| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Rectangle bounds for definite integral |
| Difficulty | Standard +0.8 This is a Further Maths question requiring students to derive Riemann sum bounds for a definite integral using upper and lower rectangles, manipulate summation formulas involving n, and prove convergence. While the curve is simple (quadratic), the algebraic manipulation of the summations and proving the given form of U_n requires careful work beyond standard A-level integration. The multi-part structure and proof elements place it moderately above average difficulty. |
| Spec | 1.08g Integration as limit of sum: Riemann sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[\int_0^1 2x-x^2\,dx <\right]\left(\frac{1}{n}\right)\!\left(2\!\left(\frac{1}{n}\right)-\!\left(\frac{1}{n}\right)^2\right)+\left(\frac{1}{n}\right)\!\left(2\!\left(\frac{2}{n}\right)-\!\left(\frac{2}{n}\right)^2\right)+\cdots+\left(\frac{1}{n}\right)\!\left(2\!\left(\frac{n}{n}\right)-\!\left(\frac{n}{n}\right)^2\right)\) | M1 A1 | Forms sum of areas of rectangles; may be in summation form. M1 for correct number of rectangles |
| \(= \frac{2}{n^2}\sum_{r=1}^n r - \frac{1}{n^3}\sum_{r=1}^n r^2 = \frac{1}{n^2}n(n+1) - \frac{1}{6n^3}n(n+1)(2n+1)\) | M1 A1 | Applies formulae from MF19. Summations must have correct limits for A1. Substituted in correctly once for M1 |
| \(= 1+\frac{1}{n}-\frac{1}{6}\!\left(1+\frac{1}{n}\right)\!\left(2+\frac{1}{n}\right) = \left(1+\frac{1}{n}\right)\!\left(\frac{2}{3}-\frac{1}{6n}\right) = \frac{2}{3}+\frac{1}{2n}-\frac{1}{6n^2}\) | A1 | AG. This A1 requires the previous A1 |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1 2x-x^2\,dx > \left(\frac{1}{n}\right)\!\left(2\!\left(\frac{1}{n}\right)-\!\left(\frac{1}{n}\right)^2\right)+\left(\frac{1}{n}\right)\!\left(2\!\left(\frac{2}{n}\right)-\!\left(\frac{2}{n}\right)^2\right)+\cdots+\left(\frac{1}{n}\right)\!\left(2\!\left(\frac{n-1}{n}\right)-\!\left(\frac{n-1}{n}\right)^2\right)\) | M1 A1 | Forms sum of areas of appropriate rectangles. M1 for seeing correct height of last rectangle |
| \(\frac{2}{n^2}\sum_{r=1}^{n-1}r - \frac{1}{n^3}\sum_{r=1}^{n-1}r^2 = \frac{1}{n^2}(n-1)n - \frac{1}{6n^3}(n-1)(n)(2n-1)\) | M1 | Applies formulae from MF19, substitutes correct limit. Substituted in correctly once for M1 |
| \(\left(1-\frac{1}{n}\right)\!\left(\frac{2}{3}+\frac{1}{6n}\right) = \frac{2}{3}-\frac{1}{2n}-\frac{1}{6n^2}\) | A1 | \(\left(1+\frac{1}{n}\right)\!\left(\frac{2}{3}-\frac{1}{6n}\right)-\frac{1}{n}\) or fully expanded or fully factorised |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(U_n - L_n = \left(1+\frac{1}{n}\right)\!\left(\frac{2}{3}-\frac{1}{6n}\right) - \left(1-\frac{1}{n}\right)\!\left(\frac{2}{3}+\frac{1}{6n}\right) \quad\left(=\frac{1}{n}\right)\) | M1 | Expresses \(U_n - L_n\) in terms of \(\frac{1}{n}\). For M1 they must take the limit of an expression that tends to a constant. Must be fully expanded if they do not take the limit at any stage |
| \(\frac{1}{n}\to 0\) as \(n\to\infty\) or \(\frac{2}{3}-\frac{2}{3}=0\) | A1 | AG, CWO. Their \(L_n\) must be correct |
| 2 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[\int_0^1 2x-x^2\,dx <\right]\left(\frac{1}{n}\right)\!\left(2\!\left(\frac{1}{n}\right)-\!\left(\frac{1}{n}\right)^2\right)+\left(\frac{1}{n}\right)\!\left(2\!\left(\frac{2}{n}\right)-\!\left(\frac{2}{n}\right)^2\right)+\cdots+\left(\frac{1}{n}\right)\!\left(2\!\left(\frac{n}{n}\right)-\!\left(\frac{n}{n}\right)^2\right)$ | M1 A1 | Forms sum of areas of rectangles; may be in summation form. M1 for correct number of rectangles |
| $= \frac{2}{n^2}\sum_{r=1}^n r - \frac{1}{n^3}\sum_{r=1}^n r^2 = \frac{1}{n^2}n(n+1) - \frac{1}{6n^3}n(n+1)(2n+1)$ | M1 A1 | Applies formulae from MF19. Summations must have correct limits for A1. Substituted in correctly once for M1 |
| $= 1+\frac{1}{n}-\frac{1}{6}\!\left(1+\frac{1}{n}\right)\!\left(2+\frac{1}{n}\right) = \left(1+\frac{1}{n}\right)\!\left(\frac{2}{3}-\frac{1}{6n}\right) = \frac{2}{3}+\frac{1}{2n}-\frac{1}{6n^2}$ | A1 | AG. This A1 requires the previous A1 |
| | **5** | |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 2x-x^2\,dx > \left(\frac{1}{n}\right)\!\left(2\!\left(\frac{1}{n}\right)-\!\left(\frac{1}{n}\right)^2\right)+\left(\frac{1}{n}\right)\!\left(2\!\left(\frac{2}{n}\right)-\!\left(\frac{2}{n}\right)^2\right)+\cdots+\left(\frac{1}{n}\right)\!\left(2\!\left(\frac{n-1}{n}\right)-\!\left(\frac{n-1}{n}\right)^2\right)$ | M1 A1 | Forms sum of areas of appropriate rectangles. M1 for seeing correct height of last rectangle |
| $\frac{2}{n^2}\sum_{r=1}^{n-1}r - \frac{1}{n^3}\sum_{r=1}^{n-1}r^2 = \frac{1}{n^2}(n-1)n - \frac{1}{6n^3}(n-1)(n)(2n-1)$ | M1 | Applies formulae from MF19, substitutes correct limit. Substituted in correctly once for M1 |
| $\left(1-\frac{1}{n}\right)\!\left(\frac{2}{3}+\frac{1}{6n}\right) = \frac{2}{3}-\frac{1}{2n}-\frac{1}{6n^2}$ | A1 | $\left(1+\frac{1}{n}\right)\!\left(\frac{2}{3}-\frac{1}{6n}\right)-\frac{1}{n}$ or fully expanded or fully factorised |
| | **4** | |
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $U_n - L_n = \left(1+\frac{1}{n}\right)\!\left(\frac{2}{3}-\frac{1}{6n}\right) - \left(1-\frac{1}{n}\right)\!\left(\frac{2}{3}+\frac{1}{6n}\right) \quad\left(=\frac{1}{n}\right)$ | M1 | Expresses $U_n - L_n$ in terms of $\frac{1}{n}$. For M1 they must take the limit of an expression that tends to a constant. Must be fully expanded if they do not take the limit at any stage |
| $\frac{1}{n}\to 0$ as $n\to\infty$ or $\frac{2}{3}-\frac{2}{3}=0$ | A1 | AG, CWO. Their $L_n$ must be correct |
| | **2** | |
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\includegraphics[max width=\textwidth, alt={}, center]{114be67d-a57f-4c36-8f1c-974a2719c1f1-08_663_1152_260_452}
The diagram shows the curve with equation $\mathrm { y } = 2 \mathrm { x } - \mathrm { x } ^ { 2 }$ for $0 \leqslant x \leqslant 1$, together with a set of $n$ rectangles of width $\frac { 1 } { n }$.
\begin{enumerate}[label=(\alph*)]
\item By considering the sum of the areas of these rectangles, show that $\int _ { 0 } ^ { 1 } \left( 2 x - x ^ { 2 } \right) d x < U _ { n }$, where
$$U _ { n } = \left( 1 + \frac { 1 } { n } \right) \left( \frac { 2 } { 3 } - \frac { 1 } { 6 n } \right) .$$
\item Use a similar method to find, in terms of $n$, a lower bound $L _ { n }$ for $\int _ { 0 } ^ { 1 } \left( 2 x - x ^ { 2 } \right) d x$.
\item Show that $\lim _ { n \rightarrow \infty } \left( \mathrm { U } _ { n } - \mathrm { L } _ { \mathrm { n } } \right) = 0$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q5 [11]}}