1. **State the problem:** Find the area bounded by the curve $y = x^2 - 2x$, the $x$-axis, and the vertical lines (ordinates) $x = -2$ and $x = 3$. 2. **Formula and rules:** The area bounded by a curve $y = f(x)$ and the $x$-axis between $x = a$ and $x = b$ is given by the definite integral $$\text{Area} = \int_a^b |f(x)| \, dx.$$ Since the curve may lie above or below the $x$-axis, we must consider the absolute value to ensure the area is positive. 3. **Find where the curve intersects the $x$-axis:** Solve $x^2 - 2x = 0$. $$x(x - 2) = 0 \implies x = 0 \text{ or } x = 2.$$ These points split the interval $[-2,3]$ into three parts: $[-2,0]$, $[0,2]$, and $[2,3]$. 4. **Determine the sign of $y$ on each interval:** - For $x$ in $[-2,0]$, pick $x = -1$: $y = (-1)^2 - 2(-1) = 1 + 2 = 3 > 0$. - For $x$ in $[0,2]$, pick $x = 1$: $y = 1 - 2 = -1 < 0$. - For $x$ in $[2,3]$, pick $x = 3$: $y = 9 - 6 = 3 > 0$. 5. **Set up the integral with absolute values:** $$\text{Area} = \int_{-2}^0 (x^2 - 2x) \, dx - \int_0^2 (x^2 - 2x) \, dx + \int_2^3 (x^2 - 2x) \, dx.$$ Note the minus sign on the middle integral because the function is negative there. 6. **Calculate each integral:** $$\int (x^2 - 2x) \, dx = \frac{x^3}{3} - x^2 + C.$$ 7. **Evaluate each definite integral:** - From $-2$ to $0$: $$\left[ \frac{x^3}{3} - x^2 \right]_{-2}^0 = \left(0 - 0\right) - \left(\frac{(-2)^3}{3} - (-2)^2\right) = 0 - \left(-\frac{8}{3} - 4\right) = \frac{8}{3} + 4 = \frac{8}{3} + \frac{12}{3} = \frac{20}{3}.$$ - From $0$ to $2$: $$\left[ \frac{x^3}{3} - x^2 \right]_0^2 = \left(\frac{8}{3} - 4\right) - (0 - 0) = \frac{8}{3} - 4 = \frac{8}{3} - \frac{12}{3} = -\frac{4}{3}.$$ - From $2$ to $3$: $$\left[ \frac{x^3}{3} - x^2 \right]_2^3 = \left(\frac{27}{3} - 9\right) - \left(\frac{8}{3} - 4\right) = (9 - 9) - \left(\frac{8}{3} - 4\right) = 0 - \left(\frac{8}{3} - 4\right) = -\frac{8}{3} + 4 = \frac{4}{3}.$$ 8. **Sum the absolute areas:** $$\text{Area} = \frac{20}{3} - \left(-\frac{4}{3}\right) + \frac{4}{3} = \frac{20}{3} + \frac{4}{3} + \frac{4}{3} = \frac{28}{3}.$$ **Final answer:** The area bounded by the curve, the $x$-axis, and the lines $x = -2$ and $x = 3$ is $$\boxed{\frac{28}{3}}.$$