| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Trapezium rule estimation |
| Difficulty | Moderate -0.8 This is a straightforward application of the trapezium rule formula with clearly specified intervals. While arctan requires calculator evaluation, the question is purely procedural with no conceptual challenges—students simply substitute values into the standard trapezium rule formula twice. Easier than average due to its routine, algorithmic nature. |
| Spec | 1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(x\) | 0 | 0.5 |
| \(\arctan x\) | 0 | 0.4636 |
| B2 | ||
| \((a)\) \(= \frac{1}{2} \times 1 \times [0 + 1.1071 + 2(0.7854)] = 1.34\) (3sf) | B1 M1 A1 | |
| \((b)\) \(= \frac{1}{2} \times 0.5 \times [0 + 1.1071 + 2(0.4636 + 0.7854 + 0.9828)] = 1.39\) (3sf) | M1 A1 | (7 marks) |
| $x$ | 0 | 0.5 | 1 | 1.5 | 2 |
| $\arctan x$ | 0 | 0.4636 | 0.7854 | 0.9828 | 1.1071 |
| B2 |
$(a)$ $= \frac{1}{2} \times 1 \times [0 + 1.1071 + 2(0.7854)] = 1.34$ (3sf) | B1 M1 A1 |
$(b)$ $= \frac{1}{2} \times 0.5 \times [0 + 1.1071 + 2(0.4636 + 0.7854 + 0.9828)] = 1.39$ (3sf) | M1 A1 | (7 marks)\begin{enumerate}
\item (a) Use the trapezium rule with two intervals of equal width to find an approximate value for the integral
\end{enumerate}
$$\int _ { 0 } ^ { 2 } \arctan x \mathrm {~d} x$$
(b) Use the trapezium rule with four intervals of equal width to find an improved approximation for the value of the integral.\\
\hfill \mbox{\textit{Edexcel C4 Q2 [7]}}