Question 12 - A-Level Maths

Edexcel PMT Mocks — Question 12 5 marks

Exam Board	Edexcel
Module	PMT Mocks (PMT Mocks)
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Integration by Parts
Type	Area under curve requiring parts
Difficulty	Standard +0.8 This requires integration by parts applied twice to x²ln(2x), careful handling of logarithmic terms and their derivatives, evaluation at non-standard limits (e²/2 and e²), and algebraic manipulation to reach a specific exact form. More demanding than a routine parts question due to the x² term, logarithm coefficient, and the need to verify a given answer precisely.
Spec	1.08e Area between curve and x-axis: using definite integrals 1.08i Integration by parts

12. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{48f9a252-61a2-491d-94d0-8470aee96942-18_1038_1271_244_440} \captionsetup{labelformat=empty} \caption{Figure 6}

\end{figure} The figure 6 shows a sketch of the curve with equation $$y = x ^ { 2 } \ln 2 x$$ The finite region $R$, shown shaded in figure 5, is bounded by the line with equation $x = \frac { e ^ { 2 } } { 2 }$, the curve $C$, the line with equation $x = e ^ { 2 }$ and the $x$-axis.
Show that the exact value of the area of region $R$ is $\frac { e ^ { 6 } } { 72 } ( 35 + 24 \ln 2 )$.

Show mark scheme Show mark scheme source

The finite region $R$, shown shaded in figure 5, is bounded by the line with equation $x = \frac{e^2}{2}$, the curve $C$, the line with equation $x = e^2$ and the $x$-axis. Show that the exact value of the area of region $R$ is $\frac{e^6}{72}(35 + 24\ln 2)$.

Answer	Marks
M1	Attempts by parts to reach a form $\int x^2\ln 2x \, dx = \pm ax^3\ln 2x \pm b\int x^3 \times \frac{1}{x} \, dx$ where $a, b \neq 0$. If a formula is stated it must be correct. E.g. $u = \ln 2x \Rightarrow \frac{du}{dx} = \frac{1}{x}$, $\frac{dv}{dx} = x^2 \Rightarrow v = \frac{x^3}{3}$. $\int x^2\ln 2x \, dx = \pm ax^3\ln 2x \pm b\int x^3 \times \frac{1}{x} \, dx$.
dM1	Integrates $\int x^3 \times \frac{1}{x} \, dx$ to reach a form $\pm bx^3$.
A1	Correct answer only $\int x^2\ln 2x \, dx = \frac{x^3}{3}\ln 2x - \frac{x^3}{9}$.
M1	Uses correct limits correct way round in an integrated function to find the area of the region and attempts to simplify by using one log law correctly. E.g. $\left[\frac{x^3}{3}\ln 2x - \frac{x^3}{9}\right]_{e^2}^{e^2} = \left(\frac{

The finite region $R$, shown shaded in figure 5, is bounded by the line with equation $x = \frac{e^2}{2}$, the curve $C$, the line with equation $x = e^2$ and the $x$-axis. Show that the exact value of the area of region $R$ is $\frac{e^6}{72}(35 + 24\ln 2)$.

| | M1 | Attempts by parts to reach a form $\int x^2\ln 2x \, dx = \pm ax^3\ln 2x \pm b\int x^3 \times \frac{1}{x} \, dx$ where $a, b \neq 0$. If a formula is stated it must be correct. E.g. $u = \ln 2x \Rightarrow \frac{du}{dx} = \frac{1}{x}$, $\frac{dv}{dx} = x^2 \Rightarrow v = \frac{x^3}{3}$. $\int x^2\ln 2x \, dx = \pm ax^3\ln 2x \pm b\int x^3 \times \frac{1}{x} \, dx$. |
| --- | --- | --- |
| | dM1 | Integrates $\int x^3 \times \frac{1}{x} \, dx$ to reach a form $\pm bx^3$. |
| | A1 | Correct answer only $\int x^2\ln 2x \, dx = \frac{x^3}{3}\ln 2x - \frac{x^3}{9}$. |
| | M1 | Uses correct limits correct way round in an integrated function to find the area of the region and attempts to simplify by using one log law correctly. E.g. $\left[\frac{x^3}{3}\ln 2x - \frac{x^3}{9}\right]_{e^2}^{e^2} = \left(\frac{

Show LaTeX source

12.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{48f9a252-61a2-491d-94d0-8470aee96942-18_1038_1271_244_440}
\captionsetup{labelformat=empty}
\caption{Figure 6}
\end{center}
\end{figure}

The figure 6 shows a sketch of the curve with equation

$$y = x ^ { 2 } \ln 2 x$$

The finite region $R$, shown shaded in figure 5, is bounded by the line with equation $x = \frac { e ^ { 2 } } { 2 }$, the curve $C$, the line with equation $x = e ^ { 2 }$ and the $x$-axis.\\
Show that the exact value of the area of region $R$ is $\frac { e ^ { 6 } } { 72 } ( 35 + 24 \ln 2 )$.\\

\hfill \mbox{\textit{Edexcel PMT Mocks  Q12 [5]}}

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Q1 4 Q2 6 Q3 6 Q4 3 Q5 6 Q6 6 Q7 5 Q8 8 Q9 8 Q10 8 Q11 7 Q12 5 Q13 4 Q14 8 Q15 10 Q16 6

Answer	Marks
M1	Attempts by parts to reach a form \(\int x^2\ln 2x \, dx = \pm ax^3\ln 2x \pm b\int x^3 \times \frac{1}{x} \, dx\) where \(a, b \neq 0\). If a formula is stated it must be correct. E.g. \(u = \ln 2x \Rightarrow \frac{du}{dx} = \frac{1}{x}\), \(\frac{dv}{dx} = x^2 \Rightarrow v = \frac{x^3}{3}\). \(\int x^2\ln 2x \, dx = \pm ax^3\ln 2x \pm b\int x^3 \times \frac{1}{x} \, dx\).
dM1	Integrates \(\int x^3 \times \frac{1}{x} \, dx\) to reach a form \(\pm bx^3\).
A1	Correct answer only \(\int x^2\ln 2x \, dx = \frac{x^3}{3}\ln 2x - \frac{x^3}{9}\).
M1	Uses correct limits correct way round in an integrated function to find the area of the region and attempts to simplify by using one log law correctly. E.g. $\left[\frac{x^3}{3}\ln 2x - \frac{x^3}{9}\right]_{e^2}^{e^2} = \left(\frac{

Edexcel PMT Mocks — Question 12 5 marks

The finite region \(R\), shown shaded in figure 5, is bounded by the line with equation \(x = \frac{e^2}{2}\), the curve \(C\), the line with equation \(x = e^2\) and the \(x\)-axis. Show that the exact value of the area of region \(R\) is \(\frac{e^6}{72}(35 + 24\ln 2)\).

This paper (16 questions)