Question 5 - A-Level Maths

CAIE FP1 2010 November — Question 5 8 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2010
Session	November
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Reduction Formulae
Type	Derive reduction formula by integration by parts
Difficulty	Challenging +1.2 This is a standard reduction formula derivation requiring two applications of integration by parts with clear substitutions, followed by routine recursive calculation. While it involves Further Maths content and multiple steps, the technique is methodical and well-practiced, making it moderately above average difficulty but not requiring novel insight.
Spec	1.08i Integration by parts 8.06a Reduction formulae: establish, use, and evaluate recursively

5 Let $I _ { n } = \int _ { 0 } ^ { 1 } ( 1 - x ) ^ { n } \sin x \mathrm {~d} x$ for $n \geqslant 0$. Show that $$I _ { n + 2 } = 1 - ( n + 1 ) ( n + 2 ) I _ { n }$$ Hence find the value of $I _ { 6 }$, correct to 4 decimal places.

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Answer	Marks	Guidance
$I_{n+2} = [-(1-x)^{n+2}\cos x] - \int(n+2)(1-x)^{n+1}\cos x dx$	M1A1
$= (1 + (n + 2)) + (n + 2)[(1-x)^{n+1}\sin x) + \int(1-x)^n\sin x dx]$	M1	integrate by parts again
$I_{n+2} = 1 - (n+1)(n+2)I_n$	A1
$I_6 = 1 - 5 \times 6I_3; I_4 = 1 - 4 \times 3I_2; I_2 = 1 - 1 \times 2I_0$	M1
$I_0 = \int_0^1 \sin x dx = 1 - \cos 1$	B1
$I_6 = 1 - 30(1 - 12(1 - 2I_0)) = 0.0177$	M1A1

OR

Answer	Marks	Guidance
$I_0 = 1 - \cos 1$	B1
$I_2 = 2\cos 1 - 1$	M1	(use of RF)
$I_4 = 13 - 24\cos 1$	A1
$I_6 = 0.0177$	A1	cao

Accept decimal versions

$I_{n+2} = [-(1-x)^{n+2}\cos x] - \int(n+2)(1-x)^{n+1}\cos x dx$ | M1A1

$= (1 + (n + 2)) + (n + 2)[(1-x)^{n+1}\sin x) + \int(1-x)^n\sin x dx]$ | M1 | integrate by parts again

$I_{n+2} = 1 - (n+1)(n+2)I_n$ | A1 | | [4]

$I_6 = 1 - 5 \times 6I_3; I_4 = 1 - 4 \times 3I_2; I_2 = 1 - 1 \times 2I_0$ | M1

$I_0 = \int_0^1 \sin x dx = 1 - \cos 1$ | B1

$I_6 = 1 - 30(1 - 12(1 - 2I_0)) = 0.0177$ | M1A1 | | [4]

**OR**

$I_0 = 1 - \cos 1$ | B1
$I_2 = 2\cos 1 - 1$ | M1 | (use of RF)
$I_4 = 13 - 24\cos 1$ | A1
$I_6 = 0.0177$ | A1 | cao
Accept decimal versions

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5 Let $I _ { n } = \int _ { 0 } ^ { 1 } ( 1 - x ) ^ { n } \sin x \mathrm {~d} x$ for $n \geqslant 0$. Show that

$$I _ { n + 2 } = 1 - ( n + 1 ) ( n + 2 ) I _ { n }$$

Hence find the value of $I _ { 6 }$, correct to 4 decimal places.

\hfill \mbox{\textit{CAIE FP1 2010 Q5 [8]}}

This paper (13 questions)

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Q1 4 Q2 5 Q3 5 Q4 5 Q5 8 Q6 8 Q7 9 Q8 10 Q9 10 Q10 10 Q11 12 Q12 EITHER Q12 OR

Answer	Marks	Guidance
\(I_{n+2} = [-(1-x)^{n+2}\cos x] - \int(n+2)(1-x)^{n+1}\cos x dx\)	M1A1
\(= (1 + (n + 2)) + (n + 2)[(1-x)^{n+1}\sin x) + \int(1-x)^n\sin x dx]\)	M1	integrate by parts again
\(I_{n+2} = 1 - (n+1)(n+2)I_n\)	A1
\(I_6 = 1 - 5 \times 6I_3; I_4 = 1 - 4 \times 3I_2; I_2 = 1 - 1 \times 2I_0\)	M1
\(I_0 = \int_0^1 \sin x dx = 1 - \cos 1\)	B1
\(I_6 = 1 - 30(1 - 12(1 - 2I_0)) = 0.0177\)	M1A1

Answer	Marks	Guidance
\(I_0 = 1 - \cos 1\)	B1
\(I_2 = 2\cos 1 - 1\)	M1	(use of RF)
\(I_4 = 13 - 24\cos 1\)	A1
\(I_6 = 0.0177\)	A1	cao