Question 9 - A-Level Maths

CAIE FP1 2015 November — Question 9

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2015
Session	November
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Reduction Formulae
Type	Logarithmic power integrals
Difficulty	Challenging +1.3 This is a standard reduction formula question requiring integration by parts to derive the recurrence relation, then applying it twice to find I₃. The integration by parts is straightforward with u=(ln x)^n, and the 'mean value' application is routine (dividing by interval length). While it requires careful algebraic manipulation and is from Further Maths, it follows a well-practiced template without requiring novel insight.
Spec	1.08i Integration by parts 4.08e Mean value of function: using integral 8.06a Reduction formulae: establish, use, and evaluate recursively

9 It is given that $I _ { n } = \int _ { 1 } ^ { \mathrm { e } } ( \ln x ) ^ { n } \mathrm {~d} x$ for $n \geqslant 0$. Show that $$I _ { n } = ( n - 1 ) \left[ I _ { n - 2 } - I _ { n - 1 } \right] \text { for } n \geqslant 2$$ Hence find, in an exact form, the mean value of $( \ln x ) ^ { 3 }$ with respect to $x$ over the interval $1 \leqslant x \leqslant \mathrm { e }$.

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Question 9:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\int \ln x \, dx = x\ln x - x$	B1
$I_n = \int_1^e (\ln x)^{n-1} \cdot \ln x \, dx$	M1
$= \left[(\ln x)^{n-1}(x\ln x - x)\right]_1^e - \int_1^e (n-1)(\ln x)^{n-2} \cdot \frac{1}{x}(x\ln x - x)dx$	M1A1
$= 0 - \int_1^e (n-1)(\ln x)^{n-2}(\ln x - 1)dx = (n-1)\left[I_{n-2} - I_{n-1}\right]$	M1A1	AG, [6]
Alternative: $I_n = \int_1^e (\ln x)^n \times 1\,dx = \left[x(\ln x)^n\right]_1^e - \int_1^e n(\ln x)^{n-1}dx$	M1A1
$\Rightarrow I_n = e - nI_{n-1}$	A1
Similarly $I_{n-1} = e - (n-1)I_{n-2}$	B1
$\Rightarrow I_n + nI_{n-1} = I_{n-1} + (n-1)I_{n-2}$	M1
$\Rightarrow I_n = (n-1)\left[I_{n-2} - I_{n-1}\right]$	A1	AG, [6]
$I_0 = \left[x\right]_1^e = e-1$	B1
$I_1 = \left[x\ln x - x\right]_1^e = 1$	B1
$I_2 = 1\times(e-1-1) = e-2$	M1
$I_3 = 2(I_1 - I_2) = 2(1-[e-2]) = 6-2e$	A1
$\text{MV} = \dfrac{I_3}{e-1} = \dfrac{6-2e}{e-1}$	M1A1	[6], Total 12

## Question 9:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \ln x \, dx = x\ln x - x$ | B1 | |
| $I_n = \int_1^e (\ln x)^{n-1} \cdot \ln x \, dx$ | M1 | |
| $= \left[(\ln x)^{n-1}(x\ln x - x)\right]_1^e - \int_1^e (n-1)(\ln x)^{n-2} \cdot \frac{1}{x}(x\ln x - x)dx$ | M1A1 | |
| $= 0 - \int_1^e (n-1)(\ln x)^{n-2}(\ln x - 1)dx = (n-1)\left[I_{n-2} - I_{n-1}\right]$ | M1A1 | AG, [6] |
| **Alternative:** $I_n = \int_1^e (\ln x)^n \times 1\,dx = \left[x(\ln x)^n\right]_1^e - \int_1^e n(\ln x)^{n-1}dx$ | M1A1 | |
| $\Rightarrow I_n = e - nI_{n-1}$ | A1 | |
| Similarly $I_{n-1} = e - (n-1)I_{n-2}$ | B1 | |
| $\Rightarrow I_n + nI_{n-1} = I_{n-1} + (n-1)I_{n-2}$ | M1 | |
| $\Rightarrow I_n = (n-1)\left[I_{n-2} - I_{n-1}\right]$ | A1 | AG, [6] |
| $I_0 = \left[x\right]_1^e = e-1$ | B1 | |
| $I_1 = \left[x\ln x - x\right]_1^e = 1$ | B1 | |
| $I_2 = 1\times(e-1-1) = e-2$ | M1 | |
| $I_3 = 2(I_1 - I_2) = 2(1-[e-2]) = 6-2e$ | A1 | |
| $\text{MV} = \dfrac{I_3}{e-1} = \dfrac{6-2e}{e-1}$ | M1A1 | [6], Total 12 |

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9 It is given that $I _ { n } = \int _ { 1 } ^ { \mathrm { e } } ( \ln x ) ^ { n } \mathrm {~d} x$ for $n \geqslant 0$. Show that

$$I _ { n } = ( n - 1 ) \left[ I _ { n - 2 } - I _ { n - 1 } \right] \text { for } n \geqslant 2$$

Hence find, in an exact form, the mean value of $( \ln x ) ^ { 3 }$ with respect to $x$ over the interval $1 \leqslant x \leqslant \mathrm { e }$.

\hfill \mbox{\textit{CAIE FP1 2015 Q9}}

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