| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Compound expressions with binomial expansion |
| Difficulty | Challenging +1.8 This is a Further Maths FP3 reduction formula question requiring integration by parts twice to establish the recurrence relation, then applying it with specific values. While the technique is standard for this module, it demands careful algebraic manipulation, proper handling of boundary terms, and systematic application of the formula—significantly harder than typical A-level questions but routine for Further Maths students who have practiced reduction formulae. |
| Spec | 1.08i Integration by parts8.06a 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 |
| \(I_n = na^{n-1} - n(n-1)I_{n-2}\) ✽ | A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_2 = 2\left(\frac{\pi}{2}\right) - 2\int_0^{\frac{\pi}{2}}\cos x\, dx\) | M1 A1 | Using reduction formula with \(n=2\), \(a=\frac{\pi}{2}\) |
| \(= \pi - 2\left[\sin x\right]_0^{\frac{\pi}{2}} = \pi - 2\) | 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 |
| $I_n = na^{n-1} - n(n-1)I_{n-2}$ ✽ | A1 | (5) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_2 = 2\left(\frac{\pi}{2}\right) - 2\int_0^{\frac{\pi}{2}}\cos x\, dx$ | M1 A1 | Using reduction formula with $n=2$, $a=\frac{\pi}{2}$ |
| $= \pi - 2\left[\sin x\right]_0^{\frac{\pi}{2}} = \pi - 2$ | A1 | (3) |
---4. $\quad I _ { n } = \int _ { 0 } ^ { a } ( a - x ) ^ { n } \cos x \mathrm {~d} x , \quad a > 0 , \quad n \geqslant 0$
\begin{enumerate}[label=(\alph*)]
\item Show that, for $n \geqslant 2$,
$$I _ { n } = n \tilde { a } ^ { - 1 } - n ( n - 1 ) I _ { n - 2 }$$
\item Hence evaluate $\int _ { 0 } ^ { \frac { \pi } { 2 } } \left( \frac { \pi } { 2 } - x \right) ^ { 2 } \cos x \mathrm {~d} x$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2010 Q4 [8]}}