$.We need to evaluate $I = \int_{-1}^2 \frac{x f(x^2)}{2+f(x+1)} dx$.For $x \

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(A) Given $f(x) = \begin{cases} [x], & x \leq 2 \ 0, & x > 2 \end{cases}$.We need to evaluate $I = \int_{-1}^2 \frac{x f(x^2)}{2+f(x+1)} dx$.For $x \in [-1, 2]$,$x^2 \in [0, 4]$.If $x^2 \leq 2$,$f(x^2) = [x^2]$. If $x^2 > 2$,$f(x^2) = 0$.$x^2 \leq 2 \implies x \in [-\sqrt{2}, \sqrt{2}]$.Since the integral is from $-1$ to $2$,we split it:For $x \in [-1, \sqrt{2}]$,$f(x^2) = [x^2]$. For $x \in (\sqrt{2}, 2]$,$f(x^2) = 0$.Also,$f(x+1) = [x+1]$ for $x+1 \leq 2 \implies x \leq 1$,and $f(x+1) = 0$ for $x > 1$.Thus,$I = \int_{-1}^1 \frac{x [x^2]}{2+[x+1]} dx + \int_1^{\sqrt{2}} \frac{x [x^2]}{2+0} dx + \int_{\sqrt{2}}^2 \frac{x \cdot 0}{2+0} dx$.For $x \in [-1, 0)$,$[x^2] = 0$,so the first part is $\int_{-1}^0 0 dx = 0$.For $x \in [0, 1)$,$[x^2] = 0$,so $\int_0^1 0 dx = 0$.For $x \in [1, \sqrt{2}]$,$[x^2] = 1$,so $\int_1^{\sqrt{2}} \frac{x \cdot 1}{2} dx = \frac{1}{2} [\frac{x^2}{2}]_1^{\sqrt{2}} = \frac{1}{4} (2-1) = \frac{1}{4}$.Thus $I = \frac{1}{4}$.Then $4I - 1 = 4(\frac{1}{4}) - 1 = 0$.

Let $f: R \rightarrow R$ be a function defined by $f(x)=\begin{cases} [x], & x \leq 2 \\ 0, & x>2 \end{cases}$,where $[x]$ is the greatest integer less than or equal to $x$. If $I=\int_{-1}^2 \frac{x f(x^2)}{2+f(x+1)} dx$,then the value of $(4I-1)$ is

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