| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Rectangle bounds for definite integral |
| Difficulty | Standard +0.8 This is a Further Maths question requiring understanding of Riemann sums and numerical integration bounds. Part (a) involves manipulating summation formulas to prove a given result, part (b) requires applying similar reasoning independently, and part (c) involves solving an inequality. While the concepts are standard for Further Maths, the algebraic manipulation of summations and the multi-step reasoning elevate this above average difficulty. |
| Spec | 1.08g Integration as limit of sum: Riemann sums1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^1 x^3\,dx < \left(\frac{1}{n}\right)\left(\frac{1}{n}\right)^3 + \left(\frac{1}{n}\right)\left(\frac{2}{n}\right)^3 + \cdots + \left(\frac{1}{n}\right)\left(\frac{n-1}{n}\right)^3 + \left(\frac{1}{n}\right)\left(\frac{n}{n}\right)^3\) | M1 A1 | Forms the sum of the areas of the \(n\) rectangles. Need last term for A1. |
| \(\frac{1}{n^4}\sum_{r=1}^{n} r^3 = \frac{n^2(n+1)^2}{4n^4} = \left(\frac{n+1}{2n}\right)^2\) | M1 A1 | Applies \(\sum_{r=1}^n r^3 = \frac{1}{4}n^2(n+1)^2\), AG. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^1 x^3\,dx > \left(\frac{1}{n}\right)\left(\frac{1}{n}\right)^3 + \left(\frac{1}{n}\right)\left(\frac{2}{n}\right)^3 + \cdots + \left(\frac{1}{n}\right)\left(\frac{n-1}{n}\right)^3\) | M1 A1 | Forms the sum of the areas of appropriate rectangles. Need the last term for A1. |
| \(\frac{1}{n^4}\sum_{r=1}^{n-1} r^3 = \frac{(n-1)^2 n^2}{4n^4} = \left(\frac{n-1}{2n}\right)^2\) | M1 A1 | Applies \(\sum_{r=1}^n r^3 = \frac{1}{4}n^2(n+1)^2\). Accept \(\left(\frac{n+1}{2n}\right)^2 - \frac{1}{n}\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\frac{n+1}{2n}\right)^2 - \left(\frac{n-1}{2n}\right)^2 = \frac{1}{n} < 10^{-3}\) leading to \(n > 10^3\) | M1 | Simplifies their \(U_n - L_n\) to \(\frac{k}{n}\). |
| Least value of \(n\) is 1001. | A1 |
## Question 3(a):
$\int_0^1 x^3\,dx < \left(\frac{1}{n}\right)\left(\frac{1}{n}\right)^3 + \left(\frac{1}{n}\right)\left(\frac{2}{n}\right)^3 + \cdots + \left(\frac{1}{n}\right)\left(\frac{n-1}{n}\right)^3 + \left(\frac{1}{n}\right)\left(\frac{n}{n}\right)^3$ | **M1 A1** | Forms the sum of the areas of the $n$ rectangles. Need last term for A1.
$\frac{1}{n^4}\sum_{r=1}^{n} r^3 = \frac{n^2(n+1)^2}{4n^4} = \left(\frac{n+1}{2n}\right)^2$ | **M1 A1** | Applies $\sum_{r=1}^n r^3 = \frac{1}{4}n^2(n+1)^2$, AG.
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## Question 3(b):
$\int_0^1 x^3\,dx > \left(\frac{1}{n}\right)\left(\frac{1}{n}\right)^3 + \left(\frac{1}{n}\right)\left(\frac{2}{n}\right)^3 + \cdots + \left(\frac{1}{n}\right)\left(\frac{n-1}{n}\right)^3$ | **M1 A1** | Forms the sum of the areas of appropriate rectangles. Need the last term for A1.
$\frac{1}{n^4}\sum_{r=1}^{n-1} r^3 = \frac{(n-1)^2 n^2}{4n^4} = \left(\frac{n-1}{2n}\right)^2$ | **M1 A1** | Applies $\sum_{r=1}^n r^3 = \frac{1}{4}n^2(n+1)^2$. Accept $\left(\frac{n+1}{2n}\right)^2 - \frac{1}{n}$.
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## Question 3(c):
$\left(\frac{n+1}{2n}\right)^2 - \left(\frac{n-1}{2n}\right)^2 = \frac{1}{n} < 10^{-3}$ leading to $n > 10^3$ | **M1** | Simplifies their $U_n - L_n$ to $\frac{k}{n}$.
Least value of $n$ is 1001. | **A1** |
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\includegraphics[max width=\textwidth, alt={}, center]{fa2213b3-480c-44cb-8ba0-ebd2b94d3d90-04_851_805_251_616}
The diagram shows the curve with equation $\mathrm { y } = \mathrm { x } ^ { 3 }$ 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 $\int _ { 0 } ^ { 1 } x ^ { 3 } d x < U _ { n }$, where
$$\mathrm { U } _ { \mathrm { n } } = \left( \frac { \mathrm { n } + 1 } { 2 \mathrm { n } } \right) ^ { 2 }$$
\item Use a similar method to find, in terms of $n$, a lower bound $L _ { n }$ for $\int _ { 0 } ^ { 1 } x ^ { 3 } d x$.
\item Find the least value of $n$ such that $\mathrm { U } _ { \mathrm { n } } - \mathrm { L } _ { \mathrm { n } } < 10 ^ { - 3 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q3 [10]}}