1. **State the problem:** Find the area of the region defined by the inequalities $$0 \leq y \leq x^2 + 1$$, $$0 \leq y \leq x + 1$$, and $$0 \leq x \leq 2$$. 2. **Understand the region:** The region is bounded below by $$y=0$$, above by the curves $$y = x^2 + 1$$ and $$y = x + 1$$, and horizontally between $$x=0$$ and $$x=2$$. 3. **Find the intersection points of the curves $$y = x^2 + 1$$ and $$y = x + 1$$:** Set $$x^2 + 1 = x + 1$$ Simplify: $$x^2 = x$$ $$x^2 - x = 0$$ $$x(x - 1) = 0$$ So, $$x=0$$ or $$x=1$$. 4. **Determine which curve is on top in each interval:** - For $$0 \leq x \leq 1$$, compare $$x^2 + 1$$ and $$x + 1$$: - At $$x=0.5$$, $$0.5^2 + 1 = 1.25$$ and $$0.5 + 1 = 1.5$$, so $$y = x + 1$$ is above. - For $$1 \leq x \leq 2$$, at $$x=1.5$$, $$1.5^2 + 1 = 3.25$$ and $$1.5 + 1 = 2.5$$, so $$y = x^2 + 1$$ is above. 5. **Set up the integral for the area:** Area $$= \int_0^1 [(x + 1) - 0] \, dx + \int_1^2 [(x^2 + 1) - 0] \, dx$$ 6. **Calculate the integrals:** $$\int_0^1 (x + 1) \, dx = \left[ \frac{x^2}{2} + x \right]_0^1 = \frac{1}{2} + 1 = \frac{3}{2}$$ $$\int_1^2 (x^2 + 1) \, dx = \left[ \frac{x^3}{3} + x \right]_1^2 = \left( \frac{8}{3} + 2 \right) - \left( \frac{1}{3} + 1 \right) = \left( \frac{8}{3} + 2 \right) - \left( \frac{1}{3} + 1 \right) = \frac{7}{3} + 1 = \frac{10}{3}$$ 7. **Sum the areas:** $$\frac{3}{2} + \frac{10}{3} = \frac{9}{6} + \frac{20}{6} = \frac{29}{6}$$ **Final answer:** The area of the region is $$\boxed{\frac{29}{6}}$$.