June 2025 Paper 1 Q14
14 The graph of \(y = 4x\sin 2x\) for \(0 \leqslant x \leqslant \pi\) is shown below.
Show that the shaded area enclosed by the \(x\)-axis and the curve is \(k\pi\), where \(k\) is an integer to be found.
Fully justify your answer. [8 marks]
| Scheme | Marks | AO |
|---|---|---|
| Begins integration by parts by writing \(u = 4x \qquad v^{\prime} = \sin 2x\) \(u^{\prime} = 4 \qquad v = A\cos 2x\) PI by \(4Ax\cos 2x - 4A\displaystyle\int (\cos 2x)\,\mathrm{d}x\) Or \(-2x\cos 2x + \sin 2x\,(+c)\) Or \(u = \sin 2x \qquad v^{\prime} = 4x\) \(u^{\prime} = A\cos 2x \qquad v = 2x^2\) PI by \(2x^2\sin 2x - 2A\displaystyle\int \left(x^2\cos 2x\right)\mathrm{d}x\) In either approach the constant 4 may be contained within the \(u\) or \(v\) terms or taken out as a factor. | M1 | 3.1a |
| Substitutes their \(u, u^{\prime}, v, v^{\prime}\) of either of the above forms into the integration by parts formula PI by \(-2x\cos 2x + \sin 2x\,(+c)\) | M1 | 1.1a |
| Applies IBP correctly to obtain \(\displaystyle\int 4x\sin 2x\,\mathrm{d}x = -\frac{4x}{2}\cos 2x - \int -\frac{4}{2}\cos 2x\,(\mathrm{d}x)\) PI by \(-2x\cos 2x + \sin 2x\,(+c)\) | A1 | 1.1b |
| Completes integration to obtain \(-2x\cos 2x + \sin 2x\,(+c)\) | A1 | 1.1b |
| Deduces \(x = \dfrac{\pi}{2}\) at intersection with \(x\)-axis. Accept \(\dfrac{\pi}{2}\) labelled on the diagram, written as a limit in their integral or seen as the solution to the equation \(4x\sin 2x = 0\) | B1 | 2.2a |
| Evaluates their integral over two separate intervals. | M1 | 3.1a |
| Demonstrates \(\big[-2x\cos 2x + \sin 2x\big]_{0}^{\frac{\pi}{2}} = \pi\) Or \(\big[-2x\cos 2x + \sin 2x\big]_{\frac{\pi}{2}}^{\pi} = -2\pi - \pi = -3\pi\) | M1 | 1.1a |
| Completes reasoned argument to show the area is \(4\pi\) | R1 | 2.1 |
| (8 marks) |
Typical solution
\[\int 4x\sin 2x\,\mathrm{d}x\]\[u = 4x \qquad u^{\prime} = 4\]\[v^{\prime} = \sin 2x \qquad v = -\frac{1}{2}\cos 2x\]\[\int 4x\sin 2x\,\mathrm{d}x = -\frac{4x}{2}\cos 2x - \int -\frac{4}{2}\cos 2x\,\mathrm{d}x\]\[= -2x\cos 2x + \sin 2x + c\]area above
\[\big[-2x\cos 2x + \sin 2x\big]_{0}^{\frac{\pi}{2}} = \pi\]area below
\[\big[-2x\cos 2x + \sin 2x\big]_{\frac{\pi}{2}}^{\pi} = -2\pi - \pi = -3\pi\]\[\pi + 3\pi = 4\pi\]