| 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 Riemann sum bounds using rectangles, summation of series involving n, algebraic manipulation to reach a specific form, and limit analysis. While the individual techniques are standard (summations, limits), the multi-step nature, need to work with general n, and requirement to derive specific algebraic forms elevates this above typical A-level integration questions. |
| 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; 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 |
| \(=1+\frac{1}{n}-\frac{1}{6}\!\left(1+\frac{1}{n}\right)\!\left(2+\frac{1}{n}\right)=\left(1+\frac{1}{n}\right)\!\left(\frac{2}{3}-\frac{1}{6n}\right)=\frac{2}{3}+\frac{1}{2n}-\frac{1}{6n^2}\) | A1 | AG; this A1 requires the previous A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1 2x-x^2\,dx > \left(\frac{1}{n}\right)\!\left(2\!\left(\frac{1}{n}\right)-\!\left(\frac{1}{n}\right)^2\right)+\cdots+\left(\frac{1}{n}\right)\!\left(2\!\left(\frac{n-1}{n}\right)-\!\left(\frac{n-1}{n}\right)^2\right)\) | M1 A1 | Forms sum of areas of appropriate rectangles; M1 for 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 |
| \(\left(1-\frac{1}{n}\right)\!\left(\frac{2}{3}+\frac{1}{6n}\right)=\frac{2}{3}-\frac{1}{2n}-\frac{1}{6n^2}\) | A1 | \(\left(1+\frac{1}{n}\right)\!\left(\frac{2}{3}-\frac{1}{6n}\right)-\frac{1}{n}\) or fully expanded or fully factorised |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(U_n - L_n = \left(1+\frac{1}{n}\right)\!\left(\frac{2}{3}-\frac{1}{6n}\right)-\left(1-\frac{1}{n}\right)\!\left(\frac{2}{3}+\frac{1}{6n}\right) \;(=\tfrac{1}{n})\) | M1 | Expresses \(U_n-L_n\) in terms of \(\frac{1}{n}\); must take limit of expression tending to a constant |
| \(\frac{1}{n}\to 0\) as \(n\to\infty\) or \(\frac{2}{3}-\frac{2}{3}=0\) | A1 | AG, CWO; their \(L_n\) must be correct |
## 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; 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 |
| $=1+\frac{1}{n}-\frac{1}{6}\!\left(1+\frac{1}{n}\right)\!\left(2+\frac{1}{n}\right)=\left(1+\frac{1}{n}\right)\!\left(\frac{2}{3}-\frac{1}{6n}\right)=\frac{2}{3}+\frac{1}{2n}-\frac{1}{6n^2}$ | A1 | AG; this A1 requires the previous A1 |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 2x-x^2\,dx > \left(\frac{1}{n}\right)\!\left(2\!\left(\frac{1}{n}\right)-\!\left(\frac{1}{n}\right)^2\right)+\cdots+\left(\frac{1}{n}\right)\!\left(2\!\left(\frac{n-1}{n}\right)-\!\left(\frac{n-1}{n}\right)^2\right)$ | M1 A1 | Forms sum of areas of appropriate rectangles; M1 for 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 |
| $\left(1-\frac{1}{n}\right)\!\left(\frac{2}{3}+\frac{1}{6n}\right)=\frac{2}{3}-\frac{1}{2n}-\frac{1}{6n^2}$ | A1 | $\left(1+\frac{1}{n}\right)\!\left(\frac{2}{3}-\frac{1}{6n}\right)-\frac{1}{n}$ or fully expanded or fully factorised |
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $U_n - L_n = \left(1+\frac{1}{n}\right)\!\left(\frac{2}{3}-\frac{1}{6n}\right)-\left(1-\frac{1}{n}\right)\!\left(\frac{2}{3}+\frac{1}{6n}\right) \;(=\tfrac{1}{n})$ | M1 | Expresses $U_n-L_n$ in terms of $\frac{1}{n}$; must take limit of expression tending to a constant |
| $\frac{1}{n}\to 0$ as $n\to\infty$ or $\frac{2}{3}-\frac{2}{3}=0$ | A1 | AG, CWO; their $L_n$ must be correct |
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\includegraphics[max width=\textwidth, alt={}, center]{bca7281b-a6a9-4b4c-94e5-3da2a561ad86-08_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]}}