1. **State the problem:** Find the average vertical height of the shaded area between the curves $y = x^2$ and $y = 6 - x$ from $x=0$ to their intersection point. 2. **Find the intersection point:** Set $x^2 = 6 - x$ to find where the curves meet. $$x^2 + x - 6 = 0$$ Factor the quadratic: $$(x + 3)(x - 2) = 0$$ So, $x = -3$ or $x = 2$. Since the shaded area is from $x=0$, the intersection relevant here is at $x=2$. 3. **Set up the integral for the area between curves:** The vertical height between the curves at any $x$ is: $$h(x) = (6 - x) - x^2$$ 4. **Calculate the area $A$ of the shaded region:** $$A = \int_0^2 [(6 - x) - x^2] \, dx = \int_0^2 (6 - x - x^2) \, dx$$ Calculate the integral: $$\int_0^2 6 \, dx = 6x \Big|_0^2 = 12$$ $$\int_0^2 x \, dx = \frac{x^2}{2} \Big|_0^2 = 2$$ $$\int_0^2 x^2 \, dx = \frac{x^3}{3} \Big|_0^2 = \frac{8}{3}$$ So, $$A = 12 - 2 - \frac{8}{3} = 10 - \frac{8}{3} = \frac{30}{3} - \frac{8}{3} = \frac{22}{3}$$ 5. **Calculate the average vertical height:** The average height $\bar{h}$ over the interval $[0,2]$ is the area divided by the length of the interval: $$\bar{h} = \frac{A}{2 - 0} = \frac{\frac{22}{3}}{2} = \frac{22}{6} = \frac{11}{3} \approx 3.67$$ **Final answer:** The average vertical height of the shaded area is $\boxed{\frac{11}{3}}$ or approximately 3.67.