OCR MEI C3 2010 June — Question 1 3 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Year2010
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard Integrals and Reverse Chain Rule
TypeDefinite integral with trigonometric functions
DifficultyEasy -1.2 This is a straightforward application of the reverse chain rule for a single trigonometric function with a linear argument. It requires only recognizing that ∫cos(3x)dx = (1/3)sin(3x) and evaluating at the given limits—a routine procedure well below average difficulty for A-level, requiring minimal problem-solving.
Spec1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)

1 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \cos 3 x \mathrm {~d} x\).

Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\int_0^{\pi/6} \cos 3x \, dx = \left[\frac{1}{3}\sin 3x\right]_0^{\pi/6}\)M1 \(k\sin 3x\), \(k > 0\), \(k \neq 3\); or M1 for \(u = 3x \Rightarrow \int \frac{1}{3}\cos u \, du\), condone 90° in limit
\(k = \frac{1}{3}\)B1 \(k = (\pm)1/3\); or M1 for \(\left[\frac{1}{3}\sin u\right]\)
\(= \frac{1}{3}\sin\frac{\pi}{2} - 0 = \frac{1}{3}\)A1cao [3] 0.33 or better; so: \(\sin 3x\): M1B0, \(-\sin 3x\): M0B0, \(\pm 3\sin 3x\): M0B0, \(-\frac{1}{3}\sin 3x\): M0B1
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^{\pi/6} \cos 3x \, dx = \left[\frac{1}{3}\sin 3x\right]_0^{\pi/6}$ | M1 | $k\sin 3x$, $k > 0$, $k \neq 3$; or M1 for $u = 3x \Rightarrow \int \frac{1}{3}\cos u \, du$, condone 90° in limit |
| $k = \frac{1}{3}$ | B1 | $k = (\pm)1/3$; or M1 for $\left[\frac{1}{3}\sin u\right]$ |
| $= \frac{1}{3}\sin\frac{\pi}{2} - 0 = \frac{1}{3}$ | A1cao [3] | 0.33 or better; so: $\sin 3x$: M1B0, $-\sin 3x$: M0B0, $\pm 3\sin 3x$: M0B0, $-\frac{1}{3}\sin 3x$: M0B1 |

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1 Evaluate $\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \cos 3 x \mathrm {~d} x$.

\hfill \mbox{\textit{OCR MEI C3 2010 Q1 [3]}}