October 2020 Paper 1 Q8
8 Fig. 8.1 shows the cross-section of a straight driveway 4 m wide made from tarmac.

The height \(h\) m of the cross-section at a displacement \(x\) m from the middle is modelled by \(h = \dfrac{0.2}{1+x^2}\) for \(-2 \leqslant x \leqslant 2\).
A lower bound of \(0.3615\,\mathrm{m^2}\) is found for the area of the cross-section using rectangles as shown in Fig. 8.2.

| Scheme | Marks | AO | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
![]()
| M1 | 3.4 | ||||||||||||||||||
| \(= 2 \times 0.5(0.2 + 0.16 + 0.1 + 0.061538)\) | M1 | 1.1a | ||||||||||||||||||
| \(= 0.522\) to 3sf | A1 | 1.1 | ||||||||||||||||||
| [3] |
Notes
M1: Finding at least 2 distinct values for \(h\)
Need not be a table of values
M1: Using rectangles forming UB for area with width 0.5. Need not be drawn
Allow both M marks if only half the area considered.
| Scheme | Marks | AO |
|---|---|---|
| Area \(\approx \dfrac{0.5}{2}\big(0.2 + 0.04 + 2(0.16 + 0.1 + 0.061538)\big)\) | M1 | 1.1a |
| \(= 0.221\) (to 3 sf) | A1 | 1.1 |
| [2] |
Notes
M1: Using trapezium rule with 5 \(h\) values
FT incorrect \(h\) values for the M mark
A1: Allow awrt 0.22
Using the definite integral functionality of calculator gives 0.2214297… So method must be seen.
| Scheme | Marks | AO |
|---|---|---|
| Volume \(= 2 \times (0.2208 \times 10)\) | M1 | 1.1a |
| \(= 4.42\,\mathrm{m^3}\) | A1 | 1.1 |
| [2] |
Notes
M1: Condone missing factor of 2 for method mark
A1: FT part (b)
