| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| 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, manipulate summation formulas (likely involving Σr² = n(n+1)(2n+1)/6), and prove a limit result. While conceptually accessible, it demands careful algebraic manipulation across multiple parts and understanding of upper/lower bounds via rectangles—significantly above average A-level difficulty but not requiring deep novel insight. |
| Spec | 1.08g Integration as limit of sum: Riemann sums |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^1(1-x)^2\, dx < \left(\frac{1}{n}\right)\left(1-\frac{0}{n}\right)^2 + \left(\frac{1}{n}\right)\left(1-\frac{1}{n}\right)^2 + \cdots + \left(\frac{1}{n}\right)\left(1-\frac{n-1}{n}\right)^2\) | M1 A1 | Forms sum of areas of rectangles. Must have first two terms and last term for A1. Accept \(\left(\frac{1}{n}\right)\left(1-\frac{n}{n}\right)^2\) for last term. Accept written in summation form with correct limits |
| \(\frac{1}{n} + \frac{1}{n^3}\sum_{r=1}^{n-1}(n-r)^2 = \frac{1}{n} + \frac{1}{n^3}\sum_{r=1}^{n-1}(n^2 - 2nr + r^2)\) \(= 1 - \frac{2}{n^2}\sum_{r=1}^{n-1}r + \frac{1}{n^3}\sum_{r=1}^{n-1}r^2\) | M1 A1 | Setting up correct series so that formulae from MF19 can be applied. Can also recognise sum as \(\frac{1}{n^3}\sum_{r=1}^{n}r^2\) |
| \(\frac{1}{n} + \frac{n^2(n-1)}{n^3} - \frac{n^2(n-1)}{n^3} + \frac{n(n-1)(2n-1)}{6n^3} = \frac{1}{n} + \frac{2n^2-3n+1}{6n^2} = \frac{2n^2+3n+1}{6n^2}\) | A1 | AG. Must have gained previous accuracy mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1(1-x)^2\,dx > \left(\frac{1}{n}\right)\left(1-\frac{1}{n}\right)^2 + \left(\frac{1}{n}\right)\left(1-\frac{2}{n}\right)^2 + \cdots + \left(\frac{1}{n}\right)\left(1-\frac{n-1}{n}\right)^2\) | M1 A1 | Forms sum of areas of appropriate rectangles. Must have first two and last terms for A1. Accept \(\left(\frac{1}{n}\right)\left(1-\frac{n}{n}\right)^2\) for last term. Accept written in summation form with correct limits. |
| \(\frac{1}{n^3}\sum_{r=1}^{n-1}(n^2-2nr+r^2) = \frac{n^2(n-1)}{n^3} - \frac{n^2(n-1)}{n^3} + \frac{n(n-1)(2n-1)}{6n^3}\) \(\left(= 1 - \frac{2}{n^2}\sum_{r=1}^{n}r + \frac{1}{n^3}\sum_{r=1}^{n}r^2\right)\) | M1 | Setting up correct series so that formulae from MF19 can be applied. Recognising sum as \(\frac{1}{n^3}\sum_{r=1}^{n-1}r^2 = \frac{1}{n^3}\sum_{r=1}^{n}r^2 - \frac{1}{n}\) without wrong working scores M1 A1 M1. |
| \(\dfrac{2n^2-3n+1}{6n^2}\) | A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(U_n - L_n = \dfrac{1}{n}\) | M1 | Simplifies \(U_n - L_n\) to \(\dfrac{c}{n}\). |
| \(\dfrac{1}{n} \to 0\) as \(n \to \infty\) | A1 | AG. |
| 2 |
## Question 6(a):
| $\int_0^1(1-x)^2\, dx < \left(\frac{1}{n}\right)\left(1-\frac{0}{n}\right)^2 + \left(\frac{1}{n}\right)\left(1-\frac{1}{n}\right)^2 + \cdots + \left(\frac{1}{n}\right)\left(1-\frac{n-1}{n}\right)^2$ | M1 A1 | Forms sum of areas of rectangles. Must have first two terms and last term for A1. Accept $\left(\frac{1}{n}\right)\left(1-\frac{n}{n}\right)^2$ for last term. Accept written in summation form with correct limits |
|---|---|---|
| $\frac{1}{n} + \frac{1}{n^3}\sum_{r=1}^{n-1}(n-r)^2 = \frac{1}{n} + \frac{1}{n^3}\sum_{r=1}^{n-1}(n^2 - 2nr + r^2)$ $= 1 - \frac{2}{n^2}\sum_{r=1}^{n-1}r + \frac{1}{n^3}\sum_{r=1}^{n-1}r^2$ | M1 A1 | Setting up correct series so that formulae from MF19 can be applied. Can also recognise sum as $\frac{1}{n^3}\sum_{r=1}^{n}r^2$ |
| $\frac{1}{n} + \frac{n^2(n-1)}{n^3} - \frac{n^2(n-1)}{n^3} + \frac{n(n-1)(2n-1)}{6n^3} = \frac{1}{n} + \frac{2n^2-3n+1}{6n^2} = \frac{2n^2+3n+1}{6n^2}$ | A1 | AG. Must have gained previous accuracy mark |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1(1-x)^2\,dx > \left(\frac{1}{n}\right)\left(1-\frac{1}{n}\right)^2 + \left(\frac{1}{n}\right)\left(1-\frac{2}{n}\right)^2 + \cdots + \left(\frac{1}{n}\right)\left(1-\frac{n-1}{n}\right)^2$ | M1 A1 | Forms sum of areas of appropriate rectangles. Must have first two and last terms for A1. Accept $\left(\frac{1}{n}\right)\left(1-\frac{n}{n}\right)^2$ for last term. Accept written in summation form with correct limits. |
| $\frac{1}{n^3}\sum_{r=1}^{n-1}(n^2-2nr+r^2) = \frac{n^2(n-1)}{n^3} - \frac{n^2(n-1)}{n^3} + \frac{n(n-1)(2n-1)}{6n^3}$ $\left(= 1 - \frac{2}{n^2}\sum_{r=1}^{n}r + \frac{1}{n^3}\sum_{r=1}^{n}r^2\right)$ | M1 | Setting up correct series so that formulae from MF19 can be applied. Recognising sum as $\frac{1}{n^3}\sum_{r=1}^{n-1}r^2 = \frac{1}{n^3}\sum_{r=1}^{n}r^2 - \frac{1}{n}$ without wrong working scores M1 A1 M1. |
| $\dfrac{2n^2-3n+1}{6n^2}$ | A1 | |
| | **4** | |
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $U_n - L_n = \dfrac{1}{n}$ | M1 | Simplifies $U_n - L_n$ to $\dfrac{c}{n}$. |
| $\dfrac{1}{n} \to 0$ as $n \to \infty$ | A1 | AG. |
| | **2** | |6\\
\includegraphics[max width=\textwidth, alt={}, center]{d421652f-576d-4843-abbf-54404e225fec-10_1015_988_260_577}
The diagram shows the curve with equation $\mathrm { y } = ( 1 - \mathrm { x } ) ^ { 2 }$ 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 $\int _ { 0 } ^ { 1 } ( 1 - x ) ^ { 2 } d x < U _ { n }$, where
$$U _ { n } = \frac { 2 n ^ { 2 } + 3 n + 1 } { 6 n ^ { 2 } }$$
\item Use a similar method to find, in terms of $n$, a lower bound $L _ { n }$ for $\int _ { 0 } ^ { 1 } ( 1 - x ) ^ { 2 } d x$.
\item Show that $\lim _ { n \rightarrow \infty } \left( U _ { n } - L _ { n } \right) = 0$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q6 [11]}}