| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 17 |
| 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 requiring standard techniques: finding x-intercepts, verifying a point on a line, differentiation with product rule, and integration by parts. While it involves several steps, each part is routine for C3 level with clear guidance through the problem structure. The integration by parts is straightforward with u=x, and the question explicitly hints at the method needed. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.07q Product and quotient rules: differentiation1.08e Area between curve and x-axis: using definite integrals1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| At P, \(x \sin 3x = 0 \Rightarrow \sin 3x = 0\) | M1 | \(x \sin 3x = 0\) |
| \(\Rightarrow 3x = \pi\) | A1 | \(3x = \pi\) or 180 |
| \(\Rightarrow x = \pi/3\) | A1cao [3] | \(x = \pi/3\) or 1.05 or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(x = \pi/6\), \(x \sin 3x = \frac{\pi}{6} \sin \frac{\pi}{2} = \frac{\pi}{6}\) | E1 | \(y = \frac{\pi}{6}\) or \(x \sin 3x = x \Rightarrow \sin 3x = 1\) etc. |
| \(\Rightarrow Q(\pi/6, \pi/6)\) lies on line \(y = x\) | [1] | Must conclude in radians, and be exact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = x \cdot 3\cos 3x + \sin 3x\) | B1, M1, A1cao | \(d/dx(\sin 3x) = 3\cos 3x\); Product rule consistent with their derivs; \(3x\cos 3x + \sin 3x\) |
| At Q, \(\frac{dy}{dx} = \frac{\pi}{6} \cdot 3\cos\frac{\pi}{2} + \sin\frac{\pi}{2} = 1\) | M1, A1ft | Substituting \(x = \pi/6\) into their derivative \(= 1\); ft dep 1st M1 |
| \(=\) gradient of \(y = x\), so line touches curve at this point | E1 [6] | \(=\) gradient of \(y = x\) (www) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area under curve \(= \int_0^{\pi/6} x \sin 3x\, dx\); parts with \(u = x\), \(dv/dx = \sin 3x \Rightarrow v = -\frac{1}{3}\cos 3x\) | M1 | Parts with \(u = x\), \(dv/dx = \sin 3x \Rightarrow v = -\frac{1}{3}\cos 3x\) [condone no negative] |
| \(\int_0^{\pi/6} x \sin 3x\, dx = \left[-\frac{1}{3}x\cos 3x\right]_0^{\pi/6} + \int_0^{\pi/6} \frac{1}{3}\cos 3x\, dx\) | A1cao | |
| \(\ldots + \left[\frac{1}{9}\sin 3x\right]_0^{\pi/6}\) | A1ft | |
| \(= -\frac{1}{3} \cdot \frac{\pi}{6}\cos\frac{\pi}{2} + \frac{1}{3} \cdot 0 \cdot \cos 0 + \left[\frac{1}{9}\sin 3x\right]_0^{\pi/6}\) | M1 | Substituting (correct) limits |
| \(= \frac{1}{9}\) | A1 | \(\frac{1}{9}\) www |
| Area under line \(= \frac{1}{2} \times \frac{\pi}{6} \times \frac{\pi}{6} = \frac{\pi^2}{72}\) | B1 | \(\frac{\pi^2}{72}\) |
| So area required \(= \frac{\pi^2}{72} - \frac{1}{9} = \frac{\pi^2 - 8}{72}\) | E1 [7] | www |
# Question 2:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| At P, $x \sin 3x = 0 \Rightarrow \sin 3x = 0$ | M1 | $x \sin 3x = 0$ |
| $\Rightarrow 3x = \pi$ | A1 | $3x = \pi$ or 180 |
| $\Rightarrow x = \pi/3$ | A1cao [3] | $x = \pi/3$ or 1.05 or better |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $x = \pi/6$, $x \sin 3x = \frac{\pi}{6} \sin \frac{\pi}{2} = \frac{\pi}{6}$ | E1 | $y = \frac{\pi}{6}$ or $x \sin 3x = x \Rightarrow \sin 3x = 1$ etc. |
| $\Rightarrow Q(\pi/6, \pi/6)$ lies on line $y = x$ | [1] | Must conclude in radians, and be exact |
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = x \cdot 3\cos 3x + \sin 3x$ | B1, M1, A1cao | $d/dx(\sin 3x) = 3\cos 3x$; Product rule consistent with their derivs; $3x\cos 3x + \sin 3x$ |
| At Q, $\frac{dy}{dx} = \frac{\pi}{6} \cdot 3\cos\frac{\pi}{2} + \sin\frac{\pi}{2} = 1$ | M1, A1ft | Substituting $x = \pi/6$ into their derivative $= 1$; ft dep 1st M1 |
| $=$ gradient of $y = x$, so line touches curve at this point | E1 [6] | $=$ gradient of $y = x$ (www) |
## Part (iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area under curve $= \int_0^{\pi/6} x \sin 3x\, dx$; parts with $u = x$, $dv/dx = \sin 3x \Rightarrow v = -\frac{1}{3}\cos 3x$ | M1 | Parts with $u = x$, $dv/dx = \sin 3x \Rightarrow v = -\frac{1}{3}\cos 3x$ [condone no negative] |
| $\int_0^{\pi/6} x \sin 3x\, dx = \left[-\frac{1}{3}x\cos 3x\right]_0^{\pi/6} + \int_0^{\pi/6} \frac{1}{3}\cos 3x\, dx$ | A1cao | |
| $\ldots + \left[\frac{1}{9}\sin 3x\right]_0^{\pi/6}$ | A1ft | |
| $= -\frac{1}{3} \cdot \frac{\pi}{6}\cos\frac{\pi}{2} + \frac{1}{3} \cdot 0 \cdot \cos 0 + \left[\frac{1}{9}\sin 3x\right]_0^{\pi/6}$ | M1 | Substituting (correct) limits |
| $= \frac{1}{9}$ | A1 | $\frac{1}{9}$ www |
| Area under line $= \frac{1}{2} \times \frac{\pi}{6} \times \frac{\pi}{6} = \frac{\pi^2}{72}$ | B1 | $\frac{\pi^2}{72}$ |
| So area required $= \frac{\pi^2}{72} - \frac{1}{9} = \frac{\pi^2 - 8}{72}$ | E1 [7] | www |
---2 Fig. 8 shows part of the curve $y = x \sin 3 x$. It crosses the $x$-axis at P . The point on the curve with $x$-coordinate $\frac { 1 } { 6 } \pi$ is Q .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{aee8da6a-7d5c-442f-9729-55d81d9a606f-2_418_769_516_673}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) Find the $x$-coordinate of P .\\
(ii) Show that Q lies on the line $y = x$.\\
(iii) Differentiate $x \sin 3 x$. Hence prove that the line $y = x$ touches the curve at Q .\\
(iv) Show that the area of the region bounded by the curve and the line $y = x$ is $\frac { 1 } { 72 } \left( \pi ^ { 2 } - 8 \right)$.
\hfill \mbox{\textit{OCR MEI C3 Q2 [17]}}