| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Session | Specimen |
| Marks | 7 |
| Topic | Integration by Substitution |
| Type | Area under curve using substitution |
| Difficulty | Standard +0.8 Part (i) is straightforward product rule differentiation. Part (ii) requires recognizing the absolute value necessitates splitting the integral at π/3 (where sin 3x changes sign), applying the given antiderivative correctly with sign changes, and careful algebraic manipulation. This is above-average difficulty due to the absolute value complication and multi-step reasoning, but remains a standard integration technique question. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.08e Area between curve and x-axis: using definite integrals1.08i Integration by parts |
**(i)** Obtain $\frac{d}{dx}(\sin 3x) = 3\cos 3x$ B1
Use of product rule to obtain form $k\cos 3x + mx\sin 3x$ M1
Obtain $\frac{d}{dx}(-3x\cos 3x) = -3\cos 3x + 9x\sin 3x$ and final correct conclusion AG A1 **[3]**
**(ii)** Recognize that the required result = area = $\int x\sin 3x\, dx$ over two distinct intervals M1
Use result for either area M1
Area $= \left[\frac{1}{9}\sin 3x - \frac{1}{3}x\cos 3x\right]$ A1
Substituting limits between $0$ and $\frac{\pi}{3}$ **and** $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ M1
Obtain value of $\frac{\pi}{9}$ A1
Obtain value of integral between $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ is $-\frac{\pi}{3}$ A1
Final answer correct AG A1 **[7]**10 (i) Show that $\frac { \mathrm { d } } { \mathrm { d } x } ( \sin 3 x - 3 x \cos 3 x ) = 9 x \sin 3 x$.
The curve shown in the figure below is part of the graph of the function $y = x \sin 3 x$.\\
\includegraphics[max width=\textwidth, alt={}, center]{3e4281d1-dbad-46a2-bbb7-97706bda2dfa-3_508_1136_1939_466}\\
(ii) Show that $\int _ { 0 } ^ { \frac { 2 \pi } { 3 } } | x \sin 3 x | \mathrm { d } x = \frac { 4 \pi } { 9 }$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 Q10 [7]}}