2 Show that \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } x \sin 2 x \mathrm {~d} x = \frac { 3 \sqrt { 3 } \pi } { 24 }\).
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Question 2:
Answer Marks
Guidance
Answer/Working Marks
Guidance
Let \(u = x\), \(\frac{dv}{dx} = \sin 2x \Rightarrow v = -\frac{1}{2}\cos 2x\) M1
Parts with \(u = x\), \(\frac{dv}{dx} = \sin 2x\)
\(\int_0^{\pi/6} x\sin 2x\,dx = \left[x \cdot -\frac{1}{2}\cos 2x\right]_0^{\pi/6} + \int_0^{\pi/6} \frac{1}{2}\cos 2x \cdot 1\,dx\) A1
\(= \frac{\pi}{6} \cdot -\frac{1}{2}\cos\frac{\pi}{3} - 0 + \left[\frac{1}{4}\sin 2x\right]_0^{\pi/6}\) B1ft
\(\ldots + \left[\frac{1}{4}\sin 2x\right]_0^{\pi/6}\)
\(= -\frac{\pi}{24} + \frac{\sqrt{3}}{8}\) M1
Substituting limits
B1 \(\cos\frac{\pi}{3} = \frac{1}{2}\), \(\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}\) soi
\(= \dfrac{3\sqrt{3}-\pi}{24}\) * E1 [6]
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## Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $u = x$, $\frac{dv}{dx} = \sin 2x \Rightarrow v = -\frac{1}{2}\cos 2x$ | M1 | Parts with $u = x$, $\frac{dv}{dx} = \sin 2x$ |
| $\int_0^{\pi/6} x\sin 2x\,dx = \left[x \cdot -\frac{1}{2}\cos 2x\right]_0^{\pi/6} + \int_0^{\pi/6} \frac{1}{2}\cos 2x \cdot 1\,dx$ | A1 | |
| $= \frac{\pi}{6} \cdot -\frac{1}{2}\cos\frac{\pi}{3} - 0 + \left[\frac{1}{4}\sin 2x\right]_0^{\pi/6}$ | B1ft | $\ldots + \left[\frac{1}{4}\sin 2x\right]_0^{\pi/6}$ |
| $= -\frac{\pi}{24} + \frac{\sqrt{3}}{8}$ | M1 | Substituting limits |
| | B1 | $\cos\frac{\pi}{3} = \frac{1}{2}$, $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ soi |
| $= \dfrac{3\sqrt{3}-\pi}{24}$ * | E1 [6] | www |
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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 Q2 [6]}}