| 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 +1.2 This is a structured Further Maths question on Riemann sums requiring geometric series summation and algebraic manipulation, but follows a standard template with clear guidance. Part (a) is scaffolded to show a given result, part (b) mirrors part (a), and part (c) is computational. While requiring more sophistication than typical A-level integration, the step-by-step structure and familiar techniques make it moderately above average difficulty. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.06a Exponential function: a^x and e^x graphs and properties1.08g Integration as limit of sum: Riemann sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1 2^x\,dx < \left(\frac{1}{N}\right)2^{\frac{1}{N}} + \left(\frac{1}{N}\right)2^{\frac{2}{N}} + \cdots + \left(\frac{1}{N}\right)2^{\frac{N-1}{N}} + \left(\frac{1}{N}\right)2^{\frac{N}{N}}\) | M1 A1 | Forms the sum of the areas of the rectangles |
| \(= \frac{1}{N}\sum_{n=1}^{N}\left(2^{\frac{1}{N}}\right)^n = \frac{2^{\frac{1}{N}}}{N\left(2^{\frac{1}{N}}-1\right)}\) | M1 A1 | Applies \(\sum_{n=1}^{N} r^n = \frac{r(r^N - 1)}{r-1}\), AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1 2^x\,dx > \left(\frac{1}{N}\right) + \left(\frac{1}{N}\right)2^{\frac{1}{N}} + \cdots + \left(\frac{1}{N}\right)2^{\frac{N-2}{N}} + \left(\frac{1}{N}\right)2^{\frac{N-1}{N}}\) | M1 A1 | Forms the sum of the areas of appropriate rectangles |
| \(= \frac{1}{N}\sum_{n=0}^{N-1}\left(2^{\frac{1}{N}}\right)^n = \frac{1}{N\left(2^{\frac{1}{N}}-1\right)}\) | M1 A1 | Applies \(\sum_{n=0}^{N-1} r^n = \frac{r^N - 1}{r-1}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{2^{\frac{1}{N}}}{N\left(2^{\frac{1}{N}}-1\right)} - \frac{1}{N\left(2^{\frac{1}{N}}-1\right)} = \frac{1}{N} < 10^{-4}\) leading to \(N > 10^4\) | M1 | Simplifies \(U_n - L_n\) to \(\frac{c}{n}\) |
| Least value of \(N\) is 10001 | A1 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 2^x\,dx < \left(\frac{1}{N}\right)2^{\frac{1}{N}} + \left(\frac{1}{N}\right)2^{\frac{2}{N}} + \cdots + \left(\frac{1}{N}\right)2^{\frac{N-1}{N}} + \left(\frac{1}{N}\right)2^{\frac{N}{N}}$ | M1 A1 | Forms the sum of the areas of the rectangles |
| $= \frac{1}{N}\sum_{n=1}^{N}\left(2^{\frac{1}{N}}\right)^n = \frac{2^{\frac{1}{N}}}{N\left(2^{\frac{1}{N}}-1\right)}$ | M1 A1 | Applies $\sum_{n=1}^{N} r^n = \frac{r(r^N - 1)}{r-1}$, AG |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 2^x\,dx > \left(\frac{1}{N}\right) + \left(\frac{1}{N}\right)2^{\frac{1}{N}} + \cdots + \left(\frac{1}{N}\right)2^{\frac{N-2}{N}} + \left(\frac{1}{N}\right)2^{\frac{N-1}{N}}$ | M1 A1 | Forms the sum of the areas of appropriate rectangles |
| $= \frac{1}{N}\sum_{n=0}^{N-1}\left(2^{\frac{1}{N}}\right)^n = \frac{1}{N\left(2^{\frac{1}{N}}-1\right)}$ | M1 A1 | Applies $\sum_{n=0}^{N-1} r^n = \frac{r^N - 1}{r-1}$ |
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{2^{\frac{1}{N}}}{N\left(2^{\frac{1}{N}}-1\right)} - \frac{1}{N\left(2^{\frac{1}{N}}-1\right)} = \frac{1}{N} < 10^{-4}$ leading to $N > 10^4$ | M1 | Simplifies $U_n - L_n$ to $\frac{c}{n}$ |
| Least value of $N$ is 10001 | A1 | |4 The diagram shows the curve with equation $\mathrm { y } = 2 ^ { \mathrm { x } }$ for $0 \leqslant x \leqslant 1$, together with a set of $N$ rectangles each of width $\frac { 1 } { N }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{114ece0d-558d-4c02-8a77-034b3681cff9-06_824_1161_376_450}
\begin{enumerate}[label=(\alph*)]
\item By considering the sum of the areas of these rectangles, show that $\int _ { 0 } ^ { 1 } 2 ^ { x } d x < U _ { N }$, where
$$\mathrm { U } _ { \mathrm { N } } = \frac { 2 ^ { \frac { 1 } { \mathrm {~N} } } } { \mathrm {~N} \left( 2 ^ { \frac { 1 } { \mathrm {~N} } } - 1 \right) }$$
\item Use a similar method to find, in terms of $N$, a lower bound $\mathrm { L } _ { \mathrm { N } }$ for $\int _ { 0 } ^ { 1 } 2 ^ { x } \mathrm {~d} x$.
\item Find the least value of $N$ such that $\mathrm { U } _ { \mathrm { N } } - \mathrm { L } _ { \mathrm { N } } < 10 ^ { - 4 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q4 [10]}}