A2 June 2019 Q1
1. Use Simpson’s rule with 4 intervals to estimate
\[\int_{0.4}^{2}\mathrm{e}^{x^2}\,\mathrm{d}x\](5)
| Scheme | Marks | AO | ||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Step length \(= 0.4\) | B1 | 1.1b | ||||||||||||||||||||||||
| M1 | 1.1b | ||||||||||||||||||||||||
| \(y_0 + 4y_1 + 2y_2 + 4y_3 + y_4 = 123.54\ldots\) | M1 | 1.1b | ||||||||||||||||||||||||
| \(\begin{aligned}&\int_{0.4}^{2}\mathrm{e}^{x^2}\,\mathrm{d}x \approx \frac{0.4}{3} \times \left\{1.173\ldots + 54.598\ldots + 4\left(1.896\ldots + 12.935\ldots\right) + 2\left(4.220\ldots\right)\right\}\\[4pt] &\approx \dfrac{0.4}{3} \times \text{“}123.54\ldots\text{”}\end{aligned}\) | dM1 | 1.1b | ||||||||||||||||||||||||
| \(= 16.5\) | A1 | 1.1b | ||||||||||||||||||||||||
| (5) | ||||||||||||||||||||||||||
| (5 marks) |
Notes
B1: Correct step length of 0.4 which may be implied e.g. by their 0.4, 0.8, etc.
M1: Attempts to find \(y\) values for their \(x\) values – may be in terms of e or numerical values. Must see an attempt to find at least 3 values.
M1: Correct structure for \(y\) values of Simpson’s rule (ends + 2evens + 4odds) (must have an odd number of ordinates). Must be \(y\) values not \(x\) values.
dM1: \(\dfrac{\text{“}0.4\text{”}}{3} \times \text{their } 123.54\ldots\) or for \(\dfrac{h}{3} \times \text{their } 123.54\ldots\) leading to a value and where \(h\) has clearly been defined earlier.
Dependent on both previous method marks
A1: Awrt 16.5
Note that a minimum we would expect to see for full marks is:
\(h = 0.4\)
| \(y_0\) | \(y_1\) | \(y_2\) | \(y_3\) | \(y_4\) | |
|---|---|---|---|---|---|
| \(x\) | \(0.4\) | \(0.8\) | \(1.2\) | \(1.6\) | \(2\) |
| \(y\) | \(\mathrm{e}^{0.16}\) | \(\mathrm{e}^{0.64}\) | \(\mathrm{e}^{1.44}\) | \(\mathrm{e}^{2.56}\) | \(\mathrm{e}^{4}\) |
| 1.173… | 1.896… | 4.220… | 12.935… | 54.598… |
(Note that a calculator gives \(16.030\ldots\) for the area)