FP3 June 2012 Q4

EdexcelOld spec11 marksIntegration

4. \[I_n = \int_0^{\frac{\pi}{4}} x^n\sin 2x\,\mathrm{d}x, \qquad n \geqslant 0\]

(a) Prove that, for \(n \geqslant 2\), \[I_n = \frac{1}{4}n\left(\frac{\pi}{4}\right)^{n-1} - \frac{1}{4}n(n - 1)I_{n-2}\] (5)

(b) Find the exact value of \(I_2\) (4)

Mark scheme (a)

Scheme	Marks
\(I_n = \left[x^n\left(-\tfrac{1}{2}\cos 2x\right)\right]_0^{\tfrac{\pi}{4}} - \displaystyle\int_0^{\tfrac{\pi}{4}} -\tfrac{1}{2}nx^{n-1}\cos 2x\,\mathrm{d}x\)	M1 A1
so \(I_n = \left\langle\left[x^n\left(-\tfrac{1}{2}\cos 2x\right)\right]_0^{\tfrac{\pi}{4}}\right\rangle + \left[\tfrac{1}{4}nx^{n-1}\sin 2x\right]_0^{\tfrac{\pi}{4}} - \displaystyle\int_0^{\tfrac{\pi}{4}} \tfrac{1}{4}n(n - 1)x^{n-2}\sin 2x\,\mathrm{d}x\)	M1 A1
i.e. \(I_n = \dfrac{1}{4}n\left(\dfrac{\pi}{4}\right)^{n-1} - \dfrac{1}{4}n(n - 1)I_{n-2}\) *	A1cso
	(5)

a1M1: Use of integration by parts, integrating \(\sin 2x\), differentiating \(x^n\).

a1A1: cao

a2M1: Second application of integration by parts, integrating \(\cos 2x\), differentiating \(x^{n-1}\).

a2A1: cao

a3A1: cso Including correct use of \(\dfrac{\pi}{4}\) and 0 as limits.

Mark scheme (b)

Scheme	Marks
\(I_0 = \displaystyle\int_0^{\tfrac{\pi}{4}} \sin 2x\,\mathrm{d}x = \left[-\tfrac{1}{2}\cos 2x\right]_0^{\tfrac{\pi}{4}} = \tfrac{1}{2}\)	M1 A1
\(I_2 = \dfrac{1}{4} \times 2 \times \left(\dfrac{\pi}{4}\right) - \dfrac{1}{4} \times 2 \times I_0\), so \(I_2 = \dfrac{\pi}{8} - \dfrac{1}{4}\)	M1 A1
	(4)

b1M1: Integrating to find \(I_0\) or setting up parts to find \(I_2\).

b1A1: cao ( Accept \(I_0 = \tfrac{1}{2}\) here for both marks)

b2M1: Finding \(I_2\) in terms of \(\pi\). If ‘n’s left in M0

b2A1: cao

Mark scheme (c)

Scheme	Marks
\(I_4 = \left(\dfrac{\pi}{4}\right)^3 - \dfrac{1}{4} \times 4 \times 3I_2 = \dfrac{\pi^3}{64} - 3\left(\dfrac{\pi}{8} - \dfrac{1}{4}\right) = \tfrac{1}{64}(\pi^3 - 24\pi + 48)\) *	M1 A1cso
	(2)
	(11 marks)

c1M1: Finding \(I_4\) in terms of \(I_2\) then in terms of \(\pi\). If ‘n’s left in M0

c1A1: cso