1. **State the problem:** Calculate the area between the curves $y = 3 - x$ and $y = x^2 + 1$ over the interval $[0, 2]$ by integrating the absolute difference of the functions. 2. **Formula and approach:** The area between two curves $y=f(x)$ and $y=g(x)$ over $[a,b]$ is given by $$\int_a^b |f(x) - g(x)| \, dx.$$ Since the curves intersect at $x=1$, split the integral at this point: $$\int_0^2 |(3 - x) - (x^2 + 1)| \, dx = \int_0^1 [(3 - x) - (x^2 + 1)] \, dx + \int_1^2 [(x^2 + 1) - (3 - x)] \, dx.$$ 3. **Simplify the integrands:** - For $0 \leq x \leq 1$: $$(3 - x) - (x^2 + 1) = 2 - x - x^2.$$ - For $1 \leq x \leq 2$: $$(x^2 + 1) - (3 - x) = x^2 + x - 2.$$ 4. **Evaluate the first integral:** $$\int_0^1 (2 - x - x^2) \, dx = \left[2x - \frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = 2(1) - \frac{1}{2} - \frac{1}{3} - 0 = 2 - 0.5 - 0.3333 = \frac{7}{6}.$$ 5. **Evaluate the second integral:** $$\int_1^2 (x^2 + x - 2) \, dx = \left[\frac{x^3}{3} + \frac{x^2}{2} - 2x\right]_1^2 = \left(\frac{8}{3} + 2 - 4\right) - \left(\frac{1}{3} + \frac{1}{2} - 2\right) = \left(\frac{8}{3} - 2\right) - \left(\frac{1}{3} - 1.5\right) = \frac{11}{6}.$$ 6. **Add the two areas:** $$\text{Area} = \frac{7}{6} + \frac{11}{6} = \frac{18}{6} = 3.$$ **Final answer:** The area between the curves over $[0,2]$ is $3$.