FP3 June 2013 Q6
6. Given that \[I_n = \int_0^4 x^n\sqrt{(16 - x^2)}\,\mathrm{d}x, \quad n \geqslant 0,\]
(a) prove that, for \(n \geqslant 2\), \[(n + 2)I_n = 16(n - 1)I_{n-2}\] (6)
(b) Hence, showing each step of your working, find the exact value of \(I_5\) (5)
| Scheme | Marks |
|---|---|
| \(I_n = \displaystyle\int_0^4 x^{n-1} \times \underline{x(16 - x^2)^{\frac{1}{2}}}\,\mathrm{d}x\) M1: Obtains \(x(16 - x^2)^{\frac{1}{2}}\) prior to integration A1: Correct underlined expression (can be implied by their integration) | M1A1 |
| \(I_n = \left[-\tfrac{1}{3}x^{n-1}(16 - x^2)^{\frac{3}{2}}\right]_0^4 + \dfrac{n - 1}{3}\displaystyle\int_0^4 x^{n-2}(16 - x^2)^{\frac{3}{2}}\,\mathrm{d}x\) dM1: Parts in the correct direction (Ignore limits) | dM1 |
| \(\therefore I_n = \dfrac{n - 1}{3}\displaystyle\int_0^4 x^{n-2}(16 - x^2)(16 - x^2)^{\frac{1}{2}}\,\mathrm{d}x\) | |
| i.e. \(I_n = \dfrac{16(n - 1)}{3}I_{n-2} - \dfrac{n - 1}{3}I_n\) Manipulates to obtain at least one integral in terms of \(I_n\) or \(I_{n-2}\) on the rhs. | M1 |
| \(I_n\left(1 + \dfrac{n - 1}{3}\right) = \dfrac{16(n - 1)}{3}I_{n-2}\) Collects terms in \(I_n\) from both sides | M1 |
| \((n + 2)I_n = 16(n - 1)I_{n-2}\ ^*\) Printed answer with no errors | A1*cso |
| (6) |
Notes
Way 2
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^4 x^n(16 - x^2)^{\frac{1}{2}}\,\mathrm{d}x = \int_0^4 x^n\frac{(16 - x^2)}{(16 - x^2)^{\frac{1}{2}}}\,\mathrm{d}x = \int_0^4 \frac{16x^n}{(16 - x^2)^{\frac{1}{2}}}\,\mathrm{d}x - \int_0^4 \frac{x^{n+2}}{(16 - x^2)^{\frac{1}{2}}}\,\mathrm{d}x\) | |
| \(= \displaystyle\int_0^4 16x^{n-1} \times x(16 - x^2)^{-\frac{1}{2}}\,\mathrm{d}x - \int_0^4 x^{n+1} \times x(16 - x^2)^{-\frac{1}{2}}\,\mathrm{d}x\) M1: Obtains \(x(16 - x^2)^{-\frac{1}{2}}\) prior to integration A1: Correct expressions | M1A1 |
| \(= \left[-16x^{n-1}(16 - x^2)^{\frac{1}{2}}\right]_0^4 + 16(n - 1)\displaystyle\int_0^4 x^{n-2}(16 - x^2)^{\frac{1}{2}}\,\mathrm{d}x\) \(-\left(\left[-x^{n+1}(16 - x^2)^{\frac{1}{2}}\right]_0^4 + (n + 1)\displaystyle\int_0^4 x^n(16 - x^2)^{\frac{1}{2}}\,\mathrm{d}x\right)\) dM1: Parts in the correct direction on both (Ignore limits) | dM1 |
| \(I_n = 16(n - 1)I_{n-2} - (n + 1)I_n\) Manipulates to obtain at least one integral in terms of \(I_n\) or \(I_{n-2}\) on the rhs. | M1 |
| \(I_n(1 + n + 1) = 16(n - 1)I_{n-2}\) Collects terms in \(I_n\) from both sides | M1 |
| \((n + 2)I_n = 16(n - 1)I_{n-2}\ ^*\) Printed answer with no errors | A1* |
Way 3
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^4 x^n(16 - x^2)^{\frac{1}{2}}\,\mathrm{d}x = \int_0^4 x \times x^{n-1}\frac{(16 - x^2)}{(16 - x^2)^{\frac{1}{2}}}\,\mathrm{d}x\) M1: Obtains \(x(16 - x^2)^{-\frac{1}{2}}\) prior to integration A1: Correct expression | M1A1 |
| \(= \left[-x^{n-1}(16 - x^2)(16 - x^2)^{\frac{1}{2}}\right]_0^4 + \displaystyle\int_0^4 (16(n - 1)x^{n-2} - (n + 1)x^n)(16 - x^2)^{\frac{1}{2}}\,\mathrm{d}x\) dM1: Parts in the correct direction (Ignore limits) | dM1 |
| \(I_n = 16(n - 1)I_{n-2} - (n + 1)I_n\) Manipulates to obtain at least one integral in terms of \(I_n\) or \(I_{n-2}\) on the rhs. | M1 |
| \(I_n(1 + n + 1) = 16(n - 1)I_{n-2}\) Collects terms in \(I_n\) from both sides | M1 |
| \((n + 2)I_n = 16(n - 1)I_{n-2}\ ^*\) Printed answer with no errors | A1* |
| Scheme | Marks |
|---|---|
| \(I_1 = \displaystyle\int_0^4 x\sqrt{(16 - x^2)}\,\mathrm{d}x = \left[-\tfrac{1}{3}(16 - x^2)^{\frac{3}{2}}\right]_0^4 = \frac{64}{3}\) M1: Correct integration to find \(I_1\) A1: \(\dfrac{64}{3}\) or equivalent (May be implied by a later work – they are not asked explicitly for \(I_1\)) \(\dfrac{64}{3}\) must come from correct work | M1 A1 |
| \(I_5 = \dfrac{64}{7}I_3,\ I_3 = \dfrac{32}{5}I_1\) Applies to apply reduction formula twice. First M1 for \(I_5\) in terms of \(I_3\), second M1 for \(I_3\) in terms of \(I_1\) (Can be implied) | M1, M1 |
| \(I_5 = \dfrac{131072}{105}\) Any exact equivalent (Depends on all previous marks having been scored) | A1 |
| (5) | |
| (11 marks) |
Notes
Using \(x = 4\sin\theta\):
\(I_1 = \displaystyle\int_0^{\frac{\pi}{2}} 4\sin\theta\sqrt{(16 - 16\sin^2\theta)}\,4\cos\theta\,\mathrm{d}\theta = \int_0^{\frac{\pi}{2}} 64\sin\theta\cos^2\theta\,\mathrm{d}\theta = \left[-\frac{64}{3}\cos^3\theta\right]_0^{\frac{\pi}{2}}\)
M1: A complete substitution and attempt to substitute changed limits
A1: \(\dfrac{64}{3}\) or equivalent