| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Session | Specimen |
| Marks | 5 |
| Topic | Integration by Parts |
| Type | Show that integral equals expression |
| Difficulty | Standard +0.3 This is a straightforward integration by parts question with a definite integral. The setup is clear (u = π - x, dv = cos 2x dx), requiring one application of integration by parts followed by routine evaluation at the limits. While it involves careful arithmetic with the bounds, it's slightly easier than average as it's a standard technique with no conceptual surprises or multi-stage reasoning. |
| Spec | 1.08i Integration by parts |
Attempt at integrating one factor, differentiating the other: M1
$\int(\pi - x)\cos 2x\, dx = \frac{1}{2}(\pi - x)\sin 2x - \int(-1)\frac{1}{2}\sin 2x\, dx$ A1
$= \frac{1}{2}(\pi - x)\sin 2x - \frac{1}{4}\cos 2x$ A1
Correct substitution of limits: M1
$\frac{1}{2}\left(\pi - \frac{\pi}{4}\right)\sin\left(\frac{\pi}{2}\right) - \frac{1}{4}\cos\left(\frac{\pi}{2}\right) - \left(0 - \frac{1}{4}\cos 0\right)$
$= \frac{\pi}{2} - \frac{\pi}{8} + \frac{1}{4}$
$= \frac{3\pi}{8} + \frac{1}{4}$ **cao, AG** A1
**Total: 5 marks**3 Show that
$$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } ( \pi - x ) \cos 2 x \mathrm {~d} x = \frac { 1 } { 4 } + \frac { 3 } { 8 } \pi$$
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 Q3 [5]}}