Let $f: R \to R$ be a function defined by $f(x) = \begin{cases} [x], & x \le 2 \\ 0, & x > 2 \end{cases}$ 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

$\text{Given: } f(x) = \begin{cases} [x], & x \le 2 \\ 0, & x > 2 \end{cases},\quad [x] = \lfloor x \rfloor$ $I = \displaystyle\int_{-1}^{2} \dfrac{x\, f(x^2)}{2 + f(x+1)}\,dx$ $\text{To find: } 4I - 1$ $1.\ \text{split } [-1,2] \text{ at the jump points}$ $2.\ \text{simplify the integrand on each piece}$ $3.\ \text{integrate, then } 4I - 1$ $\text{integrand } = \dfrac{x\, f(x^2)}{2 + f(x+1)}$ $x \in [-1,2]:\quad x^2 \in [0,4],\quad x+1 \in [0,3]$ $f(x^2) = \lfloor x^2 \rfloor\ \text{for } x^2 \le 2;\qquad f(x+1) = \lfloor x+1 \rfloor\ \text{for } x \le 1$ $\text{breakpoints}: -1,\ 0,\ 1,\ \sqrt{2},\ 2$ $\text{intervals}: [-1,0],\ [0,1],\ [1,\sqrt2],\ [\sqrt2,2]$ $[-1,0],\ [0,1]:\ x^2 < 1 \Rightarrow f(x^2) = 0 \Rightarrow \text{integrand} = 0$ $[1,\sqrt2]:\ f(x^2) = 1,\quad f(x+1) = 0 \Rightarrow \text{integrand} = \dfrac{x}{2}$ $[\sqrt2,2]:\ x^2 > 2 \Rightarrow f(x^2) = 0 \Rightarrow \text{integrand} = 0$ $I = \displaystyle\int_1^{\sqrt2} \dfrac{x}{2}\,dx$ $I = \left[\dfrac{x^2}{4}\right]_1^{\sqrt2} = \dfrac{2}{4} - \dfrac{1}{4} = \dfrac{1}{4}$ $4I - 1 = 4 \times \dfrac{1}{4} - 1 = 1 - 1 = 0$ $\boxed{0}$