| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Exponential times polynomial |
| Difficulty | Challenging +1.2 This is a standard reduction formula question requiring integration by parts with clear substitution u=(x-2)^n, dv=e^(4x)dx. Part (a) follows a routine procedure with boundary term evaluation, and part (b) is direct application. While it requires careful algebraic manipulation and is from Further Maths FP2, the technique is well-practiced and the structure is predictable, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 1.08i Integration by parts8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n = \left[(x-2)^n \times Ae^{4x}\right]_0^2 - \int_0^2 n(x-2)^{n-1} \times Ae^{4x}\,dx\) | M1 | Applies integration by parts to achieve correct form |
| \(= \left(0 - A(-2)^n\right) - A\int_0^2 n(x-2)^{n-1}e^{4x}\,dx\) | M1 | Substitutes limits 0 and 2; simplifies integral to match form of \(I_n\) |
| \(= -\dfrac{1}{(-2)^2}(-2)^n - \dfrac{n}{4}I_{n-1}\) | M1 | Writes 4 as \((-2)^2\) and replaces integral by \(I_{n-1}\) |
| \(I_n = -(-2)^{n-2} - \dfrac{n}{4}I_{n-1}\) | A1 | Correct answer following correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_2 = -(-2)^0 - \dfrac{2}{4}I_1 = -1 - \dfrac{1}{2}\left(-(-2)^{-1} - \dfrac{1}{4}I_0\right)\) | M1 | Full process of reducing integral to expression in \(I_0\) |
| \(= -1 - \dfrac{1}{4} + \dfrac{1}{8}\left(\dfrac{1}{4}e^8 - \dfrac{1}{4}\right)\) | M1 | Evaluates \(I_0\) and substitutes |
| \(= \dfrac{1}{32}e^8 - \dfrac{41}{32}\) | A1 | Correct final answer |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = \left[(x-2)^n \times Ae^{4x}\right]_0^2 - \int_0^2 n(x-2)^{n-1} \times Ae^{4x}\,dx$ | M1 | Applies integration by parts to achieve correct form |
| $= \left(0 - A(-2)^n\right) - A\int_0^2 n(x-2)^{n-1}e^{4x}\,dx$ | M1 | Substitutes limits 0 and 2; simplifies integral to match form of $I_n$ |
| $= -\dfrac{1}{(-2)^2}(-2)^n - \dfrac{n}{4}I_{n-1}$ | M1 | Writes 4 as $(-2)^2$ and replaces integral by $I_{n-1}$ |
| $I_n = -(-2)^{n-2} - \dfrac{n}{4}I_{n-1}$ | A1 | Correct answer following correct working |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_2 = -(-2)^0 - \dfrac{2}{4}I_1 = -1 - \dfrac{1}{2}\left(-(-2)^{-1} - \dfrac{1}{4}I_0\right)$ | M1 | Full process of reducing integral to expression in $I_0$ |
| $= -1 - \dfrac{1}{4} + \dfrac{1}{8}\left(\dfrac{1}{4}e^8 - \dfrac{1}{4}\right)$ | M1 | Evaluates $I_0$ and substitutes |
| $= \dfrac{1}{32}e^8 - \dfrac{41}{32}$ | A1 | Correct final answer |8.
$$I _ { n } = \int _ { 0 } ^ { 2 } ( x - 2 ) ^ { n } \mathrm { e } ^ { 4 x } \mathrm {~d} x \quad n \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Prove that for $n \geqslant 1$
$$I _ { n } = - a ^ { n - 2 } - \frac { n } { 4 } I _ { n - 1 }$$
where $a$ is a constant to be determined.
\item Hence determine the exact value of
$$\int _ { 0 } ^ { 2 } ( x - 2 ) ^ { 2 } e ^ { 4 x } d x$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2023 Q8 [7]}}