1. **Problem statement:** Find the area of the region $R$ bounded by $y = x - 1$ and $y = -(x-1)^2 + 2$. 2. **Formula and explanation:** The area between two curves $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$ is given by: $$\text{Area} = \int_a^b [f(x) - g(x)] \, dx$$ where $f(x)$ is the upper curve and $g(x)$ is the lower curve. 3. **Identify curves and limits:** Upper curve: $y = -(x-1)^2 + 2$ Lower curve: $y = x - 1$ Limits: $x = -1$ to $x = 2$ 4. **Set up the integral for area:** $$\int_{-1}^2 \left[-(x-1)^2 + 2 - (x-1)\right] dx$$ 5. **Simplify the integrand:** $$-(x-1)^2 + 2 - (x-1) = - (x^2 - 2x + 1) + 2 - x + 1 = -x^2 + 2x -1 + 2 - x + 1 = -x^2 + x + 2$$ 6. **Evaluate the integral:** $$\int_{-1}^2 (-x^2 + x + 2) dx = \left[-\frac{x^3}{3} + \frac{x^2}{2} + 2x\right]_{-1}^2$$ 7. **Calculate at bounds:** At $x=2$: $$-\frac{8}{3} + 2 + 4 = -\frac{8}{3} + 6 = \frac{10}{3}$$ At $x=-1$: $$-\frac{-1}{3} + \frac{1}{2} - 2 = \frac{1}{3} + \frac{1}{2} - 2 = \frac{5}{6} - 2 = -\frac{7}{6}$$ 8. **Subtract:** $$\frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{10}{3} + \frac{7}{6} = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = 4.5$$ **Final answer for part (a):** $$\boxed{4.5}$$ --- **Slug:** "area bounded" **Subject:** "calculus" **Desmos:** {"latex":"y=-(x-1)^2+2","features":{"intercepts":true,"extrema":true}} **q_count:** 4 (Note: Only part (a) is fully solved as per instructions.)