| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2021 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Derive reduction formula by integration by parts |
| Difficulty | Challenging +1.2 This is a standard reduction formula derivation using integration by parts with a straightforward algebraic structure. While it requires careful manipulation and is from Further Maths F3, the technique is routine for students at this level: choose u = x^(n-1), dv = x/√(x²+3) dx, apply IBP, and simplify. Part (b) is mechanical application of the formula. More routine than the geometric series proof example, but harder than basic C3 questions due to the algebraic complexity and Further Maths context. |
| Spec | 4.08f Integrate using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\frac{x^n}{\sqrt{x^2+3}}dx = \int x^{n-1}\cdot x(x^2+3)^{-1/2}dx\) | M1 | Applies \(x^n = x^{n-1}\times x\) to integral |
| \(\int x^{n-1}(x^2+3)^{-1/2}dx = x^{n-1}(x^2+3)^{1/2} - \int(n-1)x^{n-2}(x^2+3)^{1/2}dx\) | dM1 A1 | dM1: Integration by parts to obtain \(\alpha x^{n-1}(x^2+3)^{1/2} - \beta\int x^{n-2}(x^2+3)^{1/2}dx\). A1: Correct expression |
| Applies \((x^2+3)^{3/2} = (x^2+3)(x^2+3)^{1/2}\) having attempted integration by parts | M1 | Must have made attempt at IBP in correct direction |
| \(= x^{n-1}(x^2+3)^{1/2} - (n-1)I_n - 3(n-1)I_{n-2}\) | dM1 | Splits into 2 integrals involving \(I_n\) and \(I_{n-2}\). Depends on all previous method marks |
| \(I_n = \dfrac{x^{n-1}}{n}(x^2+3)^{1/2} - \dfrac{3(n-1)}{n}I_{n-2}\) | A1* | Obtains printed answer. Condone odd missing \(dx\); withhold if clear errors e.g. invisible brackets, sign errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int \frac{x^n}{\sqrt{x^2+3}}dx = \int x^{n-2}x^2(x^2+3)^{-\frac{1}{2}}dx\) | M1 | Applies \(x^n = x^{n-2} \times x^2\) |
| \(\int x^{n-2}(x^2+3)^{\frac{1}{2}}dx - \int 3x^{n-2}(x^2+3)^{-\frac{1}{2}}dx\) | dM1A1 | dM1: Writes \(x^2\) as \((x^2+3-3)\) to obtain \(\alpha\int x^{n-2}(x^2+3)^{\frac{1}{2}}dx - \beta\int x^{n-2}(x^2+3)^{-\frac{1}{2}}dx\); A1: Correct expression |
| \(\int x^{n-2}(x^2+3)^{\frac{1}{2}}dx = \frac{x^{n-1}}{n-1}(x^2+3)^{\frac{1}{2}} - \frac{1}{n-1}\int x^n(x^2+3)^{-\frac{1}{2}}dx\) | M1 | Applies integration by parts on \(\int x^{n-2}(x^2+3)^{\frac{1}{2}}dx\) to obtain \(\alpha x^{n-1}(x^2+3)^{\frac{1}{2}} - \beta\int x^n(x^2+3)^{-\frac{1}{2}}dx\); if correct formula quoted first and applied correctly, condone sign slips |
| \(I_n = \frac{x^{n-1}}{n-1}(x^2+3)^{\frac{1}{2}} - \frac{1}{n-1}I_n - 3I_{n-2}\) | dM1 | Brings all together and introduces \(I_n\) and \(I_{n-2}\); depends on all previous method marks |
| \(\Rightarrow I_n = \frac{x^{n-1}}{n}(x^2+3)^{\frac{1}{2}} - \frac{3(n-1)}{n}I_{n-2}\) * | A1* | Obtains printed answer; condone odd missing "\(dx\)" but withhold if clear errors e.g. invisible brackets not recovered, sign errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_5 = \frac{x^4}{5}(x^2+3)^{\frac{1}{2}} - \frac{12}{5}I_3\) | M1 | Applies the reduction formula once to obtain \(I_5\) in terms of \(I_3\); allow slips on coefficients only |
| \(I_5 = \frac{x^4}{5}(x^2+3)^{\frac{1}{2}} - \frac{12}{5}\left(\frac{x^2}{3}(x^2+3)^{\frac{1}{2}} - \frac{6}{3}I_1\right)\) | M1 | Applies the reduction formula again to obtain \(I_5\) in terms of \(I_1\); allow "\(I_1\)" or what they think is \(I_1\); allow slips on coefficients only |
| \(I_5 = \frac{x^4}{5}(x^2+3)^{\frac{1}{2}} - \frac{4}{5}x^2(x^2+3)^{\frac{1}{2}} + \frac{24}{5}(x^2+3)^{\frac{1}{2}}\) | A1 | Any correct expression in terms of \(x\) only |
| \(I_5 = \frac{1}{5}(x^2+3)^{\frac{1}{2}}(x^4-4x^2+24)+k\) | A1 | Must include "\(+k\)"; allow other letter e.g. \(+c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| NB \(I_1 = (x^2+3)^{\frac{1}{2}}\) | ||
| \(I_3 = \frac{x^2}{3}(x^2+3)^{\frac{1}{2}} - \frac{6}{3}I_1\) | M1 | Applies the reduction formula once to obtain \(I_3\) in terms of \(I_1\); allow "\(I_1\)" or what they think is \(I_1\); allow slips on coefficients only |
| \(I_5 = \frac{x^4}{5}(x^2+3)^{\frac{1}{2}} - \frac{12}{5}\left(\frac{x^2}{3}(x^2+3)^{\frac{1}{2}} - 2I_1\right)\) | M1 | Applies the reduction formula again to obtain \(I_5\) in terms of \(I_1\); allow slips on coefficients only |
| \(I_5 = \frac{x^4}{5}(x^2+3)^{\frac{1}{2}} - \frac{4}{5}x^2(x^2+3)^{\frac{1}{2}} + \frac{24}{5}(x^2+3)^{\frac{1}{2}}\) | A1 | Any correct expression in terms of \(x\) only |
| \(I_5 = \frac{1}{5}(x^2+3)^{\frac{1}{2}}(x^4-4x^2+24)+k\) | A1 | Must include "\(+k\)"; allow other letter e.g. \(+c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_5 = \frac{x^4}{5}(x^2+3)^{\frac{1}{2}} - \frac{12}{5}I_3\) | M1 | Applies the reduction formula once to obtain \(I_5\) in terms of \(I_3\); allow slips on coefficients only |
| \(u = x^2+3 \Rightarrow I_3 = \frac{1}{3}(x^2+3)^{\frac{3}{2}} - 6(x^2+3)^{\frac{1}{2}}\) via substitution | M1A1 | M1: A credible attempt to find \(I_3\) directly and then express \(I_5\) in terms of \(x\); A1: Any correct expression in terms of \(x\) only |
| \(I_5 = \frac{1}{5}(x^2+3)^{\frac{1}{2}}(x^4-4x^2+24)+k\) | A1 | Must include "\(+k\)"; allow other letter e.g. \(+c\) |
# Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\frac{x^n}{\sqrt{x^2+3}}dx = \int x^{n-1}\cdot x(x^2+3)^{-1/2}dx$ | M1 | Applies $x^n = x^{n-1}\times x$ to integral |
| $\int x^{n-1}(x^2+3)^{-1/2}dx = x^{n-1}(x^2+3)^{1/2} - \int(n-1)x^{n-2}(x^2+3)^{1/2}dx$ | dM1 A1 | dM1: Integration by parts to obtain $\alpha x^{n-1}(x^2+3)^{1/2} - \beta\int x^{n-2}(x^2+3)^{1/2}dx$. A1: Correct expression |
| Applies $(x^2+3)^{3/2} = (x^2+3)(x^2+3)^{1/2}$ having attempted integration by parts | M1 | Must have made attempt at IBP in correct direction |
| $= x^{n-1}(x^2+3)^{1/2} - (n-1)I_n - 3(n-1)I_{n-2}$ | dM1 | Splits into 2 integrals involving $I_n$ and $I_{n-2}$. Depends on all previous method marks |
| $I_n = \dfrac{x^{n-1}}{n}(x^2+3)^{1/2} - \dfrac{3(n-1)}{n}I_{n-2}$ | A1* | Obtains printed answer. Condone odd missing $dx$; withhold if clear errors e.g. invisible brackets, sign errors |
## Question 6(a) Way 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int \frac{x^n}{\sqrt{x^2+3}}dx = \int x^{n-2}x^2(x^2+3)^{-\frac{1}{2}}dx$ | M1 | Applies $x^n = x^{n-2} \times x^2$ |
| $\int x^{n-2}(x^2+3)^{\frac{1}{2}}dx - \int 3x^{n-2}(x^2+3)^{-\frac{1}{2}}dx$ | dM1A1 | dM1: Writes $x^2$ as $(x^2+3-3)$ to obtain $\alpha\int x^{n-2}(x^2+3)^{\frac{1}{2}}dx - \beta\int x^{n-2}(x^2+3)^{-\frac{1}{2}}dx$; A1: Correct expression |
| $\int x^{n-2}(x^2+3)^{\frac{1}{2}}dx = \frac{x^{n-1}}{n-1}(x^2+3)^{\frac{1}{2}} - \frac{1}{n-1}\int x^n(x^2+3)^{-\frac{1}{2}}dx$ | M1 | Applies integration by parts on $\int x^{n-2}(x^2+3)^{\frac{1}{2}}dx$ to obtain $\alpha x^{n-1}(x^2+3)^{\frac{1}{2}} - \beta\int x^n(x^2+3)^{-\frac{1}{2}}dx$; if correct formula quoted first and applied correctly, condone sign slips |
| $I_n = \frac{x^{n-1}}{n-1}(x^2+3)^{\frac{1}{2}} - \frac{1}{n-1}I_n - 3I_{n-2}$ | dM1 | Brings all together and introduces $I_n$ and $I_{n-2}$; depends on all previous method marks |
| $\Rightarrow I_n = \frac{x^{n-1}}{n}(x^2+3)^{\frac{1}{2}} - \frac{3(n-1)}{n}I_{n-2}$ * | A1* | Obtains printed answer; condone odd missing "$dx$" but withhold if clear errors e.g. invisible brackets not recovered, sign errors |
---
## Question 6(b) Way 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_5 = \frac{x^4}{5}(x^2+3)^{\frac{1}{2}} - \frac{12}{5}I_3$ | M1 | Applies the reduction formula once to obtain $I_5$ in terms of $I_3$; allow slips on coefficients only |
| $I_5 = \frac{x^4}{5}(x^2+3)^{\frac{1}{2}} - \frac{12}{5}\left(\frac{x^2}{3}(x^2+3)^{\frac{1}{2}} - \frac{6}{3}I_1\right)$ | M1 | Applies the reduction formula again to obtain $I_5$ in terms of $I_1$; allow "$I_1$" or what they think is $I_1$; allow slips on coefficients only |
| $I_5 = \frac{x^4}{5}(x^2+3)^{\frac{1}{2}} - \frac{4}{5}x^2(x^2+3)^{\frac{1}{2}} + \frac{24}{5}(x^2+3)^{\frac{1}{2}}$ | A1 | Any correct expression in terms of $x$ only |
| $I_5 = \frac{1}{5}(x^2+3)^{\frac{1}{2}}(x^4-4x^2+24)+k$ | A1 | Must include "$+k$"; allow other letter e.g. $+c$ |
---
## Question 6(b) Way 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| NB $I_1 = (x^2+3)^{\frac{1}{2}}$ | | |
| $I_3 = \frac{x^2}{3}(x^2+3)^{\frac{1}{2}} - \frac{6}{3}I_1$ | M1 | Applies the reduction formula once to obtain $I_3$ in terms of $I_1$; allow "$I_1$" or what they think is $I_1$; allow slips on coefficients only |
| $I_5 = \frac{x^4}{5}(x^2+3)^{\frac{1}{2}} - \frac{12}{5}\left(\frac{x^2}{3}(x^2+3)^{\frac{1}{2}} - 2I_1\right)$ | M1 | Applies the reduction formula again to obtain $I_5$ in terms of $I_1$; allow slips on coefficients only |
| $I_5 = \frac{x^4}{5}(x^2+3)^{\frac{1}{2}} - \frac{4}{5}x^2(x^2+3)^{\frac{1}{2}} + \frac{24}{5}(x^2+3)^{\frac{1}{2}}$ | A1 | Any correct expression in terms of $x$ only |
| $I_5 = \frac{1}{5}(x^2+3)^{\frac{1}{2}}(x^4-4x^2+24)+k$ | A1 | Must include "$+k$"; allow other letter e.g. $+c$ |
---
## Question 6(b) Way 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_5 = \frac{x^4}{5}(x^2+3)^{\frac{1}{2}} - \frac{12}{5}I_3$ | M1 | Applies the reduction formula once to obtain $I_5$ in terms of $I_3$; allow slips on coefficients only |
| $u = x^2+3 \Rightarrow I_3 = \frac{1}{3}(x^2+3)^{\frac{3}{2}} - 6(x^2+3)^{\frac{1}{2}}$ via substitution | M1A1 | M1: A credible attempt to find $I_3$ directly and then express $I_5$ in terms of $x$; A1: Any correct expression in terms of $x$ only |
| $I_5 = \frac{1}{5}(x^2+3)^{\frac{1}{2}}(x^4-4x^2+24)+k$ | A1 | Must include "$+k$"; allow other letter e.g. $+c$ |
> **Note:** Part (b) must involve use of the reduction formula; a direct attempt at $I_5$ scores no marks. If reduction formula used once and $I_3$ attempted directly, all marks are available but there must be a credible attempt at $I_3$.
---6.
$$I _ { n } = \int \frac { x ^ { n } } { \sqrt { x ^ { 2 } + 3 } } \mathrm {~d} x \quad n \in \mathbb { N }$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$I _ { n } = \frac { x ^ { n - 1 } } { n } \left( x ^ { 2 } + 3 \right) ^ { \frac { 1 } { 2 } } - \frac { 3 ( n - 1 ) } { n } I _ { n - 2 } \quad n \geqslant 3$$
\item Hence show that
$$\int \frac { x ^ { 5 } } { \sqrt { x ^ { 2 } + 3 } } \mathrm {~d} x = \frac { 1 } { 5 } \left( x ^ { 2 } + 3 \right) ^ { \frac { 1 } { 2 } } \left( x ^ { 4 } + p x ^ { 2 } + q \right) + k$$
where $p$ and $q$ are integers to be determined and $k$ is an arbitrary constant.\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2021 Q6 [10]}}