Challenging +1.3 This is a standard reduction formula question requiring integration by parts to establish the recurrence relation, followed by systematic application to find I₃. While it involves multiple steps and careful algebraic manipulation, the technique is well-practiced in Further Maths syllabi and follows a predictable pattern. The mean value calculation at the end is routine once I₃ is found.
It is given that \(I_n = \int_{1}^{e} (\ln x)^n \mathrm{d}x\) for \(n \geqslant 0\). Show that
$$I_n = (n - 1)[I_{n-2} - I_{n-1}] \text{ for } n \geqslant 2.$$ [6]
Hence find, in an exact form, the mean value of \((\ln x)^3\) with respect to \(x\) over the interval \(1 \leqslant x \leqslant \mathrm{e}\). [6]
= [ (lnx)n−1(xlnx−x) ] e −∫ e (n−1)(lnx)n−2. 1 (xlnx−x)dx
1 1 x
=0−∫ e (n−1)(lnx)n−2(lnx−1)dx=(n−1) [ I −I ] (AG)
n−2 n−1
1
Alternative for obtaining reduction formula:
[ ]
I =∫ e (lnx)n×1dx= x(lnx)n e −∫ e n(lnx)n−1dx
n 1
1 1
⇒I =e−nI
n n−1
Similarly I =e−(n−1)I
n−1 n−2
⇒I +nI =I +(n−1)I
n n−1 n−1 n−2
[ ]
⇒I =(n−1) I −I (AG)
n n−2 n−1
[ ]
I = x e=e−1
0 1
[ ]
I = xlnx−x e=1
1 1
I =1×(e−1−1)=e−2
2
( )
I =2 I −I =2(1−[e−2])=6−2e
3 1 2
I 6−2e
MV = 3 =
Answer
Marks
e−1 e−1
B1
M1
M1A1
M1A1
[6]
M1A1
A1
B 1
M1
A 1
[6]
B1
B1
M1
A 1
M1 A1
[6]
Total
12
Answer
Marks
Guidance
Page 8
Mark Scheme
Syllabus
Cambridge International A Level – October/November 2015
9231
11
Qn &
Answer
Marks
Guidance
Part
Solution
Marks
Question 9:
9 | ∫ l e nxdx=xlnx−x
1
I =∫ e (lnx)n−1.lnxdx
n
1
= [ (lnx)n−1(xlnx−x) ] e −∫ e (n−1)(lnx)n−2. 1 (xlnx−x)dx
1 1 x
=0−∫ e (n−1)(lnx)n−2(lnx−1)dx=(n−1) [ I −I ] (AG)
n−2 n−1
1
Alternative for obtaining reduction formula:
[ ]
I =∫ e (lnx)n×1dx= x(lnx)n e −∫ e n(lnx)n−1dx
n 1
1 1
⇒I =e−nI
n n−1
Similarly I =e−(n−1)I
n−1 n−2
⇒I +nI =I +(n−1)I
n n−1 n−1 n−2
[ ]
⇒I =(n−1) I −I (AG)
n n−2 n−1
[ ]
I = x e=e−1
0 1
[ ]
I = xlnx−x e=1
1 1
I =1×(e−1−1)=e−2
2
( )
I =2 I −I =2(1−[e−2])=6−2e
3 1 2
I 6−2e
MV = 3 =
e−1 e−1 | B1
M1
M1A1
M1A1
[6]
M1A1
A1
B 1
M1
A 1
[6]
B1
B1
M1
A 1
M1 A1
[6]
Total
12
Page 8 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2015 | 9231 | 11
Qn &
Part | Solution | Marks
It is given that $I_n = \int_{1}^{e} (\ln x)^n \mathrm{d}x$ for $n \geqslant 0$. Show that
$$I_n = (n - 1)[I_{n-2} - I_{n-1}] \text{ for } n \geqslant 2.$$ [6]
Hence find, in an exact form, the mean value of $(\ln x)^3$ with respect to $x$ over the interval $1 \leqslant x \leqslant \mathrm{e}$. [6]
\hfill \mbox{\textit{CAIE FP1 2015 Q9 [12]}}