| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Algebraic function with square root |
| Difficulty | Challenging +1.3 This is a standard Further Maths reduction formula question with clear structure: (a) requires a straightforward substitution u=25-x², (b) uses integration by parts with the hint from (a) to derive the reduction formula, and (c) applies the formula iteratively starting from I₀=π/2. While it requires multiple techniques and is from FP3, the steps are well-signposted and follow a familiar pattern for this topic, making it moderately above average difficulty but not exceptionally challenging. |
| Spec | 1.08h Integration by substitution8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-(25-x^2)^{\frac{1}{2}}\) (+c) | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_n = \int x^{n-1}\cdot\frac{x}{\sqrt{(25-x^2)}}dx = -x^{n-1}\sqrt{25-x^2} + \int(n-1)x^{n-2}\sqrt{25-x^2}\,dx\) | M1 A1ft | |
| \(I_n = \left[-x^{n-1}\sqrt{25-x^2}\right]_0^5 + \int_0^5 \frac{(n-1)x^{n-2}(25-x^2)}{\sqrt{(25-x^2)}}dx\) | M1 | |
| \(I_n = 0 + 25(n-1)\,I_{n-2} - (n-1)\,I_n\) | M1 | |
| \(\therefore nI_n = 25(n-1)I_{n-2}\) and so \(I_n = \frac{25(n-1)}{n}I_{n-2}\) ✱ | A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_0 = \int_0^5 \frac{1}{\sqrt{(25-x^2)}}dx = \left[\arcsin\left(\frac{x}{5}\right)\right]_0^5 = \frac{\pi}{2}\) | M1 A1 | |
| \(I_4 = \frac{25\times3}{4}\times\frac{25\times1}{2}I_0 = \frac{1875}{16}\pi\) | M1 A1 | (4) [11] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Using substitution \(x=5\sin\theta\): \(I_n = 5^n\int_0^{\frac{\pi}{2}}\sin^n\theta\,d\theta = \left[-5^n\sin^{n-1}\theta\cos\theta\right]_0^{\frac{\pi}{2}} + 5^n(n-1)\int_0^{\frac{\pi}{2}}\sin^{n-2}\theta\cos^2\theta\,d\theta\) | M1 A1 | |
| \(= \left[-5^n\sin^{n-1}\theta\cos\theta\right]_0^{\frac{\pi}{2}} + 5^n(n-1)\int_0^{\frac{\pi}{2}}\sin^{n-2}\theta(1-\sin^2\theta)\,d\theta\) | M1 | |
| \(I_n = 0 + 25(n-1)\,I_{n-2} - (n-1)\,I_n\) | M1 | |
| \(\therefore nI_n = 25(n-1)I_{n-2}\) and so \(I_n = \frac{25(n-1)}{n}I_{n-2}\) ✱ | A1 | (5) |
| (need to see that \(I_{n-2} = 5^{n-2}\int_0^{\frac{\pi}{2}}\sin^{n-2}\theta\,d\theta\) for final A1) |
# Question 5:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-(25-x^2)^{\frac{1}{2}}$ (+c) | M1 A1 | (2) |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_n = \int x^{n-1}\cdot\frac{x}{\sqrt{(25-x^2)}}dx = -x^{n-1}\sqrt{25-x^2} + \int(n-1)x^{n-2}\sqrt{25-x^2}\,dx$ | M1 A1ft | |
| $I_n = \left[-x^{n-1}\sqrt{25-x^2}\right]_0^5 + \int_0^5 \frac{(n-1)x^{n-2}(25-x^2)}{\sqrt{(25-x^2)}}dx$ | M1 | |
| $I_n = 0 + 25(n-1)\,I_{n-2} - (n-1)\,I_n$ | M1 | |
| $\therefore nI_n = 25(n-1)I_{n-2}$ and so $I_n = \frac{25(n-1)}{n}I_{n-2}$ ✱ | A1 | (5) |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_0 = \int_0^5 \frac{1}{\sqrt{(25-x^2)}}dx = \left[\arcsin\left(\frac{x}{5}\right)\right]_0^5 = \frac{\pi}{2}$ | M1 A1 | |
| $I_4 = \frac{25\times3}{4}\times\frac{25\times1}{2}I_0 = \frac{1875}{16}\pi$ | M1 A1 | (4) **[11]** |
## Alternative for (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Using substitution $x=5\sin\theta$: $I_n = 5^n\int_0^{\frac{\pi}{2}}\sin^n\theta\,d\theta = \left[-5^n\sin^{n-1}\theta\cos\theta\right]_0^{\frac{\pi}{2}} + 5^n(n-1)\int_0^{\frac{\pi}{2}}\sin^{n-2}\theta\cos^2\theta\,d\theta$ | M1 A1 | |
| $= \left[-5^n\sin^{n-1}\theta\cos\theta\right]_0^{\frac{\pi}{2}} + 5^n(n-1)\int_0^{\frac{\pi}{2}}\sin^{n-2}\theta(1-\sin^2\theta)\,d\theta$ | M1 | |
| $I_n = 0 + 25(n-1)\,I_{n-2} - (n-1)\,I_n$ | M1 | |
| $\therefore nI_n = 25(n-1)I_{n-2}$ and so $I_n = \frac{25(n-1)}{n}I_{n-2}$ ✱ | A1 | (5) |
| (need to see that $I_{n-2} = 5^{n-2}\int_0^{\frac{\pi}{2}}\sin^{n-2}\theta\,d\theta$ for final A1) | | |
---5.
$$I _ { n } = \int _ { 0 } ^ { 5 } \frac { x ^ { n } } { \sqrt { } \left( 25 - x ^ { 2 } \right) } d x , \quad n \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\int \frac { x } { \sqrt { } \left( 25 - x ^ { 2 } \right) } \mathrm { d } x , \quad 0 \leqslant x \leqslant 5$.
\item Using your answer to part (a), or otherwise, show that
$$I _ { n } = \frac { 25 ( n - 1 ) } { n } I _ { n - 2 } \quad n \geqslant 2$$
\item Find $I _ { 4 }$ in the form $k \pi$, where $k$ is a fraction.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2009 Q5 [11]}}