CAIE FP1 2010 June — Question 5 9 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2010
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReduction Formulae
TypeProve formula by induction
DifficultyChallenging +1.2 This question requires integration by parts to derive the reduction formula, followed by a straightforward induction proof. While it involves multiple techniques (integration by parts, reduction formulae, and induction), each step follows standard procedures without requiring novel insight. The induction is mechanical once the reduction formula is established, making this moderately above average but not exceptionally challenging for Further Maths students.
Spec1.08i Integration by parts4.01a Mathematical induction: construct proofs

5 Let $$I _ { n } = \int _ { 1 } ^ { \mathrm { e } } x ( \ln x ) ^ { n } \mathrm {~d} x$$ where \(n \geqslant 1\). Show that $$I _ { n + 1 } = \frac { 1 } { 2 } \mathrm { e } ^ { 2 } - \frac { 1 } { 2 } ( n + 1 ) I _ { n }$$ Hence prove by induction that, for all positive integers \(n , I _ { n }\) is of the form \(A _ { n } \mathrm { e } ^ { 2 } + B _ { n }\), where \(A _ { n }\) and \(B _ { n }\) are rational numbers.

AnswerMarks
\(D[(x^2/2)(\ln x)^n] = x(\ln x)^n + (nx/2)(\ln x)^{n-1}\)M1
\(\Rightarrow [(x^2/2)(\ln x)^n]_1^e = I_n + (n/2)I_{n-1}\)A1
\(\Rightarrow ... \Rightarrow I_n = e^2/2 - (n/2)I_{n-1}\) for \(n \ge 2\)A1
\(\Rightarrow I_{n+1} = e^2/2 - (n+1)2I_{n-1}\) for \(n \ge 1\)A1
[3]
OR
AnswerMarks
\(\int x(\ln x)^n dx = [(x^2/2)(\ln x)^n]_1^e - \int (nx/2)(\ln x)^{n-1}dx\)M1A1
\(\Rightarrow I_n = e^2/2 - (n/2)I_{n-1}\)A1
\(\Rightarrow I_{n+1} = \frac{e^2}{2} - \frac{(n+1)}{2}I_n\)A1
\(H_k: I_k = A_k e^2 + B_k\), where \(A_k\) and \(B_k\) are rationalB1
\(H_r \Rightarrow I_{r+1} = e^2/2 - (k+1)(A_k e^2 + B_k)/2\)M1
\(= A_{k+1}ie^2 + B_{k+1}\), where \(A_{k+1} = 1/2 - (k+1)A_k/2, B_{k+1} = -(k+1)B_k/2\)A1
\(\Rightarrow A_{k+1}\) and \(B_{k+1}\) are rationalA1
\(I_1 = e^2/4 + 1/4 \Rightarrow A_1 = 1/4, B_1 = 1/4 \Rightarrow H_1\) is trueM1A1
Completion of induction argumentA1
[6]
$D[(x^2/2)(\ln x)^n] = x(\ln x)^n + (nx/2)(\ln x)^{n-1}$ | M1 |

$\Rightarrow [(x^2/2)(\ln x)^n]_1^e = I_n + (n/2)I_{n-1}$ | A1 |

$\Rightarrow ... \Rightarrow I_n = e^2/2 - (n/2)I_{n-1}$ for $n \ge 2$ | A1 |

$\Rightarrow I_{n+1} = e^2/2 - (n+1)2I_{n-1}$ for $n \ge 1$ | A1 |

| [3] |

**OR**

$\int x(\ln x)^n dx = [(x^2/2)(\ln x)^n]_1^e - \int (nx/2)(\ln x)^{n-1}dx$ | M1A1 |

$\Rightarrow I_n = e^2/2 - (n/2)I_{n-1}$ | A1 |

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

$H_k: I_k = A_k e^2 + B_k$, where $A_k$ and $B_k$ are rational | B1 |

$H_r \Rightarrow I_{r+1} = e^2/2 - (k+1)(A_k e^2 + B_k)/2$ | M1 |

$= A_{k+1}ie^2 + B_{k+1}$, where $A_{k+1} = 1/2 - (k+1)A_k/2, B_{k+1} = -(k+1)B_k/2$ | A1 |

$\Rightarrow A_{k+1}$ and $B_{k+1}$ are rational | A1 |

$I_1 = e^2/4 + 1/4 \Rightarrow A_1 = 1/4, B_1 = 1/4 \Rightarrow H_1$ is true | M1A1 |

Completion of induction argument | A1 |

| [6] |
5 Let

$$I _ { n } = \int _ { 1 } ^ { \mathrm { e } } x ( \ln x ) ^ { n } \mathrm {~d} x$$

where $n \geqslant 1$. Show that

$$I _ { n + 1 } = \frac { 1 } { 2 } \mathrm { e } ^ { 2 } - \frac { 1 } { 2 } ( n + 1 ) I _ { n }$$

Hence prove by induction that, for all positive integers $n , I _ { n }$ is of the form $A _ { n } \mathrm { e } ^ { 2 } + B _ { n }$, where $A _ { n }$ and $B _ { n }$ are rational numbers.

\hfill \mbox{\textit{CAIE FP1 2010 Q5 [9]}}