| 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 has multiple steps (6 parts total), each individual part uses routine A-level methods with clear guidance ('show that', 'hence prove'). The integration by parts is straightforward with u=x, and the area calculation is scaffolded. Slightly above average due to length and the tangency proof, but no novel insights required. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07q 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 |
| \(\Rightarrow x = \frac{\pi}{3}\) | A1cao [3] | \(x = \frac{\pi}{3}\) or 1.05 or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(x = \frac{\pi}{6}\), \(x\sin 3x = \frac{\pi}{6}\sin\frac{\pi}{2} = \frac{\pi}{6}\) | E1 | \(y = \frac{\pi}{6}\) or \(x\sin 3x = x \Rightarrow \sin 3x = 1\) etc. |
| \(\Rightarrow Q(\frac{\pi}{6}, \frac{\pi}{6})\) lies on line \(y = x\) | [1] | Must conclude in radians, and be exact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = x\sin 3x \Rightarrow \frac{dy}{dx} = x \cdot 3\cos 3x + \sin 3x\) | B1 | \(\frac{d}{dx}(\sin 3x) = 3\cos 3x\) |
| M1 | Product rule consistent with their derivatives | |
| A1cao | \(3x\cos 3x + \sin 3x\) | |
| At Q, \(\frac{dy}{dx} = \frac{\pi}{6} \cdot 3\cos\frac{\pi}{2} + \sin\frac{\pi}{2} = 1\) | M1 | Substituting \(x = \frac{\pi}{6}\) into their derivative |
| A1ft | \(= 1\) 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 \(u = x\), \(\frac{dv}{dx} = \sin 3x \Rightarrow v = -\frac{1}{3}\cos 3x\) | M1 | Parts with \(u = x\), \(\frac{dv}{dx} = \sin 3x \Rightarrow v = -\frac{1}{3}\cos 3x\) [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 | |
| A1ft | \(\ldots + \left[\frac{1}{9}\sin 3x\right]_0^{\pi/6}\) | |
| \(= -\frac{1}{3}\cdot\frac{\pi}{6}\cos\frac{\pi}{2} + \frac{1}{3}\cdot 0\cdot\cos 0 + \left[\frac{1}{9}\sin 3x\right]_0^{\pi/6}\) | 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 \(= \frac{\pi^2}{72} - \frac{1}{9} = \dfrac{\pi^2 - 8}{72}\) * | E1 [7] | www |
## Question 3:
### 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 |
| $\Rightarrow x = \frac{\pi}{3}$ | A1cao [3] | $x = \frac{\pi}{3}$ or 1.05 or better |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $x = \frac{\pi}{6}$, $x\sin 3x = \frac{\pi}{6}\sin\frac{\pi}{2} = \frac{\pi}{6}$ | E1 | $y = \frac{\pi}{6}$ or $x\sin 3x = x \Rightarrow \sin 3x = 1$ etc. |
| $\Rightarrow Q(\frac{\pi}{6}, \frac{\pi}{6})$ lies on line $y = x$ | [1] | Must conclude in radians, and be exact |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = x\sin 3x \Rightarrow \frac{dy}{dx} = x \cdot 3\cos 3x + \sin 3x$ | B1 | $\frac{d}{dx}(\sin 3x) = 3\cos 3x$ |
| | M1 | Product rule consistent with their derivatives |
| | A1cao | $3x\cos 3x + \sin 3x$ |
| At Q, $\frac{dy}{dx} = \frac{\pi}{6} \cdot 3\cos\frac{\pi}{2} + \sin\frac{\pi}{2} = 1$ | M1 | Substituting $x = \frac{\pi}{6}$ into their derivative |
| | A1ft | $= 1$ 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 $u = x$, $\frac{dv}{dx} = \sin 3x \Rightarrow v = -\frac{1}{3}\cos 3x$ | M1 | Parts with $u = x$, $\frac{dv}{dx} = \sin 3x \Rightarrow v = -\frac{1}{3}\cos 3x$ [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 | |
| | A1ft | $\ldots + \left[\frac{1}{9}\sin 3x\right]_0^{\pi/6}$ |
| $= -\frac{1}{3}\cdot\frac{\pi}{6}\cos\frac{\pi}{2} + \frac{1}{3}\cdot 0\cdot\cos 0 + \left[\frac{1}{9}\sin 3x\right]_0^{\pi/6}$ | 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 $= \frac{\pi^2}{72} - \frac{1}{9} = \dfrac{\pi^2 - 8}{72}$ * | E1 [7] | www |
---3 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]{74cc215f-bd55-489d-aa4b-0f67c2c8de52-2_420_780_549_655}
\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 $\frac { 1 } { 72 } \left( \pi ^ { 2 } - 8 \right)$.\\
(i) Differentiate $x \cos 2 x$ with respect to $x$.\\
(ii) Integrate $x \cos 2 x$ with respect to $x$.
\hfill \mbox{\textit{OCR MEI C3 Q3 [17]}}