A2 June 2025 Q6
6.
\[I_n = \int_0^2 \left(4 - x^2\right)^n \mathrm{d}x\](Solutions relying entirely on calculator technology are not acceptable.)
(4)| Scheme | Marks | AO |
|---|---|---|
| \(\int \left(4 - x^2\right)^n\,\mathrm{d}x = \int 1 \times \left(4 - x^2\right)^n\,\mathrm{d}x\) \(u = \left(4 - x^2\right)^n \Rightarrow \tfrac{\mathrm{d}u}{\mathrm{d}x} = \lambda x\left(4 - x^2\right)^{n-1}\) and \(\tfrac{\mathrm{d}v}{\mathrm{d}x} = 1 \Rightarrow v = x\) \(= x\left(4 - x^2\right)^n - \int \lambda x^2 \left(4 - x^2\right)^{n-1}\) | M1 | 2.1 |
| \(\int \left(4 - x^2\right)^n\,\mathrm{d}x = x\left(4 - x^2\right)^n + \int 2nx^2 \left(4 - x^2\right)^{n-1}\,\mathrm{d}x\) | A1 | 1.1b |
| \(= \left[x\left(4 - x^2\right)^n\right]_0^2 + \int 2n\left\{4 - \left(4 - x^2\right)\right\}\left(4 - x^2\right)^{n-1}\,\mathrm{d}x\) \(= [0] + 8n\int \left(4 - x^2\right)^{n-1}\,\mathrm{d}x - 2n\int \left(4 - x^2\right)^n\,\mathrm{d}x\) | M1 | 2.1 |
| \(I_n = 8nI_{n-1} - 2nI_n\) \(I_n = \tfrac{8n}{2n+1}I_{n-1}\ *\) | A1* | 1.1b |
| (4) |
Notes
M1: Writes the integral as \(\int 1 \times \left(4 - x^2\right)^n\,\mathrm{d}x\) and applies integration by parts with \(u = \left(4 - x^2\right)^n\) and \(\dfrac{\mathrm{d}v}{\mathrm{d}x} = 1\) to achieve the correct form
A1: Correct integration
M1: Writes \(x^2\) as \(4 - \left(4 - x^2\right)\) to split the integral into the sum of \(I_n\) and \(I_{n-1}\)
A1*: Completes the proof by making \(I_n\) the subject with no error or omissions seen. Allow recovery of missing brackets.
Alt (a)
| Scheme | Marks | AO |
|---|---|---|
| \(\int \left(4 - x^2\right)^n\,\mathrm{d}x = \int \left(4 - x^2\right)\left(4 - x^2\right)^{n-1}\,\mathrm{d}x = 4\int \left(4 - x^2\right)^{n-1}\,\mathrm{d}x - \int x.x\left(4 - x^2\right)^{n-1}\,\mathrm{d}x\) \(u = x \Rightarrow \tfrac{\mathrm{d}u}{\mathrm{d}x} = 1\) and \(\tfrac{\mathrm{d}v}{\mathrm{d}x} = x\left(4 - x^2\right)^{n-1} \Rightarrow v = -\tfrac{1}{2n}\left(4 - x^2\right)^n\) \(\int \left(4 - x^2\right)^n\,\mathrm{d}x = 4I_{n-1} - \left[Kx\left(4 - x^2\right)^n - \int M\left(4 - x^2\right)^n\,\mathrm{d}x\right]\) | M1 | 2.1 |
| \(\int \left(4 - x^2\right)^n\,\mathrm{d}x = 4I_{n-1} - \left[-\dfrac{x}{2n}\left(4 - x^2\right)^n - \int -\dfrac{1}{2n}\left(4 - x^2\right)^n\,\mathrm{d}x\right]\) | A1 | 1.1b |
| \(= 4I_{n-1} - \left[-\dfrac{x}{2n}\left(4 - x^2\right)^n\right]_0^2 - \dfrac{1}{2n}\int_0^2 \left(4 - x^2\right)^n\,\mathrm{d}x = 4I_{n-1} + [0] - \dfrac{1}{2n}I_n\) | M1 | 2.1 |
| \(2nI_n = 8nI_{n-1} - I_n \Rightarrow I_n = \dfrac{8n}{2n+1}I_{n-1}\ *\) | A1* | 1.1b |
| (4) |
M1: Writes the integral as \(\int \left(4 - x^2\right) \times \left(4 - x^2\right)^{n-1}\,\mathrm{d}x\), splits and applies integration by parts with \(u = x\) and \(\dfrac{\mathrm{d}v}{\mathrm{d}x} = x\left(4 - x^2\right)^{n-1}\) to achieve the correct form.
A1: Correct integration.
M1: Evaluates the limits and identifies \(I_n\) and \(I_{n-1}\) within the integral.
A1*: Completes the proof by making \(I_n\) the subject with no error or omissions seen. Allow recovery of missing brackets.
NB: There is a possible method via using \(4 - x^2 = (2 - x)(2 + x)\) and integration by parts on these terms. You may score:
M1A1: \(\int_0^2 (2-x)^n(2+x)^n\,\mathrm{d}x = \left[(2-x)^n\dfrac{(2+x)^{n+1}}{n+1}\right]_0^2 - \int_0^2 -n(2-x)^{n-1}\dfrac{(2+x)^{n+1}}{n+1}\,\mathrm{d}x\) oe with \(u\), \(v\)’ reversed, then M1 for full method to reduce integral to terms of \(I\)’s and A1 if all correct. If unsure send to review. E.g. for second M
\[\begin{aligned}&= -\frac{2^{2n+1}}{n+1} + \frac{n}{n+1}\int_0^2 \left(4 - x^2\right)^{n-1}\left(4 + 4x + x^2\right)\,\mathrm{d}x\\ &= -\frac{2^{2n+1}}{n+1} + \frac{n}{n+1}\left(4I_{n-1} + 4\int_0^2 x\left(4 - x^2\right)^{n-1}\,\mathrm{d}x + \int_0^2 x^2\left(4 - x^2\right)^{n-1}\,\mathrm{d}x\right)\end{aligned}\]\[\begin{aligned}&= -\frac{2^{2n+1}}{n+1} + \frac{n}{n+1}\left(4I_{n-1} + 4\left[-\frac{\left(4 - x^2\right)^n}{2n}\right]_0^2 - \int_0^2 \left(4 - x^2 - 4\right)\left(4 - x^2\right)^{n-1}\,\mathrm{d}x\right)\\ &= -\cancel{\frac{2^{2n+1}}{n+1}} + \frac{n}{n+1}\left(4I_{n-1} + \cancel{4\frac{4^n}{2n}} - I_n + 4I_{n-1}\right)\end{aligned}\]\[\Rightarrow (n+1)I_n = 8nI_{n-1} - nI_n \Rightarrow I_n = \frac{8n}{2n+1}I_{n-1}\](corrected from the printed mark scheme: the last line is printed as \((n+1)I_{n-1} = 8nI_{n-1} - nI_n\); the left-hand side should be \((n+1)I_n\))
| Scheme | Marks | AO |
|---|---|---|
| \(I_1 = \int_0^2 \left(4 - x^2\right)\,\mathrm{d}x = \left[4x - \tfrac{x^3}{3}\right]_0^2 = \left(8 - \tfrac{8}{3}\right) - (0) = \ldots\left\{\tfrac{16}{3}\right\}\) Or \(I_0 = \int_0^2 1\,\mathrm{d}x = [x]_0^2 = \ldots \quad \left\{I_1 = \tfrac{8}{3} \times I_0 = \ldots\left\{\tfrac{8}{3} \times 2 = \tfrac{16}{3}\right\}\right\}\) | M1 | 2.1 |
| \(I_2 = \dfrac{16}{5} \times \text{their } \dfrac{16}{3} = \ldots\left\{\dfrac{256}{15}\right\}\) | M1 | 1.1b |
| \(I_3 = \dfrac{24}{7} \times \text{their } \dfrac{256}{15} = \ldots\left\{\dfrac{2048}{35}\right\},\ I_4 = \dfrac{32}{9} \times \text{their } \dfrac{2048}{35} = \ldots\left\{\dfrac{65536}{315}\right\}\) | M1 | 1.1b |
| \(n = 4\) | A1 | 2.2a |
| (4) | ||
| (8 marks) |
Notes
M1: Integrates to find the value of \(I_0 = \int_0^2 1\,\mathrm{d}x\) or \(I_1 = \int_0^2 \left(4 - x^2\right)\,\mathrm{d}x\) Need not be fully simplified at this stage.
M1: Correct method to find \(I_2\) Again, need not be simplified.
M1: Uses the reduction formula correctly to find the value of \(I_4\) Again, need not be simplified.
A1: Deduces the value of \(n\)