| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Reduction formula or recurrence |
| Difficulty | Challenging +1.8 This is a Further Maths reduction formula question requiring integration by parts with algebraic manipulation to derive the recurrence relation, then apply it to find a specific integral. The derivation involves non-trivial substitution and careful algebraic handling, and applying it requires working backwards through the formula. Significantly harder than standard A-level integration but typical for FP2 reduction formulas. |
| Spec | 8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n = \int 1 \times (4-x^2)^{-n}\,dx = x(4-x^2)^{-n} \pm \int x \cdot n(4-x^2)^{-n-1} \cdot 2x\,dx\) | M1 | Splits integrand as \(1 \times (4-x^2)^{-n}\) and attempts parts (right way round) |
| \(I_n = x(4-x^2)^{-n} - \int x \cdot (-n)(4-x^2)^{-n-1} \cdot (-2x)\,dx\) | A1 | Correct result obtained, need not be simplified |
| \(= x(4-x^2)^{-n} + 2n\int(4-x^2-4)(4-x^2)^{-n-1}\,dx\) \(= x(4-x^2)^{-n} + 2n\int(4-x^2)(4-x^2)^{-n-1} - 4(4-x^2)^{-n-1}\,dx\) | M1 | Gathers terms and writes \(x^2\) as \(-(4-x^2-4)\). Accept either line shown |
| \(I_n = x(4-x^2)^{-n} + 2n\int(4-x^2)^{-n}\,dx - 8n\int(4-x^2)^{-(n+1)}\,dx\) \(= x(4-x^2)^{-n} + 2nI_n - 8nI_{n+1}\) | M1 | Sorts out indices and replaces appropriate integrals by \(I_n\) and \(I_{n+1}\) |
| \(\Rightarrow 8nI_{n+1} = x(4-x^2)^{-n} + (2n-1)I_n \Rightarrow I_{n+1} = \dfrac{x}{8n(4-x^2)^n} + \dfrac{2n-1}{8n}I_n\) * | A1* | Completes to the correct printed result, with no errors or ambiguities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_1 = \int \dfrac{1}{4-x^2}\,dx = \dfrac{1}{2}\text{artanh}\!\left(\dfrac{x}{2}\right)\) or \(\dfrac{1}{4}\ln\left\lvert\dfrac{2+x}{2-x}\right\rvert\) | B1 | Deduces the correct result for \(I_1\) |
| \(I_2 = \dfrac{x}{8(4-x^2)} + \dfrac{1}{8}I_1\) | M1 | Applies the reduction formula correctly with \(n=1\) only |
| \(= \dfrac{x}{8(4-x^2)} + \dfrac{1}{16}\text{artanh}\!\left(\dfrac{x}{2}\right)(+c)\) oe e.g. \(\dfrac{x}{8(4-x^2)} + \dfrac{1}{32}\ln\left\lvert\dfrac{2+x}{2-x}\right\rvert(+c)\) | A1 | Correct answer; accept any equivalent form with fractions simplified. Constant of integration not needed |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = \int 1 \times (4-x^2)^{-n}\,dx = x(4-x^2)^{-n} \pm \int x \cdot n(4-x^2)^{-n-1} \cdot 2x\,dx$ | M1 | Splits integrand as $1 \times (4-x^2)^{-n}$ and attempts parts (right way round) |
| $I_n = x(4-x^2)^{-n} - \int x \cdot (-n)(4-x^2)^{-n-1} \cdot (-2x)\,dx$ | A1 | Correct result obtained, need not be simplified |
| $= x(4-x^2)^{-n} + 2n\int(4-x^2-4)(4-x^2)^{-n-1}\,dx$ $= x(4-x^2)^{-n} + 2n\int(4-x^2)(4-x^2)^{-n-1} - 4(4-x^2)^{-n-1}\,dx$ | M1 | Gathers terms and writes $x^2$ as $-(4-x^2-4)$. Accept either line shown |
| $I_n = x(4-x^2)^{-n} + 2n\int(4-x^2)^{-n}\,dx - 8n\int(4-x^2)^{-(n+1)}\,dx$ $= x(4-x^2)^{-n} + 2nI_n - 8nI_{n+1}$ | M1 | Sorts out indices and replaces appropriate integrals by $I_n$ and $I_{n+1}$ |
| $\Rightarrow 8nI_{n+1} = x(4-x^2)^{-n} + (2n-1)I_n \Rightarrow I_{n+1} = \dfrac{x}{8n(4-x^2)^n} + \dfrac{2n-1}{8n}I_n$ * | A1* | Completes to the correct printed result, with no errors or ambiguities |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_1 = \int \dfrac{1}{4-x^2}\,dx = \dfrac{1}{2}\text{artanh}\!\left(\dfrac{x}{2}\right)$ or $\dfrac{1}{4}\ln\left\lvert\dfrac{2+x}{2-x}\right\rvert$ | B1 | Deduces the correct result for $I_1$ |
| $I_2 = \dfrac{x}{8(4-x^2)} + \dfrac{1}{8}I_1$ | M1 | Applies the reduction formula correctly with $n=1$ only |
| $= \dfrac{x}{8(4-x^2)} + \dfrac{1}{16}\text{artanh}\!\left(\dfrac{x}{2}\right)(+c)$ oe e.g. $\dfrac{x}{8(4-x^2)} + \dfrac{1}{32}\ln\left\lvert\dfrac{2+x}{2-x}\right\rvert(+c)$ | A1 | Correct answer; accept any equivalent form with fractions simplified. Constant of integration not needed |
---7.
$$I _ { n } = \int \left( 4 - x ^ { 2 } \right) ^ { - n } \mathrm {~d} x \quad n > 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that, for $n > 0$
$$I _ { n + 1 } = \frac { x } { 8 n \left( 4 - x ^ { 2 } \right) ^ { n } } + \frac { 2 n - 1 } { 8 n } I _ { n }$$
\item Find $I _ { 2 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2020 Q7 [8]}}