June 2022 Paper 2 Q14
14 Fig. 14.1 shows the curve with equation \(y = \dfrac{1}{1+x^2}\), together with 5 rectangles of equal width.

Fig. 14.2 shows the coordinates of the points A, B, C, D, E and F.
| Point | A | B | C | D | E | F |
|---|---|---|---|---|---|---|
| \(x\) | 0 | 0.2 | 0.4 | 0.6 | 0.8 | 1 |
| \(y\) | 1 | 0.96154 | 0.86207 | 0.73529 | 0.60976 | 0.5 |
Fig. 14.2
Amit uses \(n\) rectangles, each of width \(\dfrac{1}{n}\), to calculate upper and lower bounds for \(\displaystyle\int_0^1 \frac{1}{1+x^2}\,\mathrm{d}x\), using different values of \(n\). His results are shown in Fig. 14.3.
| \(n\) | 10 | 20 | 40 |
|---|---|---|---|
| upper bound | 0.80998 | 0.79779 | 0.79162 |
| lower bound | 0.75998 | 0.77279 | 0.77912 |
Fig. 14.3
| Scheme | Marks | AO |
|---|---|---|
| \(0.2 \times \{0.96154 + 0.86207 + 0.73529 + 0.60976 + 0.5\}\) soi | M1 | 2.1 |
| \(0.73373\ldots \approx 0.7337\) AG | A1 | 2.4 |
| [2] |
Notes
M1: allow M1A1 for calculation of exact values using formula in parts (a) and (b)
A1: need to see 0.73373… as well as 0.7337 for A1
| Scheme | Marks | AO |
|---|---|---|
| \(0.2 \times \{1 + 0.96154 + 0.86207 + 0.73529 + 0.60976\}\) | M1 | 1.1 |
| 0.8337 correct to 4 dp | A1 | 1.1 |
| [2] |
Notes
M1: or \((3.66866 - 0.5 + 1) \times 0.2\)
| Scheme | Marks | AO |
|---|---|---|
| 0.1 | B1 | 1.1 |
| [1] |
Notes
B1: FT their 0.8337(32) – 0.7337(32), dependent on award of M1 in part (b)
| Scheme | Marks | AO |
|---|---|---|
| \(0.79162 - 0.77912\) | M1 | 3.1a |
| 0.0125 | A1 | 2.4 |
| [2] |
Notes
M1: if M0 allow SC1 for correct interval identified eg 0.77912 to 0.79162
| Scheme | Marks | AO |
|---|---|---|
| increase \(n\) oe use rectangles of smaller width oe | B1 | 2.2a |
| [1] |
Notes
B1: do not allow eg reduce interval
eg just ‘smaller’ rectangles – need to specify width reduction