OCR MEI C3 — Question 2 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
DifficultyEasy -1.2 This is a straightforward application of the reverse chain rule for a simple trigonometric function with a linear argument. Students need only recognize that ∫cos(3x)dx = (1/3)sin(3x) and evaluate at the limits. This is below average difficulty as it's a single-step standard integral with no algebraic manipulation or problem-solving required.
Spec1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)
2 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \cos 3 x \mathrm {~d} x\).
Question 2:
AnswerMarks 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
\(= \frac{1}{3}\sin\frac{\pi}{2} - 0\)B1 \(k = (\pm)\frac{1}{3}\); or M1 for \(\left[\frac{1}{3}\sin u\right]\)
\(= \frac{1}{3}\)A1cao [3] 0.33 or better; so: \(\sin 3x\): M0B0, \(-\sin 3x\): M0B0, \(\pm 3\sin 3x\): M0B0, \(-\frac{1}{3}\sin 3x\): M0B1 
## Question 2:

$$\int_0^{\pi/6} \cos 3x \, dx = \left[\frac{1}{3}\sin 3x\right]_0^{\pi/6}$$ | 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

$= \frac{1}{3}\sin\frac{\pi}{2} - 0$ | B1 | $k = (\pm)\frac{1}{3}$; or M1 for $\left[\frac{1}{3}\sin u\right]$

$= \frac{1}{3}$ | A1cao [3] | 0.33 or better; so: $\sin 3x$: M0B0, $-\sin 3x$: M0B0, $\pm 3\sin 3x$: M0B0, $-\frac{1}{3}\sin 3x$: M0B1

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

\hfill \mbox{\textit{OCR MEI C3  Q2 [3]}}
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Q1 18 Q2 3 Q3 7 Q4 18