| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Stationary points then area/volume |
| Difficulty | Standard +0.3 This is a structured multi-part question testing standard C3 techniques: product rule differentiation (routine), integration by parts (standard application with x^{-2} ln x), and applying these to find a stationary point and area. Each part scaffolds the next, requiring competent execution of familiar methods but no novel insight or complex problem-solving. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.08e Area between curve and x-axis: using definite integrals1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = x^{-2}\ln x \Rightarrow \dfrac{dy}{dx} = x^{-2}\cdot\dfrac{1}{x} - 2x^{-3}\ln x\) | M1, A1 A1 | Use of product or quotient rule; each term |
| \(= \dfrac{1-2\ln x}{x^3}\) | A1 | Convincing argument \(x^{-2}\times\dfrac{1}{x} = x^{-3}\); AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int x^{-2}\ln x\, dx\); let \(u = \ln x\), \(dv = x^{-2}\) | M1 | Attempt at integration by parts |
| \(du = \dfrac{1}{x}\), \(v = -x^{-1}\) | A1 | |
| \(\int = -\dfrac{1}{x}\ln x + \int x^{-2}\,dx\) | A1 | |
| \(= -\dfrac{1}{x}\ln x - \dfrac{1}{x}\ (+c)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| At \(A\), \(\dfrac{dy}{dx} = 0 \Rightarrow 1 - 2\ln x = 0\) | ||
| \(\ln x = \dfrac{1}{2}\) | M1 | Attempt at \(\ln x = k\) |
| \(x = e^{\frac{1}{2}}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = \left[-\dfrac{1}{x}(\ln x + 1)\right]_1^5\) | M1 | \(R = \Big[\text{Their (b)}\Big]_1^5\) |
| \(= -\dfrac{1}{5}(\ln 5 + 1) + (\ln 1 + 1)\) | A1 | OE |
| \(= \dfrac{1}{5}(4 - \ln 5)\) | A1 | Convincing argument; AG |
## Question 9:
**Part (a):**
| $y = x^{-2}\ln x \Rightarrow \dfrac{dy}{dx} = x^{-2}\cdot\dfrac{1}{x} - 2x^{-3}\ln x$ | M1, A1 A1 | Use of product or quotient rule; each term |
| $= \dfrac{1-2\ln x}{x^3}$ | A1 | Convincing argument $x^{-2}\times\dfrac{1}{x} = x^{-3}$; AG |
**Part (b):**
| $\int x^{-2}\ln x\, dx$; let $u = \ln x$, $dv = x^{-2}$ | M1 | Attempt at integration by parts |
| $du = \dfrac{1}{x}$, $v = -x^{-1}$ | A1 | |
| $\int = -\dfrac{1}{x}\ln x + \int x^{-2}\,dx$ | A1 | |
| $= -\dfrac{1}{x}\ln x - \dfrac{1}{x}\ (+c)$ | A1 | |
**Part (c)(i):**
| At $A$, $\dfrac{dy}{dx} = 0 \Rightarrow 1 - 2\ln x = 0$ | | |
| $\ln x = \dfrac{1}{2}$ | M1 | Attempt at $\ln x = k$ |
| $x = e^{\frac{1}{2}}$ | A1 | |
**Part (c)(ii):**
| $R = \left[-\dfrac{1}{x}(\ln x + 1)\right]_1^5$ | M1 | $R = \Big[\text{Their (b)}\Big]_1^5$ |
| $= -\dfrac{1}{5}(\ln 5 + 1) + (\ln 1 + 1)$ | A1 | OE |
| $= \dfrac{1}{5}(4 - \ln 5)$ | A1 | Convincing argument; AG |9
\begin{enumerate}[label=(\alph*)]
\item Given that $y = x ^ { - 2 } \ln x$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 - 2 \ln x } { x ^ { 3 } }$.
\item Using integration by parts, find $\int x ^ { - 2 } \ln x \mathrm {~d} x$.
\item The sketch shows the graph of $y = x ^ { - 2 } \ln x$.\\
\includegraphics[max width=\textwidth, alt={}, center]{908f530c-076d-47b1-90dd-38dbfe44f898-06_604_1045_687_536}
\begin{enumerate}[label=(\roman*)]
\item Using the answer to part (a), find, in terms of e, the $x$-coordinate of the stationary point $A$.
\item The region $R$ is bounded by the curve, the $x$-axis and the line $x = 5$. Using your answer to part (b), show that the area of $R$ is
$$\frac { 1 } { 5 } ( 4 - \ln 5 )$$
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2006 Q9 [14]}}