Competency-Based
Assessment Made Simple.

**1. Understand the function:** The function is \(y = x|x|\). We can rewrite this as a piecewise function: \[ y = \begin{cases} x^2 & \text{if } x \geq 0 \\ -x^2 & \text{if } x < 0 \end{cases} \]

**2. Set up the integral:** We need to find the area enclosed between the curve, the x-axis, and the lines \(x = -2\) and \(x = 2\). Since the function is negative for \(x < 0\), we need to take the absolute value of the integral over that interval. The total area is given by: \[ \text{Area} = \left| \int_{-2}^{0} -x^2 \, dx \right| + \int_{0}^{2} x^2 \, dx \]

**3. Evaluate the integrals:** \[ \int_{-2}^{0} -x^2 \, dx = \left[ -\frac{x^3}{3} \right]_{-2}^{0} = -\frac{0^3}{3} - \left( -\frac{(-2)^3}{3} \right) = 0 - \frac{8}{3} = -\frac{8}{3} \] \[ \int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} - 0 = \frac{8}{3} \]

**4. Calculate the total area:** \[ \text{Area} = \left| -\frac{8}{3} \right| + \frac{8}{3} = \frac{8}{3} + \frac{8}{3} = \frac{16}{3} \]