Question 3 - A-Level Maths

CAIE Further Paper 2 2021 June — Question 3 10 marks

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 on Riemann sums requiring students to derive upper/lower bounds using rectangles, manipulate summation formulas (likely needing Σr² = n(n+1)(2n+1)/6), and solve an inequality. While conceptually accessible, it demands algebraic facility with summations and multi-step reasoning beyond standard integration questions, placing it moderately above average difficulty.
Spec	1.08g Integration as limit of sum: Riemann sums 1.09f Trapezium rule: numerical integration

3 \includegraphics[max width=\textwidth, alt={}, center]{fd247a71-4680-45d8-89d2-fef17ed3a5e9-04_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 $\frac { 1 } { n }$.

By considering the sum of the areas of these rectangles, show that $\int _ { 0 } ^ { 1 } x ^ { 3 } d x < U _ { n }$, where $$\mathrm { U } _ { \mathrm { n } } = \left( \frac { \mathrm { n } + 1 } { 2 \mathrm { n } } \right) ^ { 2 }$$
Use a similar method to find, in terms of $n$, a lower bound $L _ { n }$ for $\int _ { 0 } ^ { 1 } x ^ { 3 } d x$.
Find the least value of $n$ such that $\mathrm { U } _ { \mathrm { n } } - \mathrm { L } _ { \mathrm { n } } < 10 ^ { - 3 }$.

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Question 3(a):

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.

Question 3(b):

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 $\left(\frac{n+1}{2n}\right)^2 - \frac{1}{n}$.

Question 3(c):

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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Show LaTeX source

3\\
\includegraphics[max width=\textwidth, alt={}, center]{fd247a71-4680-45d8-89d2-fef17ed3a5e9-04_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]}}

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Q3 10

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