| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2010 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Normal/tangent then area with parts |
| Difficulty | Standard +0.3 This is a structured multi-part question that guides students through standard techniques: finding a normal line (routine differentiation with product rule), verifying differentiation (direct calculation), and integration by parts (with the result essentially given in part ii). While it requires multiple steps, each part is straightforward with clear signposting, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Use product rule | M1 | |
| Obtain correct derivative in any form | A1 | |
| Substitute \(x = \frac{1}{2}\pi\), and obtain gradient of \(-1\) for normal | A1√ | |
| from \(y' = \sin x - x\cos x\) ONLY | ||
| Show that line through \(\left(\frac{1}{2}\pi, \frac{1}{2}\pi\right)\) with gradient \(-1\) passes through \((\pi, 0)\) | M1 | |
| A1 | [5] | |
| (ii) Differentiate \(\sin x\) and use product rule to differentiate \(x\cos x\) | M1 | |
| Obtain \(x\sin x\), or equivalent | A1 | [2] |
| (iii) State that integral is \(\sin x - x\cos x + c\) | B1 | |
| Substitute limits \(0\) and \(\frac{\pi}{2}\) correctly | M1 | |
| Obtain answer \(1\) | A1 | |
| S.R. Feeding limits into original integrand, 0/3 | [3] |
**(i)** Use product rule | M1 |
Obtain correct derivative in any form | A1 |
Substitute $x = \frac{1}{2}\pi$, and obtain gradient of $-1$ for normal | A1√ |
from $y' = \sin x - x\cos x$ ONLY | |
Show that line through $\left(\frac{1}{2}\pi, \frac{1}{2}\pi\right)$ with gradient $-1$ passes through $(\pi, 0)$ | M1 |
| A1 | [5]
**(ii)** Differentiate $\sin x$ and use product rule to differentiate $x\cos x$ | M1 |
Obtain $x\sin x$, or equivalent | A1 | [2]
**(iii)** State that integral is $\sin x - x\cos x + c$ | B1 |
Substitute limits $0$ and $\frac{\pi}{2}$ correctly | M1 |
Obtain answer $1$ | A1 |
S.R. Feeding limits into original integrand, 0/3 | | [3]8\\
\includegraphics[max width=\textwidth, alt={}, center]{2aceb797-097c-499b-99b6-cce9f287cb51-3_566_787_255_680}
The diagram shows the curve $y = x \sin x$, for $0 \leqslant x \leqslant \pi$. The point $Q \left( \frac { 1 } { 2 } \pi , \frac { 1 } { 2 } \pi \right)$ lies on the curve.\\
(i) Show that the normal to the curve at $Q$ passes through the point $( \pi , 0 )$.\\
(ii) Find $\frac { \mathrm { d } } { \mathrm { d } x } ( \sin x - x \cos x )$.\\
(iii) Hence evaluate $\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } x \sin x \mathrm {~d} x$.
\hfill \mbox{\textit{CAIE P2 2010 Q8 [10]}}