CAIE FP1 2011 June — Question 10 12 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReduction Formulae
TypeCentroid and mean value
DifficultyChallenging +1.2 This is a standard reduction formula question with a straightforward application. The first part requires integration by parts (a routine technique at this level), and the second part involves parametric mean value calculation using the established formula. While it requires multiple steps and careful algebra, the techniques are well-practiced in Further Maths syllabi with no novel insights needed.
Spec4.08e Mean value of function: using integral8.06a Reduction formulae: establish, use, and evaluate recursively
10 Let $$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \cos ^ { n } x \mathrm {~d} x$$ where \(n \geqslant 0\). Show that, for all \(n \geqslant 2\), $$I _ { n } = \frac { n - 1 } { n } I _ { n - 2 }$$ A curve has parametric equations \(x = a \sin ^ { 3 } t\) and \(y = a \cos ^ { 3 } t\), where \(a\) is a constant and \(0 \leqslant t \leqslant \frac { 1 } { 2 } \pi\). Show that the mean value \(m\) of \(y\) over the interval \(0 \leqslant x \leqslant a\) is given by $$m = 3 a \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \left( \cos ^ { 4 } t - \cos ^ { 6 } t \right) \mathrm { d } t$$ Find the exact value of \(m\), in terms of \(a\).
Question 10:
Reduction formula:
AnswerMarks Guidance
\(I_n = \left[\cos^{n-1}x\sin x\right]_0^{\frac{\pi}{2}} + (n-1)\int_0^{\frac{\pi}{2}}\cos^{n-2}x\sin^2 x\ dx\)M1A1
\(I_n = (n-1)\int_0^{\frac{\pi}{2}}(\cos^{n-2}x - \cos^n x)\ dx\)M1
\(\Rightarrow I_n = \frac{n-1}{n}I_{n-2}\ (n \geq 2)\) (AG)A1 Part mark: 4
Mean value:
AnswerMarks Guidance
Mean value \(= \frac{\int_a^b y\ dx}{b-a}\)M1
\(= \frac{\int_0^{\frac{\pi}{2}} a\cos^3 t \cdot 3a\sin^2 t\cos t\ dt}{a}\)M1A1
\(= 3a\int_0^{\frac{\pi}{2}}\cos^4 t\sin^2 t\ dt\)A1
\(= 3a\int_0^{\frac{\pi}{2}}(\cos^4 t - \cos^6 t)\ dt\) (AG) Part mark: 4
\(I_0 = \frac{\pi}{2}\) or \(I_2 = \frac{\pi}{4}\)B1
\(I_4 = \frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} = \frac{3}{16}\pi\), \(I_6 = \frac{5}{6}\cdot\frac{3}{16}\pi = \frac{5}{32}\pi\)M1A1
Mean value \(= \frac{3a}{32}\pi\)A1 Part mark: 4
Total: [12]
## Question 10:

**Reduction formula:**

$I_n = \left[\cos^{n-1}x\sin x\right]_0^{\frac{\pi}{2}} + (n-1)\int_0^{\frac{\pi}{2}}\cos^{n-2}x\sin^2 x\ dx$ | M1A1 |

$I_n = (n-1)\int_0^{\frac{\pi}{2}}(\cos^{n-2}x - \cos^n x)\ dx$ | M1 |

$\Rightarrow I_n = \frac{n-1}{n}I_{n-2}\ (n \geq 2)$ (AG) | A1 | Part mark: 4

**Mean value:**

Mean value $= \frac{\int_a^b y\ dx}{b-a}$ | M1 |

$= \frac{\int_0^{\frac{\pi}{2}} a\cos^3 t \cdot 3a\sin^2 t\cos t\ dt}{a}$ | M1A1 |

$= 3a\int_0^{\frac{\pi}{2}}\cos^4 t\sin^2 t\ dt$ | A1 |

$= 3a\int_0^{\frac{\pi}{2}}(\cos^4 t - \cos^6 t)\ dt$ (AG) | | Part mark: 4

$I_0 = \frac{\pi}{2}$ or $I_2 = \frac{\pi}{4}$ | B1 |

$I_4 = \frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} = \frac{3}{16}\pi$, $I_6 = \frac{5}{6}\cdot\frac{3}{16}\pi = \frac{5}{32}\pi$ | M1A1 |

Mean value $= \frac{3a}{32}\pi$ | A1 | Part mark: 4

**Total: [12]**

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10 Let

$$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \cos ^ { n } x \mathrm {~d} x$$

where $n \geqslant 0$. Show that, for all $n \geqslant 2$,

$$I _ { n } = \frac { n - 1 } { n } I _ { n - 2 }$$

A curve has parametric equations $x = a \sin ^ { 3 } t$ and $y = a \cos ^ { 3 } t$, where $a$ is a constant and $0 \leqslant t \leqslant \frac { 1 } { 2 } \pi$. Show that the mean value $m$ of $y$ over the interval $0 \leqslant x \leqslant a$ is given by

$$m = 3 a \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \left( \cos ^ { 4 } t - \cos ^ { 6 } t \right) \mathrm { d } t$$

Find the exact value of $m$, in terms of $a$.

\hfill \mbox{\textit{CAIE FP1 2011 Q10 [12]}}
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