Question:

Step 1: Determine the function $y = \max\{x^2, x\}$ over the given interval $x \in [-1, 2]$. The relationship between $x^2$ and $x$ changes at $x=0$ and $x=1$. For $x \in [-1, 0]$: $x^2 \ge 0$ and $x \le 0$. Thus, $x^2 \ge x$. So, $\max\{x^2, x\} = x^2$. For $x \in [0, 1]$: $x^2 \le x$ (since $x^2 - x = x(x-1) \le 0$). So, $\max\{x^2, x\} = x$. For $x \in [1, 2]$: $x^2 \ge x$ (since $x^2 - x = x(x-1) \ge 0$). So, $\max\{x^2, x\} = x^2$. Step 2: Calculate the area by integrating the function over the respective subintervals. The area $A$ is given by: $$A = \int_{-1}^0 x^2 \,dx + \int_0^1 x \,dx + \int_1^2 x^2 \,dx$$ Evaluate each integral: $$\int_{-1}^0 x^2 \,dx = \left[\frac{x^3}{3}\right]_{-1}^0 = \frac{0^3}{3} - \frac{(-1)^3}{3} = 0 - \left(-\frac{1}{3}\right) = \frac{1}{3}$$ $$\int_0^1 x \,dx = \left[\frac{x^2}{2}\right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}$$ $$\int_1^2 x^2 \,dx = \left[\frac{x^3}{3}\right]_1^2 = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$$ Sum the areas: $$A = \frac{1}{3} + \frac{1}{2} + \frac{7}{3} = \frac{1+7}{3} + \frac{1}{2} = \frac{8}{3} + \frac{1}{2}$$ To combine these fractions, find a common denominator: $$A = \frac{8 \cdot 2}{3 \cdot 2} + \frac{1 \cdot 3}{2 \cdot 3} = \frac{16}{6} + \frac{3}{6} = \frac{19}{6}$$