FP3 June 2009 Q5

EdexcelOld spec11 marksIntegration

5. \[I_n = \int_0^5 \frac{x^n}{\sqrt{(25 - x^2)}}\,\mathrm{d}x, \qquad n \geqslant 0\]

(a) Find an expression for \(\displaystyle\int \frac{x}{\sqrt{(25 - x^2)}}\,\mathrm{d}x,\ \ 0 \leqslant x \leqslant 5.\) (2)

(b) Using your answer to part (a), or otherwise, show that \[I_n = \frac{25(n - 1)}{n}I_{n-2} \qquad n \geqslant 2\] (5)

Scheme	Marks
\(-(25 - x^2)^{\frac{1}{2}}\) \((+c)\)	M1A1
	(2)

Mark scheme (b)

Scheme	Marks
\(I_n = \displaystyle\int x^{n-1} \cdot \dfrac{x}{\sqrt{(25 - x^2)}}\,\mathrm{d}x = -x^{n-1}\sqrt{25 - x^2} + \displaystyle\int (n - 1)x^{n-2}\sqrt{(25 - x^2)}\,\mathrm{d}x\)	M1 A1ft
\(I_n = \left[-x^{n-1}\sqrt{25 - x^2}\right]_0^5 + \displaystyle\int_0^5 \dfrac{(n - 1)x^{n-2}(25 - x^2)}{\sqrt{(25 - x^2)}}\,\mathrm{d}x\)	M1
\(I_n = 0 + 25(n - 1)\,I_{n-2} - (n - 1)\,I_n\)	M1
\(\therefore nI_n = 25(n - 1)I_{n-2}\) and so \(I_n = \dfrac{25(n - 1)}{n}I_{n-2}\) *	A1
	(5)

Scheme	Marks
Using substitution \(x = 5\sin\theta\) \(I_n = 5^n\displaystyle\int_0^{\frac{\pi}{2}} \sin^n\theta\,\mathrm{d}\theta = \left[-5^n\sin^{n-1}\theta\cos\theta\right]_0^{\frac{\pi}{2}} + 5^n(n - 1)\displaystyle\int_0^{\frac{\pi}{2}} \sin^{n-2}\theta\cos^2\theta\,\mathrm{d}\theta\)	M1A1
\(= \left[-5^n\sin^{n-1}\theta\cos\theta\right]_0^{\frac{\pi}{2}} + 5^n(n - 1)\displaystyle\int_0^{\frac{\pi}{2}} \sin^{n-2}\theta(1 - \sin^2\theta)\,\mathrm{d}\theta\)	M1
\(I_n = 0 + 25(n - 1)\,I_{n-2} - (n - 1)\,I_n\)	M1
\(\therefore nI_n = 25(n - 1)I_{n-2}\) and so \(I_n = \dfrac{25(n - 1)}{n}I_{n-2}\) * (need to see that \(I_{n-2} = 5^{n-2}\displaystyle\int_0^{\frac{\pi}{2}} \sin^{n-2}\theta\,\mathrm{d}\theta\) for final A1)	A1
	(5)

Mark scheme (c)

Scheme	Marks
\(I_0 = \displaystyle\int_0^5 \dfrac{1}{\sqrt{(25 - x^2)}}\,\mathrm{d}x = \left[\arcsin\left(\tfrac{x}{5}\right)\right]_0^5 = \dfrac{\pi}{2}\)	M1 A1
\(I_4 = \dfrac{25 \times 3}{4} \times \dfrac{25 \times 1}{2}I_0 = \dfrac{1875}{16}\pi\)	M1 A1
	(4)
	(11 marks)