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The function is given by $y = x|x|$. We can rewrite this function as a piecewise function:
$$y = \begin{cases} x^2, & \text{if } x \geq 0 \\ -x^2, & \text{if } x < 0 \end{cases}$$For $x \geq 0$, the graph is the parabola $y = x^2$. For $x < 0$, the graph is the parabola $y = -x^2$. The graph is symmetric about the origin.
We want to find the area bounded by the curve $y = x|x|$, the x-axis, and the ordinates $x = -2$ and $x = 2$. Since the function is negative for $x < 0$, we need to take the absolute value of the integral over the interval $[-2, 0]$. The area is given by: $$A = \int_{-2}^{2} |x|x| dx = \int_{-2}^{0} |-x^2| dx + \int_{0}^{2} |x^2| dx$$ $$A = \int_{-2}^{0} -x^2 dx + \int_{0}^{2} x^2 dx$$
Now, we evaluate the integrals: $$\int_{-2}^{0} -x^2 dx = \left[ -\frac{x^3}{3} \right]_{-2}^{0} = -\frac{0^3}{3} - \left( -\frac{(-2)^3}{3} \right) = 0 - \frac{8}{3} = -\frac{8}{3}$$ Since we are looking for area, we take the absolute value: $$|-\frac{8}{3}| = \frac{8}{3}$$ $$\int_{0}^{2} x^2 dx = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} - 0 = \frac{8}{3}$$
The total area is the sum of the absolute values of the integrals: $$A = \frac{8}{3} + \frac{8}{3} = \frac{16}{3}$$
Final Answer: 16/3
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