FP3 June 2014 (R) Q4
4. \[I_n = \int_0^{\sqrt{3}} (3 - x^2)^n\,\mathrm{d}x, \quad n \geqslant 0\]
(a) Show that, for \(n \geqslant 1\) \[I_n = \frac{6n}{2n + 1}I_{n-1}\] (6)
(b) Hence find the exact value of \(I_4\), giving your answer in the form \(k\sqrt{3}\) where \(k\) is a rational number to be found. (5)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^{\sqrt{3}} (3 - x^2)^n\,\mathrm{d}x\) | |
| \(\displaystyle\int_0^{\sqrt{3}} (3 - x^2)^n\,\mathrm{d}x = \left[x(3 - x^2)^n\right]_0^{\sqrt{3}} + \int_0^{\sqrt{3}} 2x^2n(3 - x^2)^{n-1}\,\mathrm{d}x\) M1: Integration by parts in the correct direction A1: Correct expression (Ignore limits) | M1A1 |
| \(= 0 - 2n\displaystyle\int_0^{\sqrt{3}} (3 - x^2 - 3)(3 - x^2)^{n-1}\,\mathrm{d}x\) Substitutes limits and uses \(x^2 = x^2 - 3 + 3\) Dependent on the first M | dM1 |
| \(= 0 - 2n\displaystyle\int_0^{\sqrt{3}} (3 - x^2)^n\,\mathrm{d}x + 6n\int_0^{\sqrt{3}} (3 - x^2)^{n-1}\,\mathrm{d}x\) Correct expressions | A1 |
| \(= 6nI_{n-1} - 2nI_n\) Substitutes for \(I_{n-1}\) and \(I_n\) Dependent on both M’s | ddM1 |
| \(I_n = \dfrac{6n}{2n + 1}I_{n-1}\ ^*\) Correct completion with no errors | A1* |
| (6) |
Notes
(a) Alternative
| Scheme | Marks |
|---|---|
| \(I_n = \displaystyle\int_0^{\sqrt{3}} (3 - x^2)^{n-1}(3 - x^2)\,\mathrm{d}x = 3\int_0^{\sqrt{3}} (3 - x^2)^{n-1}\,\mathrm{d}x - \int_0^{\sqrt{3}} x^2(3 - x^2)^{n-1}\,\mathrm{d}x\) M1: Writes the bracket as a product and separates into two integrals This is the second M and depends on the first M below | dM1 |
| \(= 3I_{n-1} - \displaystyle\int_0^{\sqrt{3}} x \times x(3 - x^2)^{n-1}\,\mathrm{d}x\) | |
| \(= 3I_{n-1} - \left\{\left[\dfrac{x(3 - x^2)^n}{-2n}\right]_0^{\sqrt{3}} - \displaystyle\int_0^{\sqrt{3}} \frac{(3 - x^2)^n}{-2n}\,\mathrm{d}x\right\}\) M1: Parts in the correct direction (First M1) A1: Correct expression (First A1) | M1A1 |
| \(= 3I_{n-1} - \displaystyle\int_0^{\sqrt{3}} \frac{(3 - x^2)^n}{2n}\,\mathrm{d}x\) Correct expression with no errors | A1 |
| \(= 3I_{n-1} - \dfrac{1}{2n}I_n\) Substitutes for \(I_{n-1}\) and \(I_n\) Dependent on both M’s | ddM1 |
| \(I_n = \dfrac{6n}{2n + 1}I_{n-1}\ ^*\) Correct completion with no errors | A1 |
(corrected from the printed mark scheme: the fifth line of the alternative is printed as \(= 3I_{n-1} - \dfrac{1}{2n}I_{n-1}\); the integral of \((3 - x^2)^n\) is \(I_n\))
| Scheme | Marks |
|---|---|
| \(I_0 = \sqrt{3}\) or \(I_1 = 2\sqrt{3}\) | B1 |
| \(I_4 = \dfrac{24}{9}I_3\) Attempt \(I_4\) in terms of \(I_3\) | M1 |
| \(I_4 = \dfrac{24}{9} \cdot \dfrac{18}{7}I_2 = \dfrac{24}{9} \cdot \dfrac{18}{7} \cdot \dfrac{12}{5}I_1\) M1: Attempt \(I_4\) in terms of \(I_1\) A1: Correct expression for \(I_4\) as shown or correct numerical expression | M1A1 |
| \(I_4 = \dfrac{24}{9} \cdot \dfrac{18}{7} \cdot \dfrac{12}{5} \cdot \dfrac{6}{3} \cdot \sqrt{3}\) | |
| \(I_4 = \dfrac{1152}{35}\sqrt{3}\) | A1 |
| (5) | |
| (11 marks) |