FP3 June 2011 Q4

EdexcelOld spec8 marksIntegration

4. \[I_n = \int_1^{e} x^2(\ln x)^n\,\mathrm{d}x, \qquad n \geqslant 0\]

(a) Prove that, for \(n \geqslant 1\), \[I_n = \frac{e^3}{3} - \frac{n}{3}I_{n-1}\] (4)

(b) Find the exact value of \(I_3\). (4)

Mark scheme (a)

Scheme	Marks
\(I_n = \left[\dfrac{x^3}{3}(\ln x)^n\right] - \displaystyle\int \dfrac{x^3}{3} \times \dfrac{n(\ln x)^{n-1}}{x}\,\mathrm{d}x\)	M1 A1
\(= \left[\dfrac{x^3}{3}(\ln x)^n\right]_1^{e} - \displaystyle\int_1^{e} \dfrac{nx^2(\ln x)^{n-1}}{3}\,\mathrm{d}x\)	DM1
\(\therefore I_n = \dfrac{e^3}{3} - \dfrac{n}{3}I_{n-1}\) *	A1cso
	(4)

1M1 Using integration by parts, integrating \(x^2\), differentiating \((\ln x)^n\)

1A1 CAO

2DM1 Correctly using limits 1 and e

2A1 CSO answer given

Mark scheme (b)

Scheme	Marks
\(I_0 = \displaystyle\int_1^{e} x^2\,\mathrm{d}x = \left[\dfrac{x^3}{3}\right]_1^{e} = \dfrac{e^3}{3} - \dfrac{1}{3}\) or \(I_1 = \dfrac{e^3}{3} - \dfrac{1}{3}\left(\dfrac{e^3}{3} - \dfrac{1}{3}\right) = \dfrac{2e^3}{9} + \dfrac{1}{9}\)	M1 A1
\(I_1 = \dfrac{e^3}{3} - \dfrac{1}{3}I_0,\ I_2 = \dfrac{e^3}{3} - \dfrac{2}{3}I_1\) and \(I_3 = \dfrac{e^3}{3} - \dfrac{3}{3}I_2\) so \(I_3 = \dfrac{4e^3}{27} + \dfrac{2}{27}\)	M1 A1
	(4)
	(8 marks)

1M1 Evaluating \(I_0\) or \(I_1\) by an attempt to integrate something

1A1 CAO

2M1 Finding \(I_3\) (also probably \(I_1\) and \(I_2\)) If ‘n’s left in M0

2A1 \(I_3\) CAO