| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Show that integral equals expression |
| Difficulty | Moderate -0.3 This is a straightforward integration by parts question with a standard setup (polynomial × trig function). Students apply the formula once with u=x and dv=sin(4x)dx, then evaluate at the given limits. The arithmetic is clean and the result is provided, making verification easy. Slightly easier than average due to its routine nature and single application of the technique. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(u = x\), \(v' = \sin 4x\), \(u' = 1\), \(v = A\cos 4x\) → begins integration by parts | M1 | PI by \(Ax\cos 4x - A\int(\cos 4x)\,dx\) |
| Selects correct method: \(u=x\), \(v'=\sin 4x\), \(u'=1\), \(v = A\cos 4x\); PI by \(Ax\cos 4x - A\int(\cos 4x)\,dx\) | M1 | |
| Substitutes \(u, u', v, v'\) into integration by parts formula: \(Px\cos 4x - P\int(\cos 4x)\,dx\) | M1 | |
| \(-\frac{1}{4}x\cos 4x - \frac{1}{4}\int(-\cos 4x)\,dx\) | A1 | Condone missing \(dx\); PI by \(-\frac{1}{4}x\cos 4x + \frac{1}{16}\sin 4x\) |
| Completes integration by parts to obtain \(-\frac{1}{4}x\cos 4x + B\sin 4x\) with \(B \neq \pm 1\) | M1 | |
| \(\left[-\frac{1}{4}x\cos 4x + \frac{1}{16}\sin 4x\right]_0^{\frac{\pi}{2}}\); substitutes correct limits explicitly into \(-\frac{1}{4}x\cos 4x + \frac{1}{16}\sin 4x\) | R1 | |
| \(= \left(-\frac{\pi}{8}\cos\frac{4\pi}{2} + \frac{1}{16}\sin\frac{4\pi}{2}\right) - \left(0\times\cos 0 + \frac{1}{16}\sin 0\right) = -\frac{\pi}{8}\) | Accept \(\left(-\frac{\pi}{8}\cos 2\pi + \frac{1}{16}\sin 2\pi\right) - 0\); AG |
## Question 8:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u = x$, $v' = \sin 4x$, $u' = 1$, $v = A\cos 4x$ → begins integration by parts | M1 | PI by $Ax\cos 4x - A\int(\cos 4x)\,dx$ |
| Selects correct method: $u=x$, $v'=\sin 4x$, $u'=1$, $v = A\cos 4x$; PI by $Ax\cos 4x - A\int(\cos 4x)\,dx$ | M1 | |
| Substitutes $u, u', v, v'$ into integration by parts formula: $Px\cos 4x - P\int(\cos 4x)\,dx$ | M1 | |
| $-\frac{1}{4}x\cos 4x - \frac{1}{4}\int(-\cos 4x)\,dx$ | A1 | Condone missing $dx$; PI by $-\frac{1}{4}x\cos 4x + \frac{1}{16}\sin 4x$ |
| Completes integration by parts to obtain $-\frac{1}{4}x\cos 4x + B\sin 4x$ with $B \neq \pm 1$ | M1 | |
| $\left[-\frac{1}{4}x\cos 4x + \frac{1}{16}\sin 4x\right]_0^{\frac{\pi}{2}}$; substitutes correct limits explicitly into $-\frac{1}{4}x\cos 4x + \frac{1}{16}\sin 4x$ | R1 | |
| $= \left(-\frac{\pi}{8}\cos\frac{4\pi}{2} + \frac{1}{16}\sin\frac{4\pi}{2}\right) - \left(0\times\cos 0 + \frac{1}{16}\sin 0\right) = -\frac{\pi}{8}$ | | Accept $\left(-\frac{\pi}{8}\cos 2\pi + \frac{1}{16}\sin 2\pi\right) - 0$; AG |8 Show that
$$\int _ { 0 } ^ { \frac { \pi } { 2 } } ( x \sin 4 x ) \mathrm { d } x = - \frac { \pi } { 8 }$$
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\hfill \mbox{\textit{AQA Paper 1 2023 Q8 [6]}}