The function is given by $y = x|x|$. Based on the definition of the absolute value function, we can split this into two cases: $$y = \begin{cases} x^2, & \text{if } x \ge 0 \\ -x^2, & \text{if } x < 0 \end{cases}$$
The area $A$ bounded by the curve, the x-axis, and the ordinates $x = -1$ and $x = 1$ is given by the definite integral: $$A = \int_{-1}^{1} |x|x| | dx$$ Since we are looking for the area, we integrate the absolute value of the function: $$A = \int_{-1}^{0} | -x^2 | dx + \int_{0}^{1} | x^2 | dx = \int_{-1}^{0} x^2 dx + \int_{0}^{1} x^2 dx$$
Calculate the two integrals: $$\int_{-1}^{0} x^2 dx = \left[ \frac{x^3}{3} \right]_{-1}^{0} = 0 - \left( -\frac{1}{3} \right) = \frac{1}{3}$$ $$\int_{0}^{1} x^2 dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1}{3} - 0 = \frac{1}{3}$$
Total Area $A = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$ square units.
Final Answer: 2/3
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