| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| 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 covering standard C3 techniques: product rule differentiation, integration by parts with ln x, and applying these to find stationary points and areas. Each part guides the student through the method with clear targets ('show that'), making it slightly easier than average despite requiring multiple techniques. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08i Integration by parts |
| Answer | Marks |
|---|---|
| M1 \(\pi\int x^2\,dy\), M1 correct expression for \(x^2\), M1 integrating, A1 correct integration, A1 \(\frac{160\pi}{3}\) |
## Question 9:
From $16x^2-(y-8)^2=32$: $x^2 = \frac{32+(y-8)^2}{16}$
$$V = \pi\int_0^{16} x^2\, dy = \frac{\pi}{16}\int_0^{16}\left[32+(y-8)^2\right]dy$$
$$= \frac{\pi}{16}\left[32y + \frac{(y-8)^3}{3}\right]_0^{16}$$
$$= \frac{\pi}{16}\left[\left(512 + \frac{512}{3}\right)-\left(0 - \frac{512}{3}\right)\right] = \frac{\pi}{16}\left[512 + \frac{1024}{3}\right] = \frac{\pi}{16}\cdot\frac{2560}{3} = \frac{160\pi}{3}$$
| **M1** $\pi\int x^2\,dy$, **M1** correct expression for $x^2$, **M1** integrating, **A1** correct integration, **A1** $\frac{160\pi}{3}$
---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]{9aac4ee4-2435-4315-a87d-fe9fa8e15665-007_593_1034_696_543}
\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 Q9}}