| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Algebraic function with square root |
| Difficulty | Challenging +1.8 This is a Further Maths F3 reduction formula question requiring integration by parts to derive the recurrence relation, then applying it twice with careful algebra. The derivation is methodical but requires confident manipulation of the square root term, and the definite integral evaluation demands accurate substitution at boundaries. While conceptually standard for FM students, the multi-step nature and algebraic complexity place it above average difficulty. |
| Spec | 1.08h Integration by substitution8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(I_n = \int \frac{x^n}{\sqrt{10-x^2}}dx = \int \frac{x^{n-1} \times x}{\sqrt{10-x^2}}dx\) | M1 | Writes \(x^n\) as \(x \times x^{n-1}\) |
| \(\int \frac{x^{n-1} \times x}{\sqrt{10-x^2}}dx = -x^{n-1}(10-x^2)^{\frac{1}{2}} + (n-1)\int x^{n-2}(10-x^2)^{\frac{1}{2}}dx\) i.e. \(\alpha x^{n-1}(10-x^2)^{\frac{1}{2}} + \beta \int x^{n-2}(10-x^2)^{\frac{1}{2}}dx\) | dM1A1 | dM1: Uses integration by parts; A1: Correct expression |
| \(= \ldots + (n-1)\int x^{n-2}(10-x^2)(10-x^2)^{-\frac{1}{2}}dx\) then splits: \(= \ldots + 10(n-1)\int x^{n-2}(10-x^2)^{\frac{1}{2}}dx - (n-1)\int x^n(10-x^2)^{-\frac{1}{2}}dx\) | dM1 | Applies \((10-x^2)^{\frac{1}{2}} = (10-x^2)(10-x^2)^{-\frac{1}{2}}\) and splits into 2 integrals |
| \(= \ldots + 10(n-1)I_{n-2} - (n-1)I_n \Rightarrow nI_n\) | dM1 | Introduces \(I_{n-2}\) and \(I_n\) and makes progress to given result |
| \(nI_n = 10(n-1)I_{n-2} - x^{n-1}(10-x^2)^{\frac{1}{2}}\) * | A1* | Fully correct proof with no errors (recovery of missing brackets counts as error, as does missing dx) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(I_1 = \int_0^1 \frac{x}{\sqrt{10-x^2}}dx = \left[-(10-x^2)^{\frac{1}{2}}\right]_0^1 = -3+\sqrt{10}\) | M1 | Correct method for \(I_1\); limits can be substituted later |
| \(5I_5 = 10 \times 4I_3 + \ldots\) | M1 | Applies the reduction formula at least once; allow with 3 or \(\left[-x^4(10-x^2)^{\frac{1}{2}}\right]_0^1\) |
| \(I_5 = 8I_3 - \frac{3}{5} = 8\left(\frac{20}{3}I_1 - 1\right) - \frac{3}{5} = \frac{160}{3}I_1 - \frac{43}{5}\) then \(I_5 = \frac{160}{3}(\sqrt{10}-3) - \frac{43}{5}\) | M1 | Completes process using their \(I_1\) to obtain numerical value for \(I_5\); limits must now be substituted |
| \(= \frac{1}{15}(800\sqrt{10} - 2529)\) | A1 | cao |
# Question 7:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $I_n = \int \frac{x^n}{\sqrt{10-x^2}}dx = \int \frac{x^{n-1} \times x}{\sqrt{10-x^2}}dx$ | M1 | Writes $x^n$ as $x \times x^{n-1}$ |
| $\int \frac{x^{n-1} \times x}{\sqrt{10-x^2}}dx = -x^{n-1}(10-x^2)^{\frac{1}{2}} + (n-1)\int x^{n-2}(10-x^2)^{\frac{1}{2}}dx$ i.e. $\alpha x^{n-1}(10-x^2)^{\frac{1}{2}} + \beta \int x^{n-2}(10-x^2)^{\frac{1}{2}}dx$ | dM1A1 | dM1: Uses integration by parts; A1: Correct expression |
| $= \ldots + (n-1)\int x^{n-2}(10-x^2)(10-x^2)^{-\frac{1}{2}}dx$ then splits: $= \ldots + 10(n-1)\int x^{n-2}(10-x^2)^{\frac{1}{2}}dx - (n-1)\int x^n(10-x^2)^{-\frac{1}{2}}dx$ | dM1 | Applies $(10-x^2)^{\frac{1}{2}} = (10-x^2)(10-x^2)^{-\frac{1}{2}}$ and splits into 2 integrals |
| $= \ldots + 10(n-1)I_{n-2} - (n-1)I_n \Rightarrow nI_n$ | dM1 | Introduces $I_{n-2}$ and $I_n$ and makes progress to given result |
| $nI_n = 10(n-1)I_{n-2} - x^{n-1}(10-x^2)^{\frac{1}{2}}$ * | A1* | Fully correct proof with no errors (recovery of missing brackets counts as error, as does missing dx) |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $I_1 = \int_0^1 \frac{x}{\sqrt{10-x^2}}dx = \left[-(10-x^2)^{\frac{1}{2}}\right]_0^1 = -3+\sqrt{10}$ | M1 | Correct method for $I_1$; limits can be substituted later |
| $5I_5 = 10 \times 4I_3 + \ldots$ | M1 | Applies the reduction formula at least once; allow with 3 or $\left[-x^4(10-x^2)^{\frac{1}{2}}\right]_0^1$ |
| $I_5 = 8I_3 - \frac{3}{5} = 8\left(\frac{20}{3}I_1 - 1\right) - \frac{3}{5} = \frac{160}{3}I_1 - \frac{43}{5}$ then $I_5 = \frac{160}{3}(\sqrt{10}-3) - \frac{43}{5}$ | M1 | Completes process using their $I_1$ to obtain numerical value for $I_5$; limits must now be substituted |
| $= \frac{1}{15}(800\sqrt{10} - 2529)$ | A1 | cao |
---7.
$$I _ { n } = \int \frac { x ^ { n } } { \sqrt { 10 - x ^ { 2 } } } \mathrm {~d} x \quad n \in \mathbb { N } \quad | x | < \sqrt { 10 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$n I _ { n } = 10 ( n - 1 ) I _ { n - 2 } - x ^ { n - 1 } \left( 10 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } \quad n \geqslant 2$$
\item Hence find the exact value of
$$\int _ { 0 } ^ { 1 } \frac { x ^ { 5 } } { \sqrt { 10 - x ^ { 2 } } } \mathrm {~d} x$$
giving your answer in the form $\frac { 1 } { 15 } ( p \sqrt { 10 } + q )$ where $p$ and $q$ are integers to be determined.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2022 Q7 [10]}}