Question 10 - A-Level Maths

Edexcel Paper 2 Specimen — Question 10 9 marks

Exam Board	Edexcel
Module	Paper 2 (Paper 2)
Session	Specimen
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Parametric integration
Type	Parametric area under curve
Difficulty	Standard +0.3 This is a straightforward parametric area question requiring standard technique: finding dx/dt, setting up the integral with correct limits by solving for t-values, and integrating 1/(t+1) which gives a simple logarithm. The algebra is clean and the question guides students through each step, making it slightly easier than average.
Spec	1.03g Parametric equations: of curves and conversion to cartesian 1.08h Integration by substitution

10. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{659a0479-c8c6-418b-b8a9-67ad68474023-22_554_862_260_603} \captionsetup{labelformat=empty} \caption{Figure 4}

\end{figure} Figure 4 shows a sketch of the curve $C$ with parametric equations $$x = \ln ( t + 2 ) , \quad y = \frac { 1 } { t + 1 } , \quad t > - \frac { 2 } { 3 }$$

State the domain of values of $x$ for the curve $C$. The finite region $R$, shown shaded in Figure 4, is bounded by the curve $C$, the line with equation $x = \ln 2$, the $x$-axis and the line with equation $x = \ln 4$
Use calculus to show that the area of $R$ is $\ln \left( \frac { 3 } { 2 } \right)$.

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Question 10:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$x > \ln\left(\frac{4}{3}\right)$	B1	Correct answer

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Attempts to apply $\int y\frac{dx}{dt}dt$	M1
$\left\{\int y\frac{dx}{dt}dt =\right\} = \int\left(\frac{1}{t+1}\right)\left(\frac{1}{t+2}\right)dt$	A1
$\frac{1}{(t+1)(t+2)} \equiv \frac{A}{(t+1)}+\frac{B}{(t+2)} \Rightarrow 1 \equiv A(t+2)+B(t+1)$	M1
$\{A=1, B=-1\} \Rightarrow \frac{1}{(t+1)}-\frac{1}{(t+2)}$	A1
$\left\{\int\left(\frac{1}{(t+1)}-\frac{1}{(t+2)}\right)dt =\right\} \ln(t+1)-\ln(t+2)$	M1, A1
$\text{Area}(R) = \left[\ln(t+1)-\ln(t+2)\right]_0^2 = (\ln 3 - \ln 4)-(\ln 1 - \ln 2)$	M1
$= \ln 3 - \ln 4 + \ln 2 = \ln\left(\frac{(3)(2)}{4}\right) = \ln\left(\frac{6}{4}\right)$
$= \ln\left(\frac{3}{2}\right)$ *	A1*

Alt 1 (substitution $u = e^x - 1$):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Attempts $\int ydx = \int\frac{1}{e^x-2+1}dx = \int\frac{1}{e^x-1}dx$ with substitution $u=e^x-1$	M1
$\left\{\int ydx\right\} = \int\left(\frac{1}{u}\right)\left(\frac{1}{u+1}\right)du$	A1
$\frac{1}{u(u+1)} \equiv \frac{A}{u}+\frac{B}{(u+1)} \Rightarrow 1 \equiv A(u+1)+Bu$	M1
$\{A=1, B=-1\} \Rightarrow \frac{1}{u}-\frac{1}{(u+1)}$	A1
$\left\{\int\left(\frac{1}{u}-\frac{1}{(u+1)}\right)du =\right\} \ln u - \ln(u+1)$	M1, A1
$\text{Area}(R) = \left[\ln u - \ln(u+1)\right]_1^3 = (\ln 3-\ln 4)-(\ln 1-\ln 2)$	M1
$= \ln 3 - \ln 4 + \ln 2 = \ln\left(\frac{(3)(2)}{4}\right) = \ln\left(\frac{6}{4}\right) = \ln\left(\frac{3}{2}\right)$ *	A1*

Alt 2 (substitution $v = e^x$):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Attempts $\int ydx = \int\frac{1}{e^x-2+1}dx = \int\frac{1}{e^x-1}dx$ with substitution $v=e^x$	M1
$\left\{\int ydx\right\} = \int\left(\frac{1}{v-1}\right)\left(\frac{1}{v}\right)dv$	A1
$\frac{1}{(v-1)v} \equiv \frac{A}{(v-1)}+\frac{B}{v} \Rightarrow 1 \equiv Av+B(v-1)$	M1
$\{A=1, B=-1\} \Rightarrow \frac{1}{(v-1)}-\frac{1}{v}$	A1
$\left\{\int\left(\frac{1}{(v-1)}-\frac{1}{v}\right)dv =\right\} \ln(v-1)-\ln v$	M1, A1
$\text{Area}(R) = \left[\ln(v-1)-\ln v\right]_2^4 = (\ln 3-\ln 4)-(\ln 1-\ln 2)$	M1
$= \ln 3 - \ln 4 + \ln 2 = \ln\left(\frac{(3)(2)}{4}\right) = \ln\left(\frac{6}{4}\right) = \ln\left(\frac{3}{2}\right)$ *	A1*

Question 10:

Part (a)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Uses $x = \ln(t+2)$ with $t > -\frac{2}{3}$ to deduce domain $x > \ln\left(\frac{4}{3}\right)$	B1

Part (b)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Attempts parametric process OR Cartesian process with $u = e^x - 1$ or $v = e^x$	M1
$\int\left(\frac{1}{t+1}\right)\left(\frac{1}{t+2}\right)dt$ (parametric) OR $\int\left(\frac{1}{u}\right)\left(\frac{1}{u+1}\right)du$ with $u=e^x-1$ OR $\int\left(\frac{1}{v-1}\right)\left(\frac{1}{v}\right)dv$ with $v=e^x$	A1
Attempts partial fractions on $\frac{1}{(t+1)(t+2)}$, $\frac{1}{u(u+1)}$, or $\frac{1}{(v-1)v}$	M1
Correct partial fractions for their method	A1
Integrates to give $\pm\alpha\ln(t+1) \pm \beta\ln(t+2)$ OR $\pm\alpha\ln u \pm \beta\ln(u+1)$ OR $\pm\alpha\ln(v-1) \pm \beta\ln v$; $\alpha,\beta \neq 0$	M1
Correct integration for their method	A1
Applies correct limits and subtracts correct way round (parametric: 2 and 0 in $t$; Cartesian $u$: 3 and 1; Cartesian $v$: 4 and 2)	M1
Area of $R$ is $\ln\left(\frac{3}{2}\right)$	A1*	No errors seen in working

# Question 10:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x > \ln\left(\frac{4}{3}\right)$ | B1 | Correct answer |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to apply $\int y\frac{dx}{dt}dt$ | M1 | |
| $\left\{\int y\frac{dx}{dt}dt =\right\} = \int\left(\frac{1}{t+1}\right)\left(\frac{1}{t+2}\right)dt$ | A1 | |
| $\frac{1}{(t+1)(t+2)} \equiv \frac{A}{(t+1)}+\frac{B}{(t+2)} \Rightarrow 1 \equiv A(t+2)+B(t+1)$ | M1 | |
| $\{A=1, B=-1\} \Rightarrow \frac{1}{(t+1)}-\frac{1}{(t+2)}$ | A1 | |
| $\left\{\int\left(\frac{1}{(t+1)}-\frac{1}{(t+2)}\right)dt =\right\} \ln(t+1)-\ln(t+2)$ | M1, A1 | |
| $\text{Area}(R) = \left[\ln(t+1)-\ln(t+2)\right]_0^2 = (\ln 3 - \ln 4)-(\ln 1 - \ln 2)$ | M1 | |
| $= \ln 3 - \ln 4 + \ln 2 = \ln\left(\frac{(3)(2)}{4}\right) = \ln\left(\frac{6}{4}\right)$ | | |
| $= \ln\left(\frac{3}{2}\right)$ * | A1* | |

**Alt 1** (substitution $u = e^x - 1$):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $\int ydx = \int\frac{1}{e^x-2+1}dx = \int\frac{1}{e^x-1}dx$ with substitution $u=e^x-1$ | M1 | |
| $\left\{\int ydx\right\} = \int\left(\frac{1}{u}\right)\left(\frac{1}{u+1}\right)du$ | A1 | |
| $\frac{1}{u(u+1)} \equiv \frac{A}{u}+\frac{B}{(u+1)} \Rightarrow 1 \equiv A(u+1)+Bu$ | M1 | |
| $\{A=1, B=-1\} \Rightarrow \frac{1}{u}-\frac{1}{(u+1)}$ | A1 | |
| $\left\{\int\left(\frac{1}{u}-\frac{1}{(u+1)}\right)du =\right\} \ln u - \ln(u+1)$ | M1, A1 | |
| $\text{Area}(R) = \left[\ln u - \ln(u+1)\right]_1^3 = (\ln 3-\ln 4)-(\ln 1-\ln 2)$ | M1 | |
| $= \ln 3 - \ln 4 + \ln 2 = \ln\left(\frac{(3)(2)}{4}\right) = \ln\left(\frac{6}{4}\right) = \ln\left(\frac{3}{2}\right)$ * | A1* | |

**Alt 2** (substitution $v = e^x$):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $\int ydx = \int\frac{1}{e^x-2+1}dx = \int\frac{1}{e^x-1}dx$ with substitution $v=e^x$ | M1 | |
| $\left\{\int ydx\right\} = \int\left(\frac{1}{v-1}\right)\left(\frac{1}{v}\right)dv$ | A1 | |
| $\frac{1}{(v-1)v} \equiv \frac{A}{(v-1)}+\frac{B}{v} \Rightarrow 1 \equiv Av+B(v-1)$ | M1 | |
| $\{A=1, B=-1\} \Rightarrow \frac{1}{(v-1)}-\frac{1}{v}$ | A1 | |
| $\left\{\int\left(\frac{1}{(v-1)}-\frac{1}{v}\right)dv =\right\} \ln(v-1)-\ln v$ | M1, A1 | |
| $\text{Area}(R) = \left[\ln(v-1)-\ln v\right]_2^4 = (\ln 3-\ln 4)-(\ln 1-\ln 2)$ | M1 | |
| $= \ln 3 - \ln 4 + \ln 2 = \ln\left(\frac{(3)(2)}{4}\right) = \ln\left(\frac{6}{4}\right) = \ln\left(\frac{3}{2}\right)$ * | A1* | |

## Question 10:

### Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $x = \ln(t+2)$ with $t > -\frac{2}{3}$ to deduce domain $x > \ln\left(\frac{4}{3}\right)$ | B1 | |

### Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts parametric process OR Cartesian process with $u = e^x - 1$ or $v = e^x$ | M1 | |
| $\int\left(\frac{1}{t+1}\right)\left(\frac{1}{t+2}\right)dt$ (parametric) OR $\int\left(\frac{1}{u}\right)\left(\frac{1}{u+1}\right)du$ with $u=e^x-1$ OR $\int\left(\frac{1}{v-1}\right)\left(\frac{1}{v}\right)dv$ with $v=e^x$ | A1 | |
| Attempts partial fractions on $\frac{1}{(t+1)(t+2)}$, $\frac{1}{u(u+1)}$, or $\frac{1}{(v-1)v}$ | M1 | |
| Correct partial fractions for their method | A1 | |
| Integrates to give $\pm\alpha\ln(t+1) \pm \beta\ln(t+2)$ OR $\pm\alpha\ln u \pm \beta\ln(u+1)$ OR $\pm\alpha\ln(v-1) \pm \beta\ln v$; $\alpha,\beta \neq 0$ | M1 | |
| Correct integration for their method | A1 | |
| Applies correct limits and subtracts correct way round (parametric: 2 and 0 in $t$; Cartesian $u$: 3 and 1; Cartesian $v$: 4 and 2) | M1 | |
| Area of $R$ is $\ln\left(\frac{3}{2}\right)$ | A1* | No errors seen in working |

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Show LaTeX source

10.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{659a0479-c8c6-418b-b8a9-67ad68474023-22_554_862_260_603}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}

Figure 4 shows a sketch of the curve $C$ with parametric equations

$$x = \ln ( t + 2 ) , \quad y = \frac { 1 } { t + 1 } , \quad t > - \frac { 2 } { 3 }$$
\begin{enumerate}[label=(\alph*)]
\item State the domain of values of $x$ for the curve $C$.

The finite region $R$, shown shaded in Figure 4, is bounded by the curve $C$, the line with equation $x = \ln 2$, the $x$-axis and the line with equation $x = \ln 4$
\item Use calculus to show that the area of $R$ is $\ln \left( \frac { 3 } { 2 } \right)$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel Paper 2  Q10 [9]}}

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Q1 4 Q2 5 Q3 4 Q4 6 Q5 5 Q6 9 Q7 12 Q8 7 Q9 9 Q10 9 Q11 5 Q12 7 Q13 10 Q14 8

Edexcel Paper 2 Specimen — Question 10 9 marks

Question 10:

Part (a):

Part (b):

Alt 1 (substitution \(u = e^x - 1\)):

Alt 2 (substitution \(v = e^x\)):

Question 10:

Part (a)

Part (b)

This paper (14 questions)