CAIE FP1 2019 November — Question 3 7 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2019
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeSimple recurrence evaluation
DifficultyChallenging +1.3 This is a standard integration by parts reduction formula question requiring systematic application of IBP twice to establish the recurrence relation, followed by routine substitution. While it involves multiple steps and careful algebraic manipulation, the technique is well-practiced in Further Maths and doesn't require novel insight—just methodical execution of a familiar method.
Spec1.08i Integration by parts8.06a Reduction formulae: establish, use, and evaluate recursively
The integral \(I_n\), where \(n\) is a positive integer, is defined by $$I_n = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^n \sin \pi x \, dx.$$
  1. Show that $$n(n+1)I_{n+2} = 2^{n+1}n + \pi - \pi^2 I_n.$$ [5]
  2. Find \(I_5\) in terms of \(\pi\) and \(I_1\). [2]
Question 3:
AnswerMarks
3(i) x−n−1  1 1 x−n−1
I n+2 =  sinπx −π∫ cosπx dx
−n−1 −n−1
 1
1
2
AnswerMarks Guidance
2M1 A1 Integrates by parts.
 
2n+1 π x−n  1 1 x−n 
= +  cosπx +π∫ sinπx dx
n+1 n+1 −n −n
 1 1 
 2 
AnswerMarks Guidance
2M1 Integrates by parts again.
2n+1
π 1 π 
= + − I
 n
n+1 n+1 n n 
⇒ ( n+1 ) I =2n+1+π  1 − π I 
AnswerMarks Guidance
n+2   n n n  M1 Uses I .
n
⇒n ( n+1 ) I =2n+1n+π−π2I
AnswerMarks Guidance
n+2 nA1 AG
5
AnswerMarks
3(ii)2I =4+π−π2I
3 1
π2( )
12I =48+π− 4+π−π2I
AnswerMarks Guidance
5 2 1M1 Substitutes I into reduction formula.
3
1 ( )
⇒I =4+ 2π−4π2 −π3 +π4I
AnswerMarks Guidance
5 24 1A1 AEF, must be exact with fractions simplified.
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 3:
--- 3(i) ---
3(i) |  x−n−1  1 1 x−n−1
I n+2 =  sinπx −π∫ cosπx dx
−n−1 −n−1
 1
1
2
2 | M1 A1 | Integrates by parts.
 
2n+1 π x−n  1 1 x−n 
= +  cosπx +π∫ sinπx dx
n+1 n+1 −n −n
 1 1 
 2 
2 | M1 | Integrates by parts again.
2n+1
π 1 π 
= + − I
 n
n+1 n+1 n n 
⇒ ( n+1 ) I =2n+1+π  1 − π I 
n+2   n n n   | M1 | Uses I .
n
⇒n ( n+1 ) I =2n+1n+π−π2I
n+2 n | A1 | AG
5
--- 3(ii) ---
3(ii) | 2I =4+π−π2I
3 1
π2( )
12I =48+π− 4+π−π2I
5 2 1 | M1 | Substitutes I into reduction formula.
3
1 ( )
⇒I =4+ 2π−4π2 −π3 +π4I
5 24 1 | A1 | AEF, must be exact with fractions simplified.
2
Question | Answer | Marks | Guidance
The integral $I_n$, where $n$ is a positive integer, is defined by
$$I_n = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^n \sin \pi x \, dx.$$

\begin{enumerate}[label=(\roman*)]
\item Show that
$$n(n+1)I_{n+2} = 2^{n+1}n + \pi - \pi^2 I_n.$$ [5]

\item Find $I_5$ in terms of $\pi$ and $I_1$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2019 Q3 [7]}}
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