The region R enclosed by the lines $x = 1$, $x = 6$, $y = 0$ and the curve $y = ext{ln}(8 - x)$ is shown shaded in Figure 3 below - AQA - A-Level Maths Pure - Question 11 - 2020 - Paper 1

To find the mass of Shape B, we first calculate the area using six ordinates from the function $y = ext{ln}(8 - x)$. We will evaluate at equal intervals from $x = 1$ to $x = 6$:

1. The six ordinates are:

   • $f(1) = ext{ln}(7) \approx 1.94591$
   • $f(2) = ext{ln}(6) \approx 1.79176$
   • $f(3) = ext{ln}(5) \approx 1.60944$
   • $f(4) = ext{ln}(4) \approx 1.38629$
   • $f(5) = ext{ln}(3) \approx 1.09861$
   • $f(6) = ext{ln}(2) \approx 0.69315$

2. Using the trapezium rule: $ext{Area} = \frac{h}{2} \times (f_0 + 2f_1 + 2f_2 + 2f_3 + 2f_4 + f_5)$ where $h = 5/6 = \frac{5}{6}$.

3. Substituting: $ext{Area} \approx \frac{5}{6} \times \left(1.94591 + 2(1.79176 + 1.60944 + 1.38629 + 1.09861) + 0.69315\right) \approx 7.205633$

4. Shape B consists of four copies of region R, so the total area of Shape B is: $\text{Area of Shape B} = 4 \times 7.205633 = 28.822532$

5. The volume of Shape B, considering the thickness of 2 mm (or 0.2 cm), is: $\text{Volume} = \text{Area} \times \text{Thickness} = 28.822532 \times 0.2 \approx 5.76453 \text{ cm}^3$

6. Finally, to find the mass: $\text{Mass} = \text{Volume} \times \text{Density} = 5.76453 \times 10.5 \approx 60.531565\text{ g}$ Rounding gives us approximately 61 g.

The mass found in part (b) may be an underestimate because the trapezia are all below the curve, meaning that the actual area could be larger than the area calculated using the trapezium rule.

The mass found in part (b) may be an overestimate because the trapezia might be overestimating the area by encompassing sections that are above the actual curve, leading to a higher calculated mass than what exists.