| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | June |
| 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 using Riemann sums (rectangular approximations) to bound definite integrals. While it requires understanding the relationship between areas of rectangles and integrals, the technique is well-established in Further Pure syllabi. Part (a) involves verifying a given inequality by summing a finite series, and part (b) asks students to apply the same method with rectangles positioned differently. The algebraic manipulation of the sum formula is routine for Further Maths students, making this moderately above average difficulty but not requiring novel insight. |
| Spec | 1.08g Integration as limit of sum: Riemann sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1 x^2\,dx < \left(\frac{1}{n}\right)\!\left(\frac{1}{n}\right)^2 + \left(\frac{1}{n}\right)\!\left(\frac{2}{n}\right)^2 + \cdots + \left(\frac{1}{n}\right)\!\left(\frac{n-1}{n}\right)^2 + \left(\frac{1}{n}\right)\!\left(\frac{n}{n}\right)^2\) | M1 A1 | |
| \(\frac{1}{n^3}\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6n^3} = \frac{2n^2+3n+1}{6n^2}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1 x^2\,dx > \left(\frac{1}{n}\right)\!\left(\frac{1}{n}\right)^2 + \left(\frac{1}{n}\right)\!\left(\frac{2}{n}\right)^2 + \cdots + \left(\frac{1}{n}\right)\!\left(\frac{n-1}{n}\right)^2\) | M1 A1 | |
| \(= \frac{1}{n^3}\sum_{r=1}^{n-1} r^2 = \frac{(n-1)(n)(2n-2+1)}{6n^3} = \frac{2n^2-3n+1}{6n^2}\) | M1 A1 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 x^2\,dx < \left(\frac{1}{n}\right)\!\left(\frac{1}{n}\right)^2 + \left(\frac{1}{n}\right)\!\left(\frac{2}{n}\right)^2 + \cdots + \left(\frac{1}{n}\right)\!\left(\frac{n-1}{n}\right)^2 + \left(\frac{1}{n}\right)\!\left(\frac{n}{n}\right)^2$ | M1 A1 | |
| $\frac{1}{n^3}\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6n^3} = \frac{2n^2+3n+1}{6n^2}$ | M1 A1 | |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 x^2\,dx > \left(\frac{1}{n}\right)\!\left(\frac{1}{n}\right)^2 + \left(\frac{1}{n}\right)\!\left(\frac{2}{n}\right)^2 + \cdots + \left(\frac{1}{n}\right)\!\left(\frac{n-1}{n}\right)^2$ | M1 A1 | |
| $= \frac{1}{n^3}\sum_{r=1}^{n-1} r^2 = \frac{(n-1)(n)(2n-2+1)}{6n^3} = \frac{2n^2-3n+1}{6n^2}$ | M1 A1 | |
---\begin{enumerate}[label=(\alph*)]
\item By considering the sum of the areas of these rectangles, show that
$$\int _ { 0 } ^ { 1 } x ^ { 2 } d x < \frac { 2 n ^ { 2 } + 3 n + 1 } { 6 n ^ { 2 } }$$
\item Use a similar method to find, in terms of $n$, a lower bound for $\int _ { 0 } ^ { 1 } x ^ { 2 } \mathrm {~d} x$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q4 [8]}}