Evaluate Definite Integral Involving Greatest Integer Function

The given function is:

$$f(x) = \begin{cases} [x], & x \le 2 \\ 0, & x > 2 \end{cases}$$

where $[x]$ denotes the greatest integer less than or equal to $x$.

We need to evaluate the definite integral: $I = \int_{-1}^{2} \frac{x f(x^2)}{2 + f(x+1)} \, dx$

Step 1: Analyze $f(x^2)$ on the interval $[-1, 2]$

The definition of $f$ implies that:

$$f(x^2) = \begin{cases} [x^2], & x^2 \le 2 \\ 0, & x^2 > 2 \end{cases}$$

Since $x \in [-1, 2]$:

• For $x \in [-1, \sqrt{2}]$, we have $x^2 \le 2$, so $f(x^2) = [x^2]$.
• For $x \in (\sqrt{2}, 2]$, we have $x^2 > 2$, so $f(x^2) = 0$.

Consequently, on the interval $(\sqrt{2}, 2]$, the integrand is identically $0$, meaning: $\int_{\sqrt{2}}^{2} \frac{x f(x^2)}{2 + f(x+1)} \, dx = 0$

Hence, the integral reduces to: $I = \int_{-1}^{\sqrt{2}} \frac{x f(x^2)}{2 + f(x+1)} \, dx$

Step 2: Split the integral into intervals based on the behavior of $f(x^2)$ and $f(x+1)$

We partition the interval $[-1, \sqrt{2}]$ at the critical points $x = 0$ and $x = 1$:

1. For $x \in (-1, 0)$:

   • $x^2 \in (0, 1) \implies [x^2] = 0 \implies f(x^2) = 0$
   • Thus, the integrand is $0$: $\int_{-1}^{0} \frac{x f(x^2)}{2 + f(x+1)} \, dx = 0$

2. For $x \in (0, 1)$:

   • $x^2 \in (0, 1) \implies [x^2] = 0 \implies f(x^2) = 0$
   • Thus, the integrand is $0$: $\int_{0}^{1} \frac{x f(x^2)}{2 + f(x+1)} \, dx = 0$

3. For $x \in (1, \sqrt{2})$:

   • $x^2 \in (1, 2) \implies [x^2] = 1 \implies f(x^2) = 1$
   • $x+1 \in (2, 1+\sqrt{2}) \implies x+1 > 2 \implies f(x+1) = 0$
   • Therefore, the integrand becomes: $\frac{x f(x^2)}{2 + f(x+1)} = \frac{x \cdot 1}{2 + 0} = \frac{x}{2}$
   • Evaluating this part of the integral: $\int_{1}^{\sqrt{2}} \frac{x}{2} \, dx = \left[ \frac{x^2}{4} \right]_{1}^{\sqrt{2}} = \frac{(\sqrt{2})^2 - 1^2}{4} = \frac{2 - 1}{4} = \frac{1}{4}$

Step 3: Compute the total integral $I$ and $(4I - 1)$

Adding the values across all intervals: $I = 0 + 0 + \frac{1}{4} + 0 = \frac{1}{4}$

Thus, the required value is: $4I - 1 = 4\left(\frac{1}{4}\right) - 1 = 1 - 1 = 0$