| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Normal/tangent then area with parts |
| Difficulty | Standard +0.3 This is a straightforward two-part question combining standard differentiation (product rule) with integration by parts. Part (a) requires routine application of product and chain rules to find the tangent equation. Part (b) is a textbook integration by parts problem with clear structure (x·cos 2x) requiring one application of the technique. While it involves multiple steps, each step follows standard procedures without requiring problem-solving insight or novel approaches, making it slightly easier than the average A-level question. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dy}{dx} = 4\cos 2x - 8x\sin 2x\) | M1 A1 | Product rule, both terms |
| At \(x = \frac{\pi}{4}\): \(y = 4 \cdot \frac{\pi}{4} \cdot \cos\frac{\pi}{2} = 0\) | B1 | Point on curve |
| Gradient \(= 4\cos\frac{\pi}{2} - 8\cdot\frac{\pi}{4}\cdot\sin\frac{\pi}{2} = 0 - 2\pi = -2\pi\) | A1 | Correct gradient |
| Tangent: \(y = -2\pi\left(x - \frac{\pi}{4}\right)\) i.e. \(y = -2\pi x + \frac{\pi^2}{2}\) | A1 | Correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\int 4x\cos 2x\, dx\): let \(u=4x\), \(\frac{dv}{dx}=\cos 2x\) | M1 | Integration by parts stated/used |
| \(= 4x \cdot \frac{\sin 2x}{2} - \int 4 \cdot \frac{\sin 2x}{2}\, dx\) | A1 | Correct first stage |
| \(= 2x\sin 2x + \cos 2x\) | A1 | Correct integration |
| \(\left[2x\sin 2x + \cos 2x\right]_0^{\pi/4}\) | M1 | Apply limits |
| \(= \left(\frac{\pi}{2}\cdot 1 + 0\right) - (0 + 1) = \frac{\pi}{2} - 1\) | A1 | Exact answer |
## Question 7:
**Part (a):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 4\cos 2x - 8x\sin 2x$ | M1 A1 | Product rule, both terms |
| At $x = \frac{\pi}{4}$: $y = 4 \cdot \frac{\pi}{4} \cdot \cos\frac{\pi}{2} = 0$ | B1 | Point on curve |
| Gradient $= 4\cos\frac{\pi}{2} - 8\cdot\frac{\pi}{4}\cdot\sin\frac{\pi}{2} = 0 - 2\pi = -2\pi$ | A1 | Correct gradient |
| Tangent: $y = -2\pi\left(x - \frac{\pi}{4}\right)$ i.e. $y = -2\pi x + \frac{\pi^2}{2}$ | A1 | Correct equation |
**Part (b):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int 4x\cos 2x\, dx$: let $u=4x$, $\frac{dv}{dx}=\cos 2x$ | M1 | Integration by parts stated/used |
| $= 4x \cdot \frac{\sin 2x}{2} - \int 4 \cdot \frac{\sin 2x}{2}\, dx$ | A1 | Correct first stage |
| $= 2x\sin 2x + \cos 2x$ | A1 | Correct integration |
| $\left[2x\sin 2x + \cos 2x\right]_0^{\pi/4}$ | M1 | Apply limits |
| $= \left(\frac{\pi}{2}\cdot 1 + 0\right) - (0 + 1) = \frac{\pi}{2} - 1$ | A1 | Exact answer |
---7 A curve has equation $y = 4 x \cos 2 x$.
\begin{enumerate}[label=(\alph*)]
\item Find an exact equation of the tangent to the curve at the point on the curve where
$$x = \frac { \pi } { 4 }$$
\item The region shaded on the diagram below is bounded by the curve $y = 4 x \cos 2 x$ and the $x$-axis from $x = 0$ to $x = \frac { \pi } { 4 }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{b8614dd6-2197-40c3-a673-5bef3e3653a5-8_487_878_740_591}
By using integration by parts, find the exact value of the area of the shaded region.\\
(5 marks)
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{b8614dd6-2197-40c3-a673-5bef3e3653a5-8_1275_1717_1432_150}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2013 Q7 [10]}}