FP3 June 2013 (R) Q5
5. \[I_n = \int_1^5 x^n(2x - 1)^{-\frac{1}{2}}\,\mathrm{d}x, \qquad n \geqslant 0\]
(a) Prove that, for \(n \geqslant 1\), \[(2n + 1)I_n = nI_{n-1} + 3 \times 5^n - 1\] (5)
(b) Using the reduction formula given in part (a), find the exact value of \(I_2\) (5)
| Scheme | Marks |
|---|---|
| \(I_n = \left[x^n(2x - 1)^{\frac{1}{2}}\right]_1^5 - \displaystyle\int_1^5 nx^{n-1}(2x - 1)^{\frac{1}{2}}\,\mathrm{d}x\) M1: Parts in the correct direction including a valid attempt to integrate \((2x - 1)^{-\frac{1}{2}}\) A1: Fully correct application – may be un-simplified. (Ignore limits) | M1 A1 |
| \(I_n = \underline{5^n \times 3 - 1} - \displaystyle\int_1^5 nx^{n-1}\underline{(2x - 1)(2x - 1)^{-\frac{1}{2}}}\,\mathrm{d}x\) Obtains a correct (possibly un-simplified) expression using the limits 5 and 1 and writes \((2x - 1)^{\frac{1}{2}}\) as \((2x - 1)(2x - 1)^{-\frac{1}{2}}\) | B1 |
| \(I_n = 5^n \times 3 - 1 - 2nI_n + nI_{n-1}\) Replaces \(\displaystyle\int x^n(2x - 1)^{-\frac{1}{2}}\,\mathrm{d}x\) with \(I_n\) and \(\displaystyle\int x^{n-1}(2x - 1)^{-\frac{1}{2}}\,\mathrm{d}x\) with \(I_{n-1}\) | dM1 |
| \((2n + 1)I_n = nI_{n-1} + 3 \times 5^n - 1\ ^*\) Correct completion to printed answer with no errors seen | A1cso |
| (5) |
| Scheme | Marks |
|---|---|
| \(I_0 = \displaystyle\int_1^5 (2x - 1)^{-\frac{1}{2}}\,\mathrm{d}x = \left[(2x - 1)^{\frac{1}{2}}\right]_1^5 = 2\) \(I_0 = 2\) | B1 |
| \(5I_2 = 2I_1 + 74\) and \(3I_1 = I_0 + 14\) M1: Correctly applies the given reduction formula twice A1: Correct equations for \(I_2\) and \(I_1\) (may be implied) | M1 A1 |
| So \(I_1 = \dfrac{16}{3}\) and \(I_2 = \ldots\) or \(5I_2 = 2\dfrac{I_0 + 14}{3} + 74\) and \(I_2 = \ldots\) Completes to obtains a numerical expression for \(I_2\) | dM1 |
| \(I_2 = \dfrac{254}{15}\) | B1 |
| (5) | |
| (10 marks) |