| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Rectangle bounds for definite integral |
| Difficulty | Challenging +1.8 This is a Further Maths question requiring understanding of Riemann sums with upper/lower bounds, manipulation of products into factorial form, and a limit argument. While the individual techniques are standard (rectangles for bounds, product-to-sum via logarithms), the algebraic manipulation of the product notation and factorial expressions requires careful multi-step reasoning beyond typical A-level integration questions. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.08g Integration as limit of sum: Riemann sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1 \ln(1+x)\,dx < \tfrac{1}{n}\ln\!\left(1+\tfrac{1}{n}\right) + \tfrac{1}{n}\ln\!\left(1+\tfrac{2}{n}\right) + \cdots + \tfrac{1}{n}\ln\!\left(1+\tfrac{n}{n}\right)\) | M1 A1 | Forms the sum of the areas of the rectangles |
| \(= \tfrac{1}{n}\!\left(n\ln\tfrac{1}{n} + \ln(n+1) + \ln(n+2) + \cdots + \ln(2n)\right) = \tfrac{1}{n}\ln\dfrac{(2n)!}{n!} - \ln n\) | M1 A1 | Applies laws of logarithms, AG |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1 \ln(1+x)\,dx > \tfrac{1}{n}\ln\!\left(1+\tfrac{1}{n}\right) + \tfrac{1}{n}\ln\!\left(1+\tfrac{2}{n}\right) + \cdots + \tfrac{1}{n}\ln\!\left(1+\tfrac{n-1}{n}\right)\) | M1 A1 | Forms the sum of the areas of appropriate rectangles |
| \(= \tfrac{1}{n}\!\left((n-1)\ln\tfrac{1}{n} + \ln(n+1) + \ln(n+2) + \cdots + \ln(2n-1)\right)\) | M1 | Applies laws of logarithms |
| \(\tfrac{1}{n}\ln\dfrac{(2n-1)!}{n!} - \left(\dfrac{n-1}{n}\right)\ln n\) | A1 | Equivalent answers: \(\tfrac{1}{n}\ln\dfrac{(2n)!}{n!} - \ln n - \tfrac{1}{n}\ln 2\); \(\tfrac{1}{n}\ln\dfrac{(2n-1)!}{(n-1)!} - \ln n\) |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(U_n - L_n = \dfrac{\ln 2}{n}\) | M1 | Simplifies their \(U_n - L_n\) to \(\dfrac{c}{n}\) |
| \(\dfrac{\ln 2}{n} \to 0\) as \(n \to \infty\) | A1 | AG |
| Total: 2 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 \ln(1+x)\,dx < \tfrac{1}{n}\ln\!\left(1+\tfrac{1}{n}\right) + \tfrac{1}{n}\ln\!\left(1+\tfrac{2}{n}\right) + \cdots + \tfrac{1}{n}\ln\!\left(1+\tfrac{n}{n}\right)$ | M1 A1 | Forms the sum of the areas of the rectangles |
| $= \tfrac{1}{n}\!\left(n\ln\tfrac{1}{n} + \ln(n+1) + \ln(n+2) + \cdots + \ln(2n)\right) = \tfrac{1}{n}\ln\dfrac{(2n)!}{n!} - \ln n$ | M1 A1 | Applies laws of logarithms, AG |
| **Total: 4** | | |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 \ln(1+x)\,dx > \tfrac{1}{n}\ln\!\left(1+\tfrac{1}{n}\right) + \tfrac{1}{n}\ln\!\left(1+\tfrac{2}{n}\right) + \cdots + \tfrac{1}{n}\ln\!\left(1+\tfrac{n-1}{n}\right)$ | M1 A1 | Forms the sum of the areas of appropriate rectangles |
| $= \tfrac{1}{n}\!\left((n-1)\ln\tfrac{1}{n} + \ln(n+1) + \ln(n+2) + \cdots + \ln(2n-1)\right)$ | M1 | Applies laws of logarithms |
| $\tfrac{1}{n}\ln\dfrac{(2n-1)!}{n!} - \left(\dfrac{n-1}{n}\right)\ln n$ | A1 | Equivalent answers: $\tfrac{1}{n}\ln\dfrac{(2n)!}{n!} - \ln n - \tfrac{1}{n}\ln 2$; $\tfrac{1}{n}\ln\dfrac{(2n-1)!}{(n-1)!} - \ln n$ |
| **Total: 4** | | |
---
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $U_n - L_n = \dfrac{\ln 2}{n}$ | M1 | Simplifies their $U_n - L_n$ to $\dfrac{c}{n}$ |
| $\dfrac{\ln 2}{n} \to 0$ as $n \to \infty$ | A1 | AG |
| **Total: 2** | | |6\\
\includegraphics[max width=\textwidth, alt={}, center]{23b06b1c-997f-425d-ae3d-bd4cc1295605-10_771_1146_260_497}
The diagram shows the curve with equation $\mathrm { y } = \ln ( 1 + \mathrm { x } )$ for $0 \leqslant x \leqslant 1$, together with a set of $n$ rectangles each of width $\frac { 1 } { n }$.
\begin{enumerate}[label=(\alph*)]
\item By considering the sum of the areas of these rectangles, show that $\int _ { 0 } ^ { 1 } \ln ( 1 + x ) d x < U _ { n }$, where
$$U _ { n } = \frac { 1 } { n } \ln \frac { ( 2 n ) ! } { n ! } - \ln n$$
\item Use a similar method to find, in terms of $n$, a lower bound $\mathrm { L } _ { \mathrm { n } }$ for $\int _ { 0 } ^ { 1 } \ln ( 1 + x ) \mathrm { d } x$.
\item By simplifying $\mathrm { U } _ { \mathrm { n } } - \mathrm { L } _ { \mathrm { n } }$, show that $\lim _ { \mathrm { n } \rightarrow \infty } \left( \mathrm { U } _ { \mathrm { n } } - \mathrm { L } _ { \mathrm { n } } \right) = 0$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q6 [10]}}