OCR C4 2013 January — Question 1 4 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Year2013
SessionJanuary
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeBasic integration by parts
DifficultyModerate -0.3 This is a straightforward single application of integration by parts with standard functions (polynomial times trigonometric). It requires only the basic formula with u=x, dv=cos(3x)dx, making it slightly easier than average since it's a direct textbook-style exercise with no complications or multi-step reasoning.
Spec1.08i Integration by parts
1 Find \(\int x \cos 3 x \mathrm {~d} x\).
Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(u = x\) and \(dv = \cos 3x\)M1 Integration by parts as far as \(f(x) \pm \int g(x)\,dx\); check if labelled \(v, du\)
\(x \times \frac{1}{3}\sin 3x - \int \frac{1}{3}\sin 3x\,dx\)A2 A1 for \(x \times k\sin 3x - \int k\sin 3x\,dx\); \(k \neq \frac{1}{3}\) or \(0\); \(k\) may be negative
\(\frac{x}{3}\sin 3x + \frac{1}{9}\cos 3x [+ c]\) caoA1 Not \(\frac{1}{3}\left(\frac{1}{3}\cos 3x\right)\) or \(-\frac{1}{9}\cos 3x\)
[4] 
# Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $u = x$ and $dv = \cos 3x$ | M1 | Integration by parts as far as $f(x) \pm \int g(x)\,dx$; check if labelled $v, du$ |
| $x \times \frac{1}{3}\sin 3x - \int \frac{1}{3}\sin 3x\,dx$ | A2 | **A1** for $x \times k\sin 3x - \int k\sin 3x\,dx$; $k \neq \frac{1}{3}$ or $0$; $k$ may be negative |
| $\frac{x}{3}\sin 3x + \frac{1}{9}\cos 3x [+ c]$ cao | A1 | Not $\frac{1}{3}\left(\frac{1}{3}\cos 3x\right)$ or $-\frac{1}{9}\cos 3x$ |
| **[4]** | | |

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1 Find $\int x \cos 3 x \mathrm {~d} x$.

\hfill \mbox{\textit{OCR C4 2013 Q1 [4]}}
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