CAIE Further Paper 2 2021 November — Question 3 8 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2021
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAreas by integration
TypeBounds using rectangles
DifficultyChallenging +1.2 This is a standard Riemann sum question requiring students to express rectangle areas as a sum, use standard summation formulas (Σr and Σr²), and compare to the integral. While it involves multiple steps and algebraic manipulation, it follows a well-established template taught in Further Maths courses. The two-part structure (upper and lower bounds) is predictable, making it moderately above average difficulty but not requiring novel insight.
Spec1.08g Integration as limit of sum: Riemann sums
\includegraphics{figure_3} The diagram shows the curve with equation \(y = 1 - x^2\) for \(0 \leq x \leq 1\), together with a set of \(n\) rectangles of width \(\frac{1}{n}\).
  1. By considering the sum of the areas of the rectangles, show that $$\int_0^1 (1 - x^2) \, dx < \frac{4n^2 + 3n - 1}{6n^2}.$$ [4]
  2. Use a similar method to find, in terms of \(n\), a lower bound for \(\int_0^1 (1 - x^2) \, dx\). [4]
Question 3:
AnswerMarks
3(a)( ) ( )
 1 1−x2dx< 1 + 1 1− 1 ++ 1 1−(n−1)2
n n n2 n n2
AnswerMarks Guidance
0M1 A1 Forms the sum of the areas of the rectangles. Three
terms, including last for A1. Allow use of sigma.
1 n−1 (n−1)n(2n−1)
1− r2 =1−
n3 6n3
AnswerMarks Guidance
r=1M1 n−1
Applies r2 =1(n−1)n(2n−1).
6
r=1
6n2 −(n−1)(2n−1) 4n2 +3n−1
=
AnswerMarks Guidance
6n2 6n2A1 AG. [Can get this A1 without previous A1.]
4
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
3(b)( )
( ) ( )
 1 1−x2dx> 1 1− 1 + 1 1−22 ++ 1 1− ( n−1 )2
n n2 n n2 n n2
AnswerMarks Guidance
0M1 A1 Forms the sum of the areas of appropriate rectangles.
n ( )
Three terms, including last for A1. Allow 1 1− r2 .
n n2
r=1
n−1 1 n−1 n−1 (n−1)n(2n−1)
− r2 = −
n n3 n 6n3
AnswerMarks Guidance
r=1M1 n−1
Applies r2 =1(n−1)n(2n−1).
6
r=1
6n(n−1)−(n−1)(2n−1) 4n2 −3n−1
=
AnswerMarks Guidance
6n2 6n2A1 4n2 +3n−1 1
SC: − scores 2/4.
6n2 n
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 3:
--- 3(a) ---
3(a) | ( ) ( )
 1 1−x2dx< 1 + 1 1− 1 ++ 1 1−(n−1)2
n n n2 n n2
0 | M1 A1 | Forms the sum of the areas of the rectangles. Three
terms, including last for A1. Allow use of sigma.
1 n−1 (n−1)n(2n−1)
1− r2 =1−
n3 6n3
r=1 | M1 | n−1
Applies r2 =1(n−1)n(2n−1).
6
r=1
6n2 −(n−1)(2n−1) 4n2 +3n−1
=
6n2 6n2 | A1 | AG. [Can get this A1 without previous A1.]
4
Question | Answer | Marks | Guidance
--- 3(b) ---
3(b) | ( )
( ) ( )
 1 1−x2dx> 1 1− 1 + 1 1−22 ++ 1 1− ( n−1 )2
n n2 n n2 n n2
0 | M1 A1 | Forms the sum of the areas of appropriate rectangles.
n ( )
Three terms, including last for A1. Allow 1 1− r2 .
n n2
r=1
n−1 1 n−1 n−1 (n−1)n(2n−1)
− r2 = −
n n3 n 6n3
r=1 | M1 | n−1
Applies r2 =1(n−1)n(2n−1).
6
r=1
6n(n−1)−(n−1)(2n−1) 4n2 −3n−1
=
6n2 6n2 | A1 | 4n2 +3n−1 1
SC: − scores 2/4.
6n2 n
4
Question | Answer | Marks | Guidance
\includegraphics{figure_3}

The diagram shows the curve with equation $y = 1 - x^2$ for $0 \leq x \leq 1$, 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 the rectangles, show that

$$\int_0^1 (1 - x^2) \, dx < \frac{4n^2 + 3n - 1}{6n^2}.$$ [4]

\item Use a similar method to find, in terms of $n$, a lower bound for $\int_0^1 (1 - x^2) \, dx$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q3 [8]}}
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