# Question 8(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = x^n$, $u' = nx^{n-1}$, $v' = (k-x)^{\frac{1}{2}}$, $v = -\frac{2}{3}(k-x)^{\frac{3}{2}}$ $I_n = \left[-\frac{2}{3}x^n(k-x)^{\frac{3}{2}}\right]_0^k - \int_0^k -\frac{2}{3}nx^{n-1}(k-x)^{\frac{3}{2}}\,dx$ | M1 A1 | M1: Uses parts in correct direction to obtain $\pm x^n(k-x)^{\frac{3}{2}} \pm \int x^{n-1}(k-x)^{\frac{3}{2}}(dx)$. A1: Correct expression (limits not required, $dx$ may be missing) |
| $(I_n =)\, 0 + \frac{2}{3}n\int_0^k x^{n-1}(k-x)(k-x)^{\frac{1}{2}}\,dx$ | dM1 | Applies $(k-x)^{\frac{3}{2}} = (k-x)(k-x)^{\frac{1}{2}}$. Could be implied if work correct. Do not accept going straight to $\frac{2}{3}nkI_{n-1} - \frac{2}{3}nI_n$. Requires previous M mark |
| $\frac{2}{3}n\int_0^k\left(kx^{n-1}(k-x)^{\frac{1}{2}} - x^n(k-x)^{\frac{1}{2}}\right)dx$ $\Rightarrow \frac{2}{3}n(kI_{n-1} - I_n)$ | ddM1 | Expands and writes RHS in terms of both $I_n$ and $I_{n-1}$. Not available until $\left[x^n(k-x)^{\frac{3}{2}}\right]_0^k$ disappears. Requires both previous M marks |
| $\left(1+\frac{2}{3}n\right)I_n = \frac{2}{3}knI_{n-1} \Rightarrow I_n = \frac{2kn}{3+2n}I_{n-1}$* | A1* | Reaches given answer with no mathematical errors. At least one non-trivial intermediate line where LHS $= f(n)I_n$. Allow minor variations e.g. $I_n = \frac{2nkI_{n-1}}{2n+3}$. Condone missing $dx$s but $\left[-\frac{2}{3}x^n(k-x)^{\frac{3}{2}}\right]_0^k$ must be replaced by "0" or better |

# Question 8(a):

**Split first method:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = \int_0^k x^n(k-x)^{\frac{1}{2}}\,dx = \int_0^k x^n(k-x)(k-x)^{-\frac{1}{2}}\,dx = \int_0^k kx^n(k-x)^{-\frac{1}{2}}\,dx - \int_0^k x^{n+1}(k-x)^{-\frac{1}{2}}\,dx$ | | Split shown |
| $= \left[-2kx^n(k-x)^{\frac{1}{2}}\right]_0^k + \int_0^k 2knx^{n-1}(k-x)^{\frac{1}{2}}\,dx + \left[2x^{n+1}(k-x)^{\frac{1}{2}}\right]_0^k - \int_0^k 2(n+1)x^n(k-x)^{\frac{1}{2}}\,dx$ | | Integration by parts applied to both parts |
| $\Rightarrow 0 + 2knI_{n-1} + 0 - 2(n+1)I_n \Rightarrow (3+2n)I_n = 2knI_{n-1} \Rightarrow I_n = \dfrac{2kn}{3+2n}I_{n-1}$ | **(5)** | First 2 method marks awarded together for split + parts giving correct form; accuracy mark for correct expression (limits not required, $dx$ may be missing) |

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# Question 8(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_2 = \dfrac{4k}{7}I_1 = \dfrac{4k}{7}\left(\dfrac{2k}{5}I_0\right)$ or $I_2 = \dfrac{4k}{7}I_1,\quad I_1 = \dfrac{2k}{5}I_0$ | **M1** | Attempts $I_2$ in terms of $I_0$, or $I_2$ in terms of $I_1$ and $I_1$ in terms of $I_0$. Accept with $I_0$ substituted. Allow $I_0 = 1$ to be used (i.e. $I_0$ lost) |
| $I_0 = \int_0^k (k-x)^{\frac{1}{2}}\,dx = \left[-\dfrac{2}{3}(k-x)^{\frac{3}{2}}\right]_0^k$ | **M1** | $I_0 = \ldots(k-x)^{\frac{3}{2}}$. Limits do not have to be seen or applied |
| $I_2 = \dfrac{8k^2}{35} \times \dfrac{2}{3}k^{\frac{3}{2}} \Rightarrow \dfrac{16}{105}k^{\frac{7}{2}} = \dfrac{9\sqrt{3}}{280} \Rightarrow k = \ldots$ | **ddM1** | Solves equation of form $\dfrac{a}{b}k^c = \dfrac{9\sqrt{3}}{280}$ where $a,b \in \mathbb{Z}^*$, $\dfrac{a}{b} \notin \mathbb{Z}$, $c = 5$ or $7$. LHS must be their $I_2$. Requires both previous M marks |
| $k^{\frac{7}{2}} = \dfrac{27\sqrt{3}}{128} \Rightarrow k^7 = \dfrac{2187}{16384} \Rightarrow k = \dfrac{3}{4}$ | **A1** | Correct exact value for $k$ from correct equation. Not $\sqrt[7]{\dfrac{2187}{16384}}$ nor $\pm\dfrac{3}{4}$ |

**Total: (4) marks for 8(b), Total 9**

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