CAIE Further Paper 2 2020 June — Question 2 6 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2020
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeReduction formula or recurrence
DifficultyChallenging +1.2 This is a standard reduction formula question requiring integration by parts with definite integrals. Part (a) involves routine application of integration by parts to establish the recurrence relation (3 marks suggests straightforward execution). Part (b) requires using the formula twice with careful arithmetic involving exponentials. While it demands precision and multiple steps, the technique is well-practiced in Further Maths syllabi and doesn't require novel insight—just systematic application of a standard method.
Spec1.08i Integration by parts8.06a Reduction formulae: establish, use, and evaluate recursively
Let \(I_n = \int_0^1 (1+3x)^n e^{-3x} dx\), where \(n\) is an integer.
  1. Show that \(3I_n = 1 - 4^n e^{-3} + 3nI_{n-1}\). [3]
  2. Find the exact value of \(I_2\). [3]
Question 2:
AnswerMarks
2(a)I = −1e−3x ( 1+3x )n 1 +n 1 e−3x ( 1+3x )n−1 dx
n  3  0 0M1 A1
−1e −34n+1+nI 3I =1−4ne −3+3nI
AnswerMarks
3 3 n−1 n n−1A1
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
2(b)( )
I 0 = 1 e −3x dx=−1 3   e −3x  1 = 1 3 1−e −3 .
AnswerMarks
0 0B1
( ) ( )
I = 1 1−4e −3 +1−e −3 = 1 2−5e −3
1 3 3
( ( ))
I = 1 1−16e−3 +2 2−5e−3
AnswerMarks
2 3M1
( )
I = 1 5−26e −3
AnswerMarks
2 3A1
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 2:
--- 2(a) ---
2(a) | I = −1e−3x ( 1+3x )n 1 +n 1 e−3x ( 1+3x )n−1 dx
n  3  0 0 | M1 A1
−1e −34n+1+nI 3I =1−4ne −3+3nI
3 3 n−1 n n−1 | A1
3
Question | Answer | Marks
--- 2(b) ---
2(b) | ( )
I 0 = 1 e −3x dx=−1 3   e −3x  1 = 1 3 1−e −3 .
0 0 | B1
( ) ( )
I = 1 1−4e −3 +1−e −3 = 1 2−5e −3
1 3 3
( ( ))
I = 1 1−16e−3 +2 2−5e−3
2 3 | M1
( )
I = 1 5−26e −3
2 3 | A1
3
Question | Answer | Marks
Let $I_n = \int_0^1 (1+3x)^n e^{-3x} dx$, where $n$ is an integer.

\begin{enumerate}[label=(\alph*)]
\item Show that $3I_n = 1 - 4^n e^{-3} + 3nI_{n-1}$. [3]

\item Find the exact value of $I_2$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q2 [6]}}
This paper (8 questions)
View full paper
Q1 6 Q2 6 Q3 8 Q4 8 Q5 9 Q6 12 Q7 11 Q8 15