OCR MEI C3 — Question 3 3 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard Integrals and Reverse Chain Rule
TypeDefinite integral with trigonometric functions
DifficultyModerate -0.8 This is a straightforward definite integral requiring basic knowledge of standard trigonometric integrals and the reverse chain rule for sin(3x). The integration is direct (x - (-1/3)cos(3x)), followed by routine substitution of limits. No problem-solving or insight required, just mechanical application of standard techniques.
Spec1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits
3 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } ( 1 - \sin 3 x ) \mathrm { d } x\), giving your answer in exact form.
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(\int_0^{\pi/6}(1 - \sin 3x)\,dx = \left[x + \tfrac{1}{3}\cos 3x\right]_0^{\pi/6}\)M1 \(\pm\tfrac{1}{3}\cos 3x\) seen or \(\tfrac{1}{3}(1-\sin u)[du]\); i.e. condone sign error
A1\(\left[x + \tfrac{1}{3}\cos 3x\right]\) or \(\left[\tfrac{1}{3}(u + \cos u)\right]\); condone \(+c\)
\(= \pi/6 - 1/3\)A1cao [3] o.e., must be exact; isw after correct answer seen 
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^{\pi/6}(1 - \sin 3x)\,dx = \left[x + \tfrac{1}{3}\cos 3x\right]_0^{\pi/6}$ | M1 | $\pm\tfrac{1}{3}\cos 3x$ seen or $\tfrac{1}{3}(1-\sin u)[du]$; i.e. condone sign error |
| | A1 | $\left[x + \tfrac{1}{3}\cos 3x\right]$ or $\left[\tfrac{1}{3}(u + \cos u)\right]$; condone $+c$ |
| $= \pi/6 - 1/3$ | A1cao **[3]** | o.e., must be exact; isw after correct answer seen |

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3 Evaluate $\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } ( 1 - \sin 3 x ) \mathrm { d } x$, giving your answer in exact form.

\hfill \mbox{\textit{OCR MEI C3  Q3 [3]}}
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Q1 4 Q2 18 Q3 3 Q4 18 Q5 5