1. **State the problem:** Find the total area of the shaded regions between the curves $y=2-x$ and $y=4-x^2$ over the interval where they intersect. 2. **Find the points of intersection:** Set the two functions equal: $$2 - x = 4 - x^2$$ Rearrange: $$x^2 - x - 2 = 0$$ Factor: $$ (x-2)(x+1) = 0 $$ So, the points of intersection are at $x=-1$ and $x=2$. 3. **Determine which function is on top between the intersection points:** Test a point between $-1$ and $2$, for example $x=0$: $$y_1 = 2 - 0 = 2$$ $$y_2 = 4 - 0^2 = 4$$ Since $4 > 2$, $y=4-x^2$ is above $y=2-x$ on $[-1,2]$. 4. **Set up the integral for the area:** The area $A$ is given by $$A = \int_{-1}^{2} \big[(4 - x^2) - (2 - x)\big] \, dx = \int_{-1}^{2} (4 - x^2 - 2 + x) \, dx = \int_{-1}^{2} (2 + x - x^2) \, dx$$ 5. **Integrate:** $$\int (2 + x - x^2) \, dx = 2x + \frac{x^2}{2} - \frac{x^3}{3} + C$$ 6. **Evaluate the definite integral:** $$A = \left[2x + \frac{x^2}{2} - \frac{x^3}{3}\right]_{-1}^{2}$$ Calculate at $x=2$: $$2(2) + \frac{2^2}{2} - \frac{2^3}{3} = 4 + 2 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{18}{3} - \frac{8}{3} = \frac{10}{3}$$ Calculate at $x=-1$: $$2(-1) + \frac{(-1)^2}{2} - \frac{(-1)^3}{3} = -2 + \frac{1}{2} + \frac{1}{3} = -2 + \frac{3}{6} + \frac{2}{6} = -2 + \frac{5}{6} = -\frac{12}{6} + \frac{5}{6} = -\frac{7}{6}$$ 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 total area of the shaded regions is $\boxed{\frac{9}{2}}$ or 4.5.