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 application of the reverse chain rule for integrating sin(kx), requiring only the standard formula ∫sin(kx)dx = -1/k·cos(kx) + C, then substituting limits. It's simpler than average A-level questions as it's a single-step definite integral with no algebraic manipulation or problem-solving required.
Spec1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits
3 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \sin 3 x \mathrm {~d} x\).
[0pt] [3]
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(\int_0^{\pi/6} \sin 3x\, dx = \left[-\frac{1}{3}\cos 3x\right]_0^{\pi/6}\)B1 \(\left[-\frac{1}{3}\cos 3x\right]\) or \(\left[-\frac{1}{3}\cos u\right]\)
\(= -\frac{1}{3}\cos\frac{\pi}{2} + \frac{1}{3}\cos 0\)M1 substituting correct limits in \(\pm k\cos\ldots\)
\(= \frac{1}{3}\)A1cao 0.33 or better
[3] 
# Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^{\pi/6} \sin 3x\, dx = \left[-\frac{1}{3}\cos 3x\right]_0^{\pi/6}$ | B1 | $\left[-\frac{1}{3}\cos 3x\right]$ or $\left[-\frac{1}{3}\cos u\right]$ |
| $= -\frac{1}{3}\cos\frac{\pi}{2} + \frac{1}{3}\cos 0$ | M1 | substituting correct limits in $\pm k\cos\ldots$ |
| $= \frac{1}{3}$ | A1cao | 0.33 or better |
| **[3]** | | |

---
3 Evaluate $\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \sin 3 x \mathrm {~d} x$.\\[0pt]
[3]

\hfill \mbox{\textit{OCR MEI C3  Q3 [3]}}
This paper (4 questions)
View full paper
Q1 5 Q2 23 Q3 3 Q4 18