| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| 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 iterative application with careful algebraic manipulation. While the technique is standard for this module, the multi-step derivation and exact evaluation demand precision and systematic working beyond typical A-level questions. |
| Spec | 1.08d Evaluate definite integrals: between limits8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(I_n = \int x^{n-1}\cdot x(x^2+k^2)^{-\frac{1}{2}}\,dx\) | B1 | Separates correctly (without this there will be no progress) |
| \(I_n = x^{n-1}(x^2+k^2)^{\frac{1}{2}} - \int(n-1)x^{n-2}(x^2+k^2)^{\frac{1}{2}}\,dx\) | M1, A1 | M1: Parts in the correct direction; A1: Correct expression |
| \(= \ldots-(n-1)\int\frac{x^{n-2}(x^2+k^2)}{\sqrt{(x^2+k^2)}}\,dx\) | dM1 | Writes \((x^2+k^2)^{\frac{1}{2}}\) as \(\frac{(x^2+k^2)}{\sqrt{(x^2+k^2)}}\) |
| \(= \ldots-(n-1)\int\frac{x^n}{\sqrt{(x^2+k^2)}}\,dx - (n-1)\int\frac{k^2 x^{n-2}}{\sqrt{(x^2+k^2)}}\,dx\) | A1 | Correct separation |
| \(I_n = x^{n-1}(x^2+k^2)^{\frac{1}{2}} - (n-1)I_n - (n-1)k^2 I_{n-2}\) | ddM1 | Introduces \(I_n\) and \(I_{n-2}\) on rhs; depends on both M marks above |
| \(I_n = \frac{x^{n-1}}{n}(x^2+k^2)^{\frac{1}{2}} - \frac{(n-1)}{n}k^2 I_{n-2}\) * | A1* | Cso (given answer) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(I_5 = \int\frac{x^5}{\sqrt{(x^2+1)}}\,dx = \frac{x^4}{5}(x^2+1)^{\frac{1}{2}} - \frac{4}{5}I_3\) | M1 | Correct first application of reduction formula (can have \(k^2\) instead of 1) |
| \(I_3 = \frac{x^2}{3}(x^2+1)^{\frac{1}{2}} - \frac{2}{3}I_1\) | M1 | Correct second application of reduction formula (can have \(k^2\) instead of 1) |
| \(I_1 = \int\frac{x}{\sqrt{(x^2+1)}}\,dx = \left[\sqrt{x^2+1}\right] \Rightarrow I_5 = \ldots\) | ddM1 | \(\int\frac{x}{\sqrt{(x^2+1)}}\,dx = a\sqrt{x^2+1}\) and attempt \(I_5\) using correct limits (\(k^2\) or 1) |
| \(\int_0^1\frac{x^5}{\sqrt{(x^2+1)}}\,dx = \frac{7}{15}\sqrt{2} - \frac{8}{15}\) | A1, A1 | A1: Either term correct; A1: Both terms correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(I_1 = \int\frac{x}{\sqrt{(x^2+1)}}\,dx = \sqrt{x^2+1}\) | M1 | \(\int\frac{x}{\sqrt{(x^2+1)}}\,dx = a\sqrt{x^2+1}\) (\(k^2\) or 1) |
| \(I_3 = \frac{x^2}{3}(x^2+1)^{\frac{1}{2}} - \frac{2}{3}I_1\) | M1 | Attempt \(I_3\) by using reduction formula (\(k^2\) or 1) |
| \(I_5 = \frac{x^4}{5}(x^2+1)^{\frac{1}{2}} - \frac{4}{5}\!\left(\frac{x^2}{3}(x^2+1)^{\frac{1}{2}} - \frac{2}{3}(x^2+1)^{\frac{1}{2}}\right)\) | ddM1 | Form complete statement for \(I_5\) and use correct limits (\(k^2\) or 1) |
| \(\int_0^1\frac{x^5}{\sqrt{(x^2+1)}}\,dx = \frac{7}{15}\sqrt{2} - \frac{8}{15}\) | A1, A1 | A1: Either term correct; A1: Both terms correct |
## Question 8:
**Given:** $I_n = \int \frac{x^n}{\sqrt{(x^2+k^2)}}\,dx$
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $I_n = \int x^{n-1}\cdot x(x^2+k^2)^{-\frac{1}{2}}\,dx$ | B1 | Separates correctly (without this there will be no progress) |
| $I_n = x^{n-1}(x^2+k^2)^{\frac{1}{2}} - \int(n-1)x^{n-2}(x^2+k^2)^{\frac{1}{2}}\,dx$ | M1, A1 | M1: Parts in the correct direction; A1: Correct expression |
| $= \ldots-(n-1)\int\frac{x^{n-2}(x^2+k^2)}{\sqrt{(x^2+k^2)}}\,dx$ | dM1 | Writes $(x^2+k^2)^{\frac{1}{2}}$ as $\frac{(x^2+k^2)}{\sqrt{(x^2+k^2)}}$ |
| $= \ldots-(n-1)\int\frac{x^n}{\sqrt{(x^2+k^2)}}\,dx - (n-1)\int\frac{k^2 x^{n-2}}{\sqrt{(x^2+k^2)}}\,dx$ | A1 | Correct separation |
| $I_n = x^{n-1}(x^2+k^2)^{\frac{1}{2}} - (n-1)I_n - (n-1)k^2 I_{n-2}$ | ddM1 | Introduces $I_n$ and $I_{n-2}$ on rhs; depends on both M marks above |
| $I_n = \frac{x^{n-1}}{n}(x^2+k^2)^{\frac{1}{2}} - \frac{(n-1)}{n}k^2 I_{n-2}$ * | A1* | Cso (given answer) |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $I_5 = \int\frac{x^5}{\sqrt{(x^2+1)}}\,dx = \frac{x^4}{5}(x^2+1)^{\frac{1}{2}} - \frac{4}{5}I_3$ | M1 | Correct first application of reduction formula (can have $k^2$ instead of 1) |
| $I_3 = \frac{x^2}{3}(x^2+1)^{\frac{1}{2}} - \frac{2}{3}I_1$ | M1 | Correct second application of reduction formula (can have $k^2$ instead of 1) |
| $I_1 = \int\frac{x}{\sqrt{(x^2+1)}}\,dx = \left[\sqrt{x^2+1}\right] \Rightarrow I_5 = \ldots$ | ddM1 | $\int\frac{x}{\sqrt{(x^2+1)}}\,dx = a\sqrt{x^2+1}$ and attempt $I_5$ using correct limits ($k^2$ or 1) |
| $\int_0^1\frac{x^5}{\sqrt{(x^2+1)}}\,dx = \frac{7}{15}\sqrt{2} - \frac{8}{15}$ | A1, A1 | A1: Either term correct; A1: Both terms correct |
### Part (b) Way 2:
| Answer | Mark | Guidance |
|--------|------|----------|
| $I_1 = \int\frac{x}{\sqrt{(x^2+1)}}\,dx = \sqrt{x^2+1}$ | M1 | $\int\frac{x}{\sqrt{(x^2+1)}}\,dx = a\sqrt{x^2+1}$ ($k^2$ or 1) |
| $I_3 = \frac{x^2}{3}(x^2+1)^{\frac{1}{2}} - \frac{2}{3}I_1$ | M1 | Attempt $I_3$ by using reduction formula ($k^2$ or 1) |
| $I_5 = \frac{x^4}{5}(x^2+1)^{\frac{1}{2}} - \frac{4}{5}\!\left(\frac{x^2}{3}(x^2+1)^{\frac{1}{2}} - \frac{2}{3}(x^2+1)^{\frac{1}{2}}\right)$ | ddM1 | Form complete statement for $I_5$ and use correct limits ($k^2$ or 1) |
| $\int_0^1\frac{x^5}{\sqrt{(x^2+1)}}\,dx = \frac{7}{15}\sqrt{2} - \frac{8}{15}$ | A1, A1 | A1: Either term correct; A1: Both terms correct |
The image appears to be essentially blank/white with only a footer showing Pearson Education Limited's registration details and the "PMT" watermark. There is no mark scheme content visible on this page to extract.
Could you please share the actual mark scheme pages that contain the questions, answers, mark allocations, and guidance notes? This appears to be either a blank page or the back cover of a mark scheme document.8.
$$I _ { n } = \int \frac { x ^ { n } } { \sqrt { \left( x ^ { 2 } + k ^ { 2 } \right) } } \mathrm { d } x \quad \text { where } k \text { is a constant and } n \in \mathbb { Z } ^ { + }$$
\begin{enumerate}[label=(\alph*)]
\item Show that, for $n \geqslant 2$
$$I _ { n } = \frac { x ^ { n - 1 } } { n } \left( x ^ { 2 } + k ^ { 2 } \right) ^ { \frac { 1 } { 2 } } - \frac { ( n - 1 ) } { n } k ^ { 2 } I _ { n - 2 }$$
\item Hence find the exact value of
$$\int _ { 0 } ^ { 1 } \frac { x ^ { 5 } } { \sqrt { \left( x ^ { 2 } + 1 \right) } } \mathrm { d } x$$
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2018 Q8 [12]}}