1. **Problem statement:** Find the area of the region bounded by the graphs of the functions $f(x) = 2x - x^2$ and $g(x) = x - 2$. 2. **Find the points of intersection:** Set $f(x) = g(x)$ to find the limits of integration: $$2x - x^2 = x - 2$$ Rearranged: $$-x^2 + 2x - x + 2 = 0$$ $$-x^2 + x + 2 = 0$$ Multiply both sides by $-1$: $$x^2 - x - 2 = 0$$ Factor: $$(x - 2)(x + 1) = 0$$ So, $x = 2$ or $x = -1$. 3. **Set up the integral for the area:** The area $A$ between two curves $f(x)$ and $g(x)$ from $a$ to $b$ is: $$A = \int_a^b |f(x) - g(x)| \, dx$$ Check which function is on top between $-1$ and $2$: At $x=0$, $f(0) = 0$, $g(0) = -2$, so $f(x) > g(x)$. 4. **Calculate the integral:** $$A = \int_{-1}^2 (f(x) - g(x)) \, dx = \int_{-1}^2 \big( (2x - x^2) - (x - 2) \big) \, dx$$ Simplify the integrand: $$2x - x^2 - x + 2 = -x^2 + x + 2$$ 5. **Integrate:** $$\int (-x^2 + x + 2) \, dx = -\frac{x^3}{3} + \frac{x^2}{2} + 2x + C$$ 6. **Evaluate definite integral:** $$A = \left[-\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{-1}^2$$ Calculate at $x=2$: $$-\frac{8}{3} + 2 + 4 = -\frac{8}{3} + 6 = \frac{10}{3}$$ Calculate at $x=-1$: $$-\frac{-1}{3} + \frac{1}{2} - 2 = \frac{1}{3} + \frac{1}{2} - 2 = \frac{2}{6} + \frac{3}{6} - \frac{12}{6} = -\frac{7}{6}$$ 7. **Subtract:** $$A = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{10}{3} + \frac{7}{6} = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} = 4.5$$ **Final answer:** The area bounded by the graphs of $f$ and $g$ is $\boxed{4.5}$.