OCR MEI C3 2006 June — Question 2 6 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Year2006
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeShow that integral equals expression
DifficultyStandard +0.3 This is a straightforward integration by parts question with a definite integral. Students must apply the standard technique (u=x, dv=sin 2x dx), evaluate the resulting integral, and substitute limits. While it requires careful arithmetic with the π/6 limit involving √3, it's a direct application of a core C3 technique with no conceptual challenges—slightly easier than average.
Spec1.08i Integration by parts
2 Show that \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } x \sin 2 x \mathrm {~d} x = \frac { 3 \sqrt { 3 } - \pi } { 24 }\).
AnswerMarks Guidance
let \(u = x, dv/dx = \sin 2x \Rightarrow v = -\frac{1}{2}\cos 2x\)M1 A1 parts with \(u = x, dv/dx = \sin 2x\)
\(\Rightarrow \int_0^{\pi/6} x\sin 2x dx = \left[-\frac{x}{2}\cos 2x\right]_0^{\pi/6} + \int_0^{\pi/6} \frac{1}{2}\cos 2x \cdot 1 dx\)B1ft \(\ldots + \left[\frac{1}{4}\sin 2x\right]_0^{\pi/6}\)
\(= \frac{\pi}{6} - \frac{1}{2}\cos \frac{\pi}{3} + 0 + \left[\frac{1}{4}\sin 2x\right]_0^{\pi/6}\)M1 B1 substituting limits, \(\cos \pi/3 = \frac{1}{2}\), \(\sin \pi/3 = \frac{\sqrt{3}}{2}\) soi
\(= -\frac{\pi}{24} + \frac{\sqrt{3}}{8}\)E1
\(= \frac{3\sqrt{3} - \pi}{24}\)[6] 
let $u = x, dv/dx = \sin 2x \Rightarrow v = -\frac{1}{2}\cos 2x$ | M1 A1 | parts with $u = x, dv/dx = \sin 2x$
$\Rightarrow \int_0^{\pi/6} x\sin 2x dx = \left[-\frac{x}{2}\cos 2x\right]_0^{\pi/6} + \int_0^{\pi/6} \frac{1}{2}\cos 2x \cdot 1 dx$ | B1ft | $\ldots + \left[\frac{1}{4}\sin 2x\right]_0^{\pi/6}$
$= \frac{\pi}{6} - \frac{1}{2}\cos \frac{\pi}{3} + 0 + \left[\frac{1}{4}\sin 2x\right]_0^{\pi/6}$ | M1 B1 | substituting limits, $\cos \pi/3 = \frac{1}{2}$, $\sin \pi/3 = \frac{\sqrt{3}}{2}$ soi
$= -\frac{\pi}{24} + \frac{\sqrt{3}}{8}$ | E1 | 
$= \frac{3\sqrt{3} - \pi}{24}$ | [6] |
2 Show that $\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } x \sin 2 x \mathrm {~d} x = \frac { 3 \sqrt { 3 } - \pi } { 24 }$.

\hfill \mbox{\textit{OCR MEI C3 2006 Q2 [6]}}
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