Question 3 - A-Level Maths

Edexcel C4 — Question 3 10 marks

Exam Board	Edexcel
Module	C4 (Core Mathematics 4)
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Numerical integration
Type	Complete table then apply trapezium rule
Difficulty	Standard +0.3 This is a straightforward C4 trapezium rule question with standard bookwork. Part (a) requires simple calculator evaluation, parts (b)(i-iii) are routine applications of the trapezium rule formula with different strip numbers, and part (c) asks for pattern recognition to estimate the limit. The volume of revolution in question 4 is also standard C4 content. Slightly easier than average due to the mechanical, procedural nature with no conceptual challenges.
Spec	1.03g Parametric equations: of curves and conversion to cartesian 1.07s Parametric and implicit differentiation 1.09f Trapezium rule: numerical integration

3. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{3056ad22-f87b-46c3-86cf-d46939927465-04_560_1059_146_406} \captionsetup{labelformat=empty} \caption{Figure 1}

\end{figure} Figure 1 shows the curve with equation $y = \ln ( 2 + \cos x ) , 0 \leq x \leq \pi$.

Complete the table below for points on the curve, giving the $y$ values to 4 decimal places.
Giving your answers to 3 decimal places, find estimates for the area of the region bounded by the curve and the coordinate axes using the trapezium rule with
1. 1 strip,
2. 2 strips,
3. 4 strips.
Making your reasoning clear, suggest a value to 2 decimal places for the actual area of the region bounded by the curve and the coordinate axes.
$x$ 0 $\frac { \pi } { 4 }$ $\frac { \pi } { 2 }$ $\frac { 3 \pi } { 4 }$ $\pi$
$y$ 1.0986 0
1. continued
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3056ad22-f87b-46c3-86cf-d46939927465-06_563_983_146_379} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} Figure 2 shows the curve with parametric equations $$x = \tan \theta , \quad y = \cos ^ { 2 } \theta , \quad - \frac { \pi } { 2 } < \theta < \frac { \pi } { 2 }$$ The shaded region bounded by the curve, the $x$-axis and the lines $x = - 1$ and $x = 1$ is rotated through $2 \pi$ radians about the $x$-axis.

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
(a) $0.9959, 0.6931, 0.2569$ (4dp)	B2
(b)(i) $\frac{1}{2} \times \pi \times (1.0986 + 0) = 1.726$ (3dp)	B1 M1 A1
(ii) $\frac{1}{2} \times \frac{2}{3} \times [1.0986 + 0 + 2(0.6931)] = 1.952$ (3dp)	M1 A1
(iii) $\frac{1}{2} \times \frac{2}{3} \times [1.0986 + 0 + 2(0.9959 + 0.6931 + 0.2569)] = 1.960$ (3dp)	A1
(c) $1.96;$ large change from 1 to 2 strips but from 2 to 4 strips the change is less than 0.01 so the error in 4 strip value is likely to be less than 0.005	B2	(10)

**(a)** $0.9959, 0.6931, 0.2569$ (4dp) | B2 |

**(b)(i)** $\frac{1}{2} \times \pi \times (1.0986 + 0) = 1.726$ (3dp) | B1 M1 A1 |

**(ii)** $\frac{1}{2} \times \frac{2}{3} \times [1.0986 + 0 + 2(0.6931)] = 1.952$ (3dp) | M1 A1 |

**(iii)** $\frac{1}{2} \times \frac{2}{3} \times [1.0986 + 0 + 2(0.9959 + 0.6931 + 0.2569)] = 1.960$ (3dp) | A1 |

**(c)** $1.96;$ large change from 1 to 2 strips but from 2 to 4 strips the change is less than 0.01 so the error in 4 strip value is likely to be less than 0.005 | B2 | (10)

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Show LaTeX source

3.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{3056ad22-f87b-46c3-86cf-d46939927465-04_560_1059_146_406}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 shows the curve with equation $y = \ln ( 2 + \cos x ) , 0 \leq x \leq \pi$.\\
(a) Complete the table below for points on the curve, giving the $y$ values to 4 decimal places.\\
(b) Giving your answers to 3 decimal places, find estimates for the area of the region bounded by the curve and the coordinate axes using the trapezium rule with\\
(i) 1 strip,\\
(ii) 2 strips,\\
(iii) 4 strips.\\
(c) Making your reasoning clear, suggest a value to 2 decimal places for the actual area of the region bounded by the curve and the coordinate axes.

\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 0 & $\frac { \pi } { 4 }$ & $\frac { \pi } { 2 }$ & $\frac { 3 \pi } { 4 }$ & $\pi$ \\
\hline
$y$ & 1.0986 &  &  &  & 0 \\
\hline
\end{tabular}
\end{center}

\begin{enumerate}
  \item continued
  \item 
\end{enumerate}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{3056ad22-f87b-46c3-86cf-d46939927465-06_563_983_146_379}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}

Figure 2 shows the curve with parametric equations

$$x = \tan \theta , \quad y = \cos ^ { 2 } \theta , \quad - \frac { \pi } { 2 } < \theta < \frac { \pi } { 2 }$$

The shaded region bounded by the curve, the $x$-axis and the lines $x = - 1$ and $x = 1$ is rotated through $2 \pi$ radians about the $x$-axis.\\

\hfill \mbox{\textit{Edexcel C4  Q3 [10]}}

This paper (7 questions)

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Q1 8 Q2 9 Q3 10 Q4 11 Q5 11 Q6 11 Q7 15

\(x\)	0	\(\frac { \pi } { 4 }\)	\(\frac { \pi } { 2 }\)	\(\frac { 3 \pi } { 4 }\)	\(\pi\)
\(y\)	1.0986				0

Answer	Marks	Guidance
(a) \(0.9959, 0.6931, 0.2569\) (4dp)	B2
(b)(i) \(\frac{1}{2} \times \pi \times (1.0986 + 0) = 1.726\) (3dp)	B1 M1 A1
(ii) \(\frac{1}{2} \times \frac{2}{3} \times [1.0986 + 0 + 2(0.6931)] = 1.952\) (3dp)	M1 A1
(iii) \(\frac{1}{2} \times \frac{2}{3} \times [1.0986 + 0 + 2(0.9959 + 0.6931 + 0.2569)] = 1.960\) (3dp)	A1
(c) \(1.96;\) large change from 1 to 2 strips but from 2 to 4 strips the change is less than 0.01 so the error in 4 strip value is likely to be less than 0.005	B2	(10)