| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Integral bounds for series |
| Difficulty | Challenging +1.2 This is a standard Further Maths question on integral bounds using Riemann sums. Part (a) requires recognizing that rectangles with heights at left endpoints give an upper bound, then summing a cubic series. Part (b) mirrors this with right endpoints for a lower bound. While it requires knowledge of summation formulas and careful algebraic manipulation, the technique is well-established in Further Maths syllabi and follows a predictable pattern. The multi-step nature and algebraic complexity place it slightly above average difficulty. |
| Spec | 1.08g Integration as limit of sum: Riemann sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1 1-x^3\,dx \leqslant \frac{1}{n} + \frac{1}{n}\left(1 - \frac{1}{n^3}\right) + \ldots + \frac{1}{n}\left(1 - \frac{(n-1)^3}{n^3}\right)\) | M1 A1 | Forms the sum of the areas of the rectangles |
| \(= 1 - \frac{1}{n^4}\sum_{r=1}^{n-1} r^3 = 1 - \frac{(n-1)^2 n^2}{4n^4}\) | M1 | Applies \(\sum_{r=1}^{n-1} r^3 = \frac{1}{4}(n-1)^2 n^2\) |
| \(= \frac{4n^2 - (n-1)^2}{4n^2} = \frac{3n^2 + 2n - 1}{4n^2}\) | A1 | AG |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1 1-x^3\,dx \geqslant \frac{1}{n}\left(1 - \frac{1}{n^3}\right) + \frac{1}{n}\left(1 - \frac{2^3}{n^3}\right) + \ldots + \frac{1}{n}\left(1 - \frac{(n-1)^3}{n^3}\right)\) | M1 A1 | Forms the sum of the areas of appropriate rectangles |
| \(= \frac{n-1}{n} - \frac{1}{n^4}\sum_{r=1}^{n-1} r^3 = \frac{n-1}{n} - \frac{(n-1)^2 n^2}{4n^4}\) | M1 | Applies \(\sum_{r=1}^{n-1} r^3 = \frac{1}{4}(n-1)^2 n^2\) |
| \(= \frac{4n(n-1) - (n-1)^2}{4n^2} = \frac{3n^2 - 2n - 1}{4n^2}\) | A1 | |
| Total: 4 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 1-x^3\,dx \leqslant \frac{1}{n} + \frac{1}{n}\left(1 - \frac{1}{n^3}\right) + \ldots + \frac{1}{n}\left(1 - \frac{(n-1)^3}{n^3}\right)$ | M1 A1 | Forms the sum of the areas of the rectangles |
| $= 1 - \frac{1}{n^4}\sum_{r=1}^{n-1} r^3 = 1 - \frac{(n-1)^2 n^2}{4n^4}$ | M1 | Applies $\sum_{r=1}^{n-1} r^3 = \frac{1}{4}(n-1)^2 n^2$ |
| $= \frac{4n^2 - (n-1)^2}{4n^2} = \frac{3n^2 + 2n - 1}{4n^2}$ | A1 | AG |
| **Total: 4** | | |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 1-x^3\,dx \geqslant \frac{1}{n}\left(1 - \frac{1}{n^3}\right) + \frac{1}{n}\left(1 - \frac{2^3}{n^3}\right) + \ldots + \frac{1}{n}\left(1 - \frac{(n-1)^3}{n^3}\right)$ | M1 A1 | Forms the sum of the areas of appropriate rectangles |
| $= \frac{n-1}{n} - \frac{1}{n^4}\sum_{r=1}^{n-1} r^3 = \frac{n-1}{n} - \frac{(n-1)^2 n^2}{4n^4}$ | M1 | Applies $\sum_{r=1}^{n-1} r^3 = \frac{1}{4}(n-1)^2 n^2$ |
| $= \frac{4n(n-1) - (n-1)^2}{4n^2} = \frac{3n^2 - 2n - 1}{4n^2}$ | A1 | |
| **Total: 4** | | |\begin{enumerate}[label=(\alph*)]
\item By considering the sum of the areas of the rectangles, show that
$$\int _ { 0 } ^ { 1 } \left( 1 - x ^ { 3 } \right) d x \leqslant \frac { 3 n ^ { 2 } + 2 n - 1 } { 4 n ^ { 2 } }$$
\item Use a similar method to find, in terms of $n$, a lower bound for $\int _ { 0 } ^ { 1 } \left( 1 - x ^ { 3 } \right) \mathrm { dx }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q4 [8]}}