1. **State the problem:** We have the function $$y = -(x-3)(x+1)$$ and a table with values of $$x$$ and corresponding $$y$$ values. We need to fill in the missing values (Box 1, Box 2, Box 3) for $$x=0,1,2$$. 2. **Calculate the missing values:** - For $$x=0$$: $$y = -(0-3)(0+1) = -(-3)(1) = 3$$ (Box 1 = 3) - For $$x=1$$: $$y = -(1-3)(1+1) = -(-2)(2) = 4$$ (Box 2 = 4) - For $$x=2$$: $$y = -(2-3)(2+1) = -(-1)(3) = 3$$ (Box 3 = 3) 3. **Use the trapezium rule to approximate the area of region R between $$x=0$$ and $$x=3$$:** The trapezium rule formula for interval $$[a,b]$$ with equal subintervals of width $$h$$ is: $$\text{Area} \approx \frac{h}{2} \left(y_0 + 2y_1 + 2y_2 + \cdots + 2y_{n-1} + y_n\right)$$ Here, $$a=0$$, $$b=3$$, and the points are at $$x=0,1,2,3$$ with $$y$$ values $$3,4,3,0$$ respectively. - Number of subintervals $$n=3$$ - Width $$h = \frac{3-0}{3} = 1$$ Calculate: $$\text{Area} \approx \frac{1}{2} (3 + 2\times4 + 2\times3 + 0) = \frac{1}{2} (3 + 8 + 6 + 0) = \frac{1}{2} (17) = 8.5$$ 4. **Calculate the exact area by integration:** The area $$A$$ is the integral of $$y$$ from $$x=0$$ to $$x=3$$: $$A = \int_0^3 -(x-3)(x+1) \, dx$$ Expand the integrand: $$-(x-3)(x+1) = -(x^2 + x - 3x - 3) = -(x^2 - 2x - 3) = -x^2 + 2x + 3$$ Integrate term-by-term: $$\int_0^3 (-x^2 + 2x + 3) \, dx = \left[-\frac{x^3}{3} + x^2 + 3x \right]_0^3$$ Evaluate at $$x=3$$: $$-\frac{27}{3} + 9 + 9 = -9 + 9 + 9 = 9$$ Evaluate at $$x=0$$: $$0$$ So, $$A = 9 - 0 = 9$$ **Final answers:** - Box 1 = 3 - Box 2 = 4 - Box 3 = 3 - Approximate area using trapezium rule = 8.5 - Exact area by integration = 9