| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Simpson's rule application |
| Difficulty | Moderate -0.8 This is a straightforward application of Simpson's rule with clearly specified intervals and a simple function to evaluate. It requires only mechanical substitution into the Simpson's rule formula with no problem-solving or conceptual insight, making it easier than average despite being from Further Maths. |
| Spec | 1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Step length \(= 0.4\) | B1 | Correct step length of 0.4, may be implied by values 0.4, 0.8, etc. |
| Table of values: \(x = 0.4, 0.8, 1.2, 1.6, 2\); \(y = e^{0.16}, e^{0.64}, e^{1.44}, e^{2.56}, e^4\); \(y = 1.173\ldots, 1.896\ldots, 4.220\ldots, 12.935\ldots, 54.598\ldots\) | M1 | Attempts to find \(y\) values for their \(x\) values. Must see at least 3 values. |
| \(y_0 + 4y_1 + 2y_2 + 4y_3 + y_4 = 123.54\ldots\) | M1 | Correct structure for Simpson's rule (ends + 2evens + 4odds). Must use \(y\) values not \(x\) values. |
| \(\int_{0.4}^{2} e^{x^2}\,dx \approx \frac{0.4}{3} \times \{1.173\ldots + 54.598\ldots + 4(1.896\ldots + 12.935\ldots) + 2(4.220\ldots)\} \approx \frac{0.4}{3} \times 123.54\ldots\) | dM1 | \(\frac{0.4}{3} \times \text{their } 123.54\ldots\) or \(\frac{h}{3} \times \text{their } 123.54\ldots\); dependent on both previous M marks |
| \(= 16.5\) | A1 | Awrt 16.5 |
# Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Step length $= 0.4$ | B1 | Correct step length of 0.4, may be implied by values 0.4, 0.8, etc. |
| Table of values: $x = 0.4, 0.8, 1.2, 1.6, 2$; $y = e^{0.16}, e^{0.64}, e^{1.44}, e^{2.56}, e^4$; $y = 1.173\ldots, 1.896\ldots, 4.220\ldots, 12.935\ldots, 54.598\ldots$ | M1 | Attempts to find $y$ values for their $x$ values. Must see at least 3 values. |
| $y_0 + 4y_1 + 2y_2 + 4y_3 + y_4 = 123.54\ldots$ | M1 | Correct structure for Simpson's rule (ends + 2evens + 4odds). Must use $y$ values not $x$ values. |
| $\int_{0.4}^{2} e^{x^2}\,dx \approx \frac{0.4}{3} \times \{1.173\ldots + 54.598\ldots + 4(1.896\ldots + 12.935\ldots) + 2(4.220\ldots)\} \approx \frac{0.4}{3} \times 123.54\ldots$ | dM1 | $\frac{0.4}{3} \times \text{their } 123.54\ldots$ or $\frac{h}{3} \times \text{their } 123.54\ldots$; dependent on both previous M marks |
| $= 16.5$ | A1 | Awrt 16.5 |
---\begin{enumerate}
\item Use Simpson's rule with 4 intervals to estimate
\end{enumerate}
$$\int _ { 0.4 } ^ { 2 } e ^ { x ^ { 2 } } d x$$
\hfill \mbox{\textit{Edexcel FP1 2019 Q1 [5]}}