CAIE FP1 2010 June — Question 9 10 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2010
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReduction Formulae
TypeCentroid and mean value
DifficultyChallenging +1.2 This is a standard Further Maths reduction formula question with a centroid application. The reduction formula derivation uses routine integration by parts, and the centroid calculation requires applying the formula with straightforward substitution. While it involves multiple steps and Further Maths content, it follows well-established techniques without requiring novel insight.
Spec4.08e Mean value of function: using integral8.06a Reduction formulae: establish, use, and evaluate recursively
9 Let $$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \sin ^ { n } \theta \mathrm {~d} \theta$$ where \(n\) is a non-negative integer. Show that \(I _ { n + 2 } = \frac { n + 1 } { n + 2 } I _ { n }\). The region \(R\) of the \(x - y\) plane is bounded by the \(x\)-axis, the line \(x = \frac { \pi } { 2 m }\) and the curve whose equation is \(y = \sin ^ { 4 } m x\), where \(m > 0\). Find the \(y\)-coordinate of the centroid of \(R\).
AnswerMarks Guidance
\(D[s^{n+1}c] = -x^{n+2} + (n+1)k^n c^2\)M1
\(= \ldots = -(n+2)s^{n+2} + (n+1)k^n\)A1
Integrates w.r.t. \(x\) over the range \([0, \pi/2]\) to obtain \(0 = -(n+2)I_{n+2} + (n+1)I_n\)M1
Result (AG)A1 [4]
OR \(I_n = [-cs^{n+1}]_0^{\pi/2} + (n+1) \int_0^{\pi/2} c^2 s^{n-2} dx\)M1
\(\Rightarrow (n-1)I_{n-2} - (n-1)I_n\) (\(n \geq 2\))M1A1
\(\Rightarrow I_n = \left[\frac{(n-1)}{n}\right]I_{n-2} \Rightarrow I_{n+2} = \left[\frac{(n+1)}{(n+2)}\right]I_n\) for \(n \geq 0\)A1
OR Starts with \(I_{n+2}\) and relates to \(I_n\) directly: mark as above
AnswerMarks
OR \(I_n = \int_0^{\pi/2} \sin^{n+2} \theta \cos \sec^2 \theta \, d\theta\)M1
\(= \left[-\sin^{n+2} \theta \cot \theta\right]_0^{\pi/2} + \int_0^{\pi/2} (n+2) \sin^{n+1} \theta \cos \theta \cot \theta \, d\theta\) (LNR)M1
\(= 0 + (n+2) \int_0^{\pi/2} \sin^n \theta (1 - \sin^2 \theta) d\theta\) (LR)M1
\(= (n+2)I_n - (n+2)I_{n+2}\)A1
\(\Rightarrow I_{n+2} = \frac{(n+1)}{(n+2)} I_n\)A1
Question 9 (continued)
AnswerMarks Guidance
\(\bar{y} = \frac{\frac{1}{2} \int_0^{2\pi} \sin^8 mx \, dx}{\int_0^{2\pi} \sin^4 mx \, dx}\)M1 (LR)
let \(u = mx\)M1
\(\bar{y} = \frac{\frac{1}{2} \int_0^{\pi/2} \sin^8 u \, du}{\int_0^{\pi/2} \sin^4 u \, du}\)A1
\(I_0 = \frac{\pi}{2}\), \(I_2 = \frac{1}{2} \times \frac{\pi}{4}\), \(I_4 = \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2} \left(= \frac{3\pi}{16}\right)\)B1
\(I_6 = \frac{5}{6} \times \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}\), \(I_8 = \frac{7}{8} \times \frac{5}{6} \times \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2} \left(= \frac{105\pi}{768}\right)\)B1
\(\bar{y} = \frac{1}{2} \frac{I_8}{I_4} = \frac{35}{96}\) (or 0.365)B1 [6]
OR for last 3 marks:
AnswerMarks
\(\frac{I_8}{I_4} = \frac{35}{48}\) (oe)M1A1
\(\therefore \bar{y} = \frac{35}{96}\)A1 
$D[s^{n+1}c] = -x^{n+2} + (n+1)k^n c^2$ | M1 |

$= \ldots = -(n+2)s^{n+2} + (n+1)k^n$ | A1 |

Integrates w.r.t. $x$ over the range $[0, \pi/2]$ to obtain $0 = -(n+2)I_{n+2} + (n+1)I_n$ | M1 |

Result (AG) | A1 | [4] |

**OR** $I_n = [-cs^{n+1}]_0^{\pi/2} + (n+1) \int_0^{\pi/2} c^2 s^{n-2} dx$ | M1 |

$\Rightarrow (n-1)I_{n-2} - (n-1)I_n$ ($n \geq 2$) | M1A1 |

$\Rightarrow I_n = \left[\frac{(n-1)}{n}\right]I_{n-2} \Rightarrow I_{n+2} = \left[\frac{(n+1)}{(n+2)}\right]I_n$ for $n \geq 0$ | A1 |

**OR** Starts with $I_{n+2}$ and relates to $I_n$ directly: mark as above

**OR** $I_n = \int_0^{\pi/2} \sin^{n+2} \theta \cos \sec^2 \theta \, d\theta$ | M1 |

$= \left[-\sin^{n+2} \theta \cot \theta\right]_0^{\pi/2} + \int_0^{\pi/2} (n+2) \sin^{n+1} \theta \cos \theta \cot \theta \, d\theta$ (LNR) | M1 |

$= 0 + (n+2) \int_0^{\pi/2} \sin^n \theta (1 - \sin^2 \theta) d\theta$ (LR) | M1 |

$= (n+2)I_n - (n+2)I_{n+2}$ | A1 |

$\Rightarrow I_{n+2} = \frac{(n+1)}{(n+2)} I_n$ | A1 |

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## Question 9 (continued)

$\bar{y} = \frac{\frac{1}{2} \int_0^{2\pi} \sin^8 mx \, dx}{\int_0^{2\pi} \sin^4 mx \, dx}$ | M1 (LR) |

let $u = mx$ | M1 |

$\bar{y} = \frac{\frac{1}{2} \int_0^{\pi/2} \sin^8 u \, du}{\int_0^{\pi/2} \sin^4 u \, du}$ | A1 |

$I_0 = \frac{\pi}{2}$, $I_2 = \frac{1}{2} \times \frac{\pi}{4}$, $I_4 = \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2} \left(= \frac{3\pi}{16}\right)$ | B1 |

$I_6 = \frac{5}{6} \times \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}$, $I_8 = \frac{7}{8} \times \frac{5}{6} \times \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2} \left(= \frac{105\pi}{768}\right)$ | B1 |

$\bar{y} = \frac{1}{2} \frac{I_8}{I_4} = \frac{35}{96}$ (or 0.365) | B1 | [6] |

**OR for last 3 marks:**

$\frac{I_8}{I_4} = \frac{35}{48}$ (oe) | M1A1 |

$\therefore \bar{y} = \frac{35}{96}$ | A1 |

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9 Let

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

where $n$ is a non-negative integer. Show that $I _ { n + 2 } = \frac { n + 1 } { n + 2 } I _ { n }$.

The region $R$ of the $x - y$ plane is bounded by the $x$-axis, the line $x = \frac { \pi } { 2 m }$ and the curve whose equation is $y = \sin ^ { 4 } m x$, where $m > 0$. Find the $y$-coordinate of the centroid of $R$.

\hfill \mbox{\textit{CAIE FP1 2010 Q9 [10]}}
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