| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2024 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Rectangle bounds for definite integral |
| Difficulty | Challenging +1.2 This is a structured multi-part question on numerical integration using upper/lower Riemann sums. Parts (a)-(c) involve standard geometric series summation and limit evaluation. Part (d) requires Maclaurin series manipulation but the series is provided. While it requires careful algebraic manipulation across multiple steps, each individual technique is well-practiced in Further Maths, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 1.08g Integration as limit of sum: Riemann sums4.08a Maclaurin series: find series for function4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
| Answer | Marks | Guidance |
|---|---|---|
| 6(a) | 1 e1−xdx (1) e 1− n 0 + (1) e 1−1 n + + (1) e 1−n n −1 | |
| 0 n n n | M1A1 | Forms the sum of the areas of the rectangles. |
| Answer | Marks | Guidance |
|---|---|---|
| n r=0 n 1−e −1 n n ( 1−e −1 n ) | M1A1 | Applies sum of geometric progression, AG. |
| Answer | Marks |
|---|---|
| 6(b) | 1 e1−x dx(1) e 1−1 n +(1) e 1− n 2 +(1) e 1−n n −1 +(1) e 1− n n |
| Answer | Marks | Guidance |
|---|---|---|
| 0 | M1A1 | Forms the sum of the areas of appropriate |
| Answer | Marks | Guidance |
|---|---|---|
| n 1−e n n e n −1 | M1A1 | Applies sum of geometric progression. |
| Answer | Marks |
|---|---|
| 6(c) | e−1 |
| Answer | Marks | Guidance |
|---|---|---|
| n n n | M1 | c |
| Answer | Marks | Guidance |
|---|---|---|
| n | A1 | AG. CWO. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 6(d) | z ( 1−e −1 z ) =z ( 1− ( 1−1+ 1 − 1 + )) =1− 1 + 1 + | |
| z 2z2 6z3 2z 6z2 | M1A1 | Substitutes power series for e −1 z. A1 for |
| Answer | Marks |
|---|---|
| n→ | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 6:
--- 6(a) ---
6(a) | 1 e1−xdx (1) e 1− n 0 + (1) e 1−1 n + + (1) e 1−n n −1
0 n n n | M1A1 | Forms the sum of the areas of the rectangles.
M1 for correct number of rectangles.
= e n−1 e − n r = e 1−e−1 = e−1
n r=0 n 1−e −1 n n ( 1−e −1 n ) | M1A1 | Applies sum of geometric progression, AG.
4
--- 6(b) ---
6(b) | 1 e1−x dx(1) e 1−1 n +(1) e 1− n 2 +(1) e 1−n n −1 +(1) e 1− n n
n n n n
0 | M1A1 | Forms the sum of the areas of appropriate
rectangles. M1 for correct number rectangles.
e−1 e 1 e−1
= − + =
( ) ( )
−1 n n 1
n 1−e n n e n −1 | M1A1 | Applies sum of geometric progression.
e −1 n (e−1) e 1−1 n ( 1−e−1)
OE. E.g. or .
n ( 1−e −1 n ) n ( 1−e −1 n )
4
--- 6(c) ---
6(c) | e−1
U −L =
n n n | M1 | c
Simplifies U −L to , where c is a
n n n
nonzero constant.
e−1
→0 as n→
n | A1 | AG. CWO.
2
Question | Answer | Marks | Guidance
--- 6(d) ---
6(d) | z ( 1−e −1 z ) =z ( 1− ( 1−1+ 1 − 1 + )) =1− 1 + 1 +
z 2z2 6z3 2z 6z2 | M1A1 | Substitutes power series for e −1 z. A1 for
enough terms.
lim(U )=e−1
n
n→ | B1
3
Question | Answer | Marks | Guidance\includegraphics{figure_6}
The diagram shows the curve with equation $y = e^{1-x}$ 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 $\int_0^1 e^{1-x} \, dx < U_n$, where
$$U_n = \frac{e-1}{n(1-e^{-1})}.$$ [4]
\item Use a similar method to find, in terms of $n$, a lower bound $L_n$ for $\int_0^1 e^{1-x} \, dx$. [4]
\item Show that $\lim_{n \to \infty}(U_n - L_n) = 0$. [2]
\item Use the Maclaurin's series for $e^x$ given in the list of formulae (MF19) to find the first three terms of the series expansion of $z(1-e^{-z})$, in ascending powers of $z$, and deduce the value of $\lim_{n \to \infty}(U_n)$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q6 [13]}}