| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Reduction formula or recurrence |
| Difficulty | Challenging +1.2 This is a standard Further Maths reduction formula question requiring two applications of integration by parts to derive the recurrence relation, followed by straightforward substitution. While it involves multiple steps and careful algebraic manipulation, the technique is well-practiced in F3 and follows a predictable pattern with no novel insight required. |
| Spec | 8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int (a-x)^n \cos x\,dx = (a-x)^n \sin x + \int n(a-x)^{n-1}\sin x\,dx\) | M1 A1 | Integration by parts |
| \(\left[(a-x)^n \sin x\right]_0^a = 0\) | A1 | Boundary terms vanish |
| \(= -n(a-x)^{n-1}\cos x - \int n(n-1)(a-x)^{n-2}\cos x\,dx\) | dM1 | Second integration by parts |
| \(I_n = na^{n-1} - n(n-1)I_{n-2}\) \(\ast\) | A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_2 = 2\left(\dfrac{\pi}{2}\right) - 2\int_0^{\pi/2}\cos x\,dx\) | M1 A1 | Using reduction formula with \(n=2\), \(a=\frac{\pi}{2}\) |
| \(= \pi - 2[\sin x]_0^{\pi/2} = \pi - 2\) | A1 | (3) |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int (a-x)^n \cos x\,dx = (a-x)^n \sin x + \int n(a-x)^{n-1}\sin x\,dx$ | M1 A1 | Integration by parts |
| $\left[(a-x)^n \sin x\right]_0^a = 0$ | A1 | Boundary terms vanish |
| $= -n(a-x)^{n-1}\cos x - \int n(n-1)(a-x)^{n-2}\cos x\,dx$ | dM1 | Second integration by parts |
| $I_n = na^{n-1} - n(n-1)I_{n-2}$ $\ast$ | A1 | (5) | Completion to given result |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_2 = 2\left(\dfrac{\pi}{2}\right) - 2\int_0^{\pi/2}\cos x\,dx$ | M1 A1 | Using reduction formula with $n=2$, $a=\frac{\pi}{2}$ |
| $= \pi - 2[\sin x]_0^{\pi/2} = \pi - 2$ | A1 | (3) | |
---4. $I _ { n } = \int _ { 0 } ^ { a } ( a - x ) ^ { n } \cos x \mathrm {~d} x , \quad a > 0 , \quad n \geqslant 0$
\begin{enumerate}[label=(\alph*)]
\item Show that, for $n \geqslant 2$,
$$I _ { n } = n a ^ { n - 1 } - n ( n - 1 ) I _ { n - 2 }$$
\item Hence evaluate $\int _ { 0 } ^ { \frac { \pi } { 2 } } \left( \frac { \pi } { 2 } - x \right) ^ { 2 } \cos x d x$
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 Q4 [8]}}