| Exam Board | Edexcel |
|---|---|
| Module | FP3 (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, then working backwards from base cases. While it requires careful algebraic manipulation and is beyond standard A-level, it follows a well-established template that FP3 students practice extensively. The 8-mark allocation and straightforward structure place it moderately above average difficulty but not exceptionally challenging for the Further Maths cohort. |
| Spec | 1.08i Integration by parts4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_n = \int_0^{\frac{\pi}{2}} x^n \sin x \, dx\) | ||
| \(= \left[x^n(-\cos x)\right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} nx^{n-1}(-\cos x) dx\) | M1A1 | |
| \(= 0 + n\left\{x^{n-1}\sin x\Big | _0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}}(n-1)x^{n-2}\sin x \, dx\right\}\) | A1 |
| \(= n\left(\frac{\pi}{2}\right)^{n-1} - (n-1)I_{n-2}\) | ||
| \(\text{So } I_n = n\left(\frac{\pi}{2}\right)^{n-1} - n(n-1)I_{n-2}\) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_3 = 3\left(\frac{\pi}{2}\right)^2 - 3.2I_1\) | ||
| \(I_1 = \int_0^{\frac{\pi}{2}} x\sin x \, dx = [x(-\cos x)]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}}\cos x \, dx\) | M1 | |
| \(= [\sin x]_0^{\frac{\pi}{2}} = 1\) | A1 | |
| \(I_3 = 3\left(\frac{\pi}{2}\right)^2 - 6 = \frac{3\pi^2}{4} - 6\) | M1A1 | (4) |
## (a)
$I_n = \int_0^{\frac{\pi}{2}} x^n \sin x \, dx$ | |
$= \left[x^n(-\cos x)\right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} nx^{n-1}(-\cos x) dx$ | M1A1 |
$= 0 + n\left\{x^{n-1}\sin x\Big|_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}}(n-1)x^{n-2}\sin x \, dx\right\}$ | A1 |
$= n\left(\frac{\pi}{2}\right)^{n-1} - (n-1)I_{n-2}$ | |
$\text{So } I_n = n\left(\frac{\pi}{2}\right)^{n-1} - n(n-1)I_{n-2}$ | A1 | (4)
## (b)
$I_3 = 3\left(\frac{\pi}{2}\right)^2 - 3.2I_1$ | |
$I_1 = \int_0^{\frac{\pi}{2}} x\sin x \, dx = [x(-\cos x)]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}}\cos x \, dx$ | M1 |
$= [\sin x]_0^{\frac{\pi}{2}} = 1$ | A1 |
$I_3 = 3\left(\frac{\pi}{2}\right)^2 - 6 = \frac{3\pi^2}{4} - 6$ | M1A1 | (4)
---$$I_n = \int_0^{\pi} x^n \sin x \, dx$$
\begin{enumerate}[label=(\alph*)]
\item Show that for $n \geq 2$
$$I_n = n \left( \frac{\pi}{2} \right)^{n-1} - n(n-1)I_{n-2}$$ [4]
\item Hence obtain $I_3$, giving your answers in terms of $\pi$. [4]
\end{enumerate}
(Total 8 marks)
\hfill \mbox{\textit{Edexcel FP3 Q6 [8]}}