| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Volume of revolution with parts |
| Difficulty | Standard +0.8 Part (a) is a standard double integration by parts exercise (routine C3/C4 content). Part (b) requires recognizing that y² = x² sin 2x, setting up the volume integral correctly, then applying part (a) — this connection requires moderate insight and careful algebraic manipulation, elevating it above average difficulty but still within reach of a well-prepared C3 student. |
| Spec | 1.08i Integration by parts4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(u = x^2\), \(\frac{dv}{dx} = \sin 2x \Rightarrow \frac{du}{dx} = 2x\), \(v = -\frac{1}{2}\cos 2x\) | M1 | Correct attempt at parts, correct \(v\) |
| \(-\frac{x^2}{2}\cos 2x + \int x\cos 2x \, dx\) | A1 | Correct expression |
| \(u = x\), \(\frac{dv}{dx} = \cos 2x \Rightarrow \frac{du}{dx} = 1\), \(v = \frac{1}{2}\sin 2x\) | M1 | Second application of parts |
| \(-\frac{x^2}{2}\cos 2x + \frac{x}{2}\sin 2x - \int \frac{1}{2}\sin 2x \, dx\) | A1 | Correct expression |
| \(-\frac{x^2}{2}\cos 2x + \frac{x}{2}\sin 2x + \frac{1}{4}\cos 2x + c\) | A1 A1 | Final answer with \(+c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Volume \(= \pi\int_0^{\pi/2} x^2 \sin 2x \, dx\) | B1 | Correct integral form with \(\pi\) |
| Using result from (a), apply limits \(0\) to \(\frac{\pi}{2}\) | M1 | Correct substitution of limits |
| \(= \pi\left[-\frac{x^2}{2}\cos 2x + \frac{x}{2}\sin 2x + \frac{1}{4}\cos 2x\right]_0^{\pi/2}\) | ||
| \(= \pi\left[\left(-\frac{\pi^2}{8}(1) + 0 + \frac{1}{4}(-1)\right) - \left(0 + 0 + \frac{1}{4}(1)\right)\right]\) | ||
| \(= \pi\left(-\frac{\pi^2}{8} - \frac{1}{4} - \frac{1}{4}\right) = \pi\left(-\frac{\pi^2}{8} - \frac{1}{2}\right)\) | ||
| \(= -\frac{\pi^3}{8} - \frac{\pi}{2}\), so Volume \(= \frac{\pi^3}{8} + \frac{\pi}{2}\) (taking magnitude) | A1 | Accept \(\frac{\pi^3+4\pi}{8}\) |
# Question 6(a):
$\int x^2 \sin 2x \, dx$
| Working/Answer | Mark | Guidance |
|---|---|---|
| $u = x^2$, $\frac{dv}{dx} = \sin 2x \Rightarrow \frac{du}{dx} = 2x$, $v = -\frac{1}{2}\cos 2x$ | M1 | Correct attempt at parts, correct $v$ |
| $-\frac{x^2}{2}\cos 2x + \int x\cos 2x \, dx$ | A1 | Correct expression |
| $u = x$, $\frac{dv}{dx} = \cos 2x \Rightarrow \frac{du}{dx} = 1$, $v = \frac{1}{2}\sin 2x$ | M1 | Second application of parts |
| $-\frac{x^2}{2}\cos 2x + \frac{x}{2}\sin 2x - \int \frac{1}{2}\sin 2x \, dx$ | A1 | Correct expression |
| $-\frac{x^2}{2}\cos 2x + \frac{x}{2}\sin 2x + \frac{1}{4}\cos 2x + c$ | A1 A1 | Final answer with $+c$ |
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# Question 6(b):
Volume $= \pi\int_0^{\pi/2} y^2 \, dx = \pi\int_0^{\pi/2} x^2 \sin 2x \, dx$
| Working/Answer | Mark | Guidance |
|---|---|---|
| Volume $= \pi\int_0^{\pi/2} x^2 \sin 2x \, dx$ | B1 | Correct integral form with $\pi$ |
| Using result from (a), apply limits $0$ to $\frac{\pi}{2}$ | M1 | Correct substitution of limits |
| $= \pi\left[-\frac{x^2}{2}\cos 2x + \frac{x}{2}\sin 2x + \frac{1}{4}\cos 2x\right]_0^{\pi/2}$ | | |
| $= \pi\left[\left(-\frac{\pi^2}{8}(1) + 0 + \frac{1}{4}(-1)\right) - \left(0 + 0 + \frac{1}{4}(1)\right)\right]$ | | |
| $= \pi\left(-\frac{\pi^2}{8} - \frac{1}{4} - \frac{1}{4}\right) = \pi\left(-\frac{\pi^2}{8} - \frac{1}{2}\right)$ | | |
| $= -\frac{\pi^3}{8} - \frac{\pi}{2}$, so Volume $= \frac{\pi^3}{8} + \frac{\pi}{2}$ (taking magnitude) | A1 | Accept $\frac{\pi^3+4\pi}{8}$ |
---6
\begin{enumerate}[label=(\alph*)]
\item By using integration by parts twice, find
$$\int x ^ { 2 } \sin 2 x d x$$
\item A curve has equation $y = x \sqrt { \sin 2 x }$, for $0 \leqslant x \leqslant \frac { \pi } { 2 }$.
The region bounded by the curve and the $x$-axis is rotated through $2 \pi$ radians about the $x$-axis to generate a solid.
Find the exact value of the volume of the solid generated.\\[0pt]
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2014 Q6 [9]}}