| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Algebraic function with square root |
| Difficulty | Challenging +1.8 This is a Further Maths FP3 reduction formula question requiring integration by parts to derive the recurrence relation, then recursive application to find I_5. While the technique is standard for FP3, it demands careful algebraic manipulation with the square root term, proper handling of boundary terms (which vanish at both limits), and multiple recursive steps. The derivation requires insight into choosing u and dv appropriately, making it significantly harder than typical A-level questions but still a recognizable FP3 exercise type. |
| Spec | 8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_n = \int_0^4 x^{n-1} \times x(16-x^2)^{\frac{1}{2}}\,dx\) | M1A1 | M1: Obtains \(x(16-x^2)^{\frac{1}{2}}\) prior to integration. A1: Correct underlined expression |
| \(I_n = \left[-\frac{1}{3}x^{n-1}(16-x^2)^{\frac{3}{2}}\right]_0^4 + \frac{n-1}{3}\int_0^4 x^{n-2}(16-x^2)^{\frac{3}{2}}\,dx\) | dM1 | Parts in correct direction (ignore limits) |
| \(\therefore I_n = \frac{n-1}{3}\int_0^4 x^{n-2}(16-x^2)(16-x^2)^{\frac{1}{2}}\,dx\) | ||
| \(I_n = \frac{16(n-1)}{3}I_{n-2} - \frac{n-1}{3}I_n\) | M1 | Manipulates to obtain at least one integral in terms of \(I_n\) or \(I_{n-2}\) on rhs |
| \(I_n\!\left(1+\frac{n-1}{3}\right) = \frac{16(n-1)}{3}I_{n-2}\) | M1 | Collects terms in \(I_n\) from both sides |
| \((n+2)I_n = 16(n-1)I_{n-2}\) * | A1*cso | Printed answer with no errors |
| (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_0^4 x^n(16-x^2)^{\frac{1}{2}}\,dx = \int_0^4 x^n\frac{(16-x^2)}{(16-x^2)^{\frac{1}{2}}}\,dx = \int_0^4\frac{16x^n}{(16-x^2)^{\frac{1}{2}}}\,dx - \int_0^4\frac{x^{n+2}}{(16-x^2)^{\frac{1}{2}}}\,dx\) | ||
| \(= \int_0^4 16x^{n-1}\cdot x(16-x^2)^{-\frac{1}{2}}\,dx - \int_0^4 x^{n+1}\cdot x(16-x^2)^{-\frac{1}{2}}\,dx\) | M1A1 | M1: Obtains \(x(16-x^2)^{-\frac{1}{2}}\) prior to integration. A1: Correct expressions |
| \(=\left[-16x^{n-1}(16-x^2)^{\frac{1}{2}}\right]_0^4 + 16(n-1)\int_0^4 x^{n-2}(16-x^2)^{\frac{1}{2}}\,dx\) \(-\left(\left[-x^{n+1}(16-x^2)^{\frac{1}{2}}\right]_0^4 + (n+1)\int_0^4 x^n(16-x^2)^{\frac{1}{2}}\,dx\right)\) | dM1 | Parts in correct direction on both (ignore limits) |
| \(I_n = 16(n-1)I_{n-2}-(n+1)I_n\) | M1 | Manipulates to at least one integral in terms of \(I_n\) or \(I_{n-2}\) on rhs |
| \(I_n(1+n+1)=16(n-1)I_{n-2}\) | M1 | Collects terms in \(I_n\) from both sides |
| \((n+2)I_n=16(n-1)I_{n-2}\) * | A1* | Printed answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_0^4 x^n(16-x^2)^{\frac{1}{2}}\,dx = \int_0^4 x\times x^{n-1}\frac{(16-x^2)}{(16-x^2)^{\frac{1}{2}}}\,dx\) | M1A1 | M1: Obtains \(x(16-x^2)^{-\frac{1}{2}}\) prior to integration. A1: Correct expression |
| \(=\left[-x^{n-1}(16-x^2)(16-x^2)^{\frac{1}{2}}\right]_0^4 + \int_0^4(16(n-1)x^{n-2}-(n+1)x^n)(16-x^2)^{\frac{1}{2}}\,dx\) | dM1 | Parts in correct direction (ignore limits) |
| \(I_n = 16(n-1)I_{n-2}-(n+1)I_n\) | M1 | Manipulates to at least one integral in terms of \(I_n\) or \(I_{n-2}\) on rhs |
| \(I_n(1+n+1)=16(n-1)I_{n-2}\) | M1 | Collects terms in \(I_n\) from both sides |
| \((n+2)I_n=16(n-1)I_{n-2}\) * | A1* | Printed answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_1=\int_0^4 x\sqrt{16-x^2}\,dx=\left[-\frac{1}{3}(16-x^2)^{\frac{3}{2}}\right]_0^4=\frac{64}{3}\) | M1 A1 | M1: Correct integration to find \(I_1\). A1: \(\frac{64}{3}\) or equivalent. (May be implied by later work) |
| Using \(x=4\sin\theta\): \(I_1=\int_0^{\frac{\pi}{2}}4\sin\theta\sqrt{16-16\sin^2\theta}\,4\cos\theta\,d\theta=\int_0^{\frac{\pi}{2}}64\sin\theta\cos^2\theta\,d\theta=\left[-\frac{64}{3}\cos^3\theta\right]_0^{\frac{\pi}{2}}\) | M1 A1 | M1: Complete substitution and attempt to substitute changed limits. A1: \(\frac{64}{3}\) or equivalent |
| \(I_5=\frac{64}{7}I_3\), \(I_3=\frac{32}{5}I_1\) | M1, M1 | Applies reduction formula twice. First M1 for \(I_5\) in terms of \(I_3\), second M1 for \(I_3\) in terms of \(I_1\) (can be implied) |
| \(I_5=\frac{131072}{105}\) | A1 | Any exact equivalent (depends on all previous marks) |
| (5) | ||
| Total 11 |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_n = \int_0^4 x^{n-1} \times x(16-x^2)^{\frac{1}{2}}\,dx$ | M1A1 | M1: Obtains $x(16-x^2)^{\frac{1}{2}}$ prior to integration. A1: Correct underlined expression |
| $I_n = \left[-\frac{1}{3}x^{n-1}(16-x^2)^{\frac{3}{2}}\right]_0^4 + \frac{n-1}{3}\int_0^4 x^{n-2}(16-x^2)^{\frac{3}{2}}\,dx$ | dM1 | Parts in correct direction (ignore limits) |
| $\therefore I_n = \frac{n-1}{3}\int_0^4 x^{n-2}(16-x^2)(16-x^2)^{\frac{1}{2}}\,dx$ | | |
| $I_n = \frac{16(n-1)}{3}I_{n-2} - \frac{n-1}{3}I_n$ | M1 | Manipulates to obtain at least one integral in terms of $I_n$ or $I_{n-2}$ on rhs |
| $I_n\!\left(1+\frac{n-1}{3}\right) = \frac{16(n-1)}{3}I_{n-2}$ | M1 | Collects terms in $I_n$ from both sides |
| $(n+2)I_n = 16(n-1)I_{n-2}$ * | A1*cso | Printed answer with no errors |
| | | (6) |
**Way 2:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^4 x^n(16-x^2)^{\frac{1}{2}}\,dx = \int_0^4 x^n\frac{(16-x^2)}{(16-x^2)^{\frac{1}{2}}}\,dx = \int_0^4\frac{16x^n}{(16-x^2)^{\frac{1}{2}}}\,dx - \int_0^4\frac{x^{n+2}}{(16-x^2)^{\frac{1}{2}}}\,dx$ | | |
| $= \int_0^4 16x^{n-1}\cdot x(16-x^2)^{-\frac{1}{2}}\,dx - \int_0^4 x^{n+1}\cdot x(16-x^2)^{-\frac{1}{2}}\,dx$ | M1A1 | M1: Obtains $x(16-x^2)^{-\frac{1}{2}}$ prior to integration. A1: Correct expressions |
| $=\left[-16x^{n-1}(16-x^2)^{\frac{1}{2}}\right]_0^4 + 16(n-1)\int_0^4 x^{n-2}(16-x^2)^{\frac{1}{2}}\,dx$ $-\left(\left[-x^{n+1}(16-x^2)^{\frac{1}{2}}\right]_0^4 + (n+1)\int_0^4 x^n(16-x^2)^{\frac{1}{2}}\,dx\right)$ | dM1 | Parts in correct direction on both (ignore limits) |
| $I_n = 16(n-1)I_{n-2}-(n+1)I_n$ | M1 | Manipulates to at least one integral in terms of $I_n$ or $I_{n-2}$ on rhs |
| $I_n(1+n+1)=16(n-1)I_{n-2}$ | M1 | Collects terms in $I_n$ from both sides |
| $(n+2)I_n=16(n-1)I_{n-2}$ * | A1* | Printed answer with no errors |
**Way 3:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^4 x^n(16-x^2)^{\frac{1}{2}}\,dx = \int_0^4 x\times x^{n-1}\frac{(16-x^2)}{(16-x^2)^{\frac{1}{2}}}\,dx$ | M1A1 | M1: Obtains $x(16-x^2)^{-\frac{1}{2}}$ prior to integration. A1: Correct expression |
| $=\left[-x^{n-1}(16-x^2)(16-x^2)^{\frac{1}{2}}\right]_0^4 + \int_0^4(16(n-1)x^{n-2}-(n+1)x^n)(16-x^2)^{\frac{1}{2}}\,dx$ | dM1 | Parts in correct direction (ignore limits) |
| $I_n = 16(n-1)I_{n-2}-(n+1)I_n$ | M1 | Manipulates to at least one integral in terms of $I_n$ or $I_{n-2}$ on rhs |
| $I_n(1+n+1)=16(n-1)I_{n-2}$ | M1 | Collects terms in $I_n$ from both sides |
| $(n+2)I_n=16(n-1)I_{n-2}$ * | A1* | Printed answer with no errors |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_1=\int_0^4 x\sqrt{16-x^2}\,dx=\left[-\frac{1}{3}(16-x^2)^{\frac{3}{2}}\right]_0^4=\frac{64}{3}$ | M1 A1 | M1: Correct integration to find $I_1$. A1: $\frac{64}{3}$ or equivalent. (May be implied by later work) |
| Using $x=4\sin\theta$: $I_1=\int_0^{\frac{\pi}{2}}4\sin\theta\sqrt{16-16\sin^2\theta}\,4\cos\theta\,d\theta=\int_0^{\frac{\pi}{2}}64\sin\theta\cos^2\theta\,d\theta=\left[-\frac{64}{3}\cos^3\theta\right]_0^{\frac{\pi}{2}}$ | M1 A1 | M1: Complete substitution and attempt to substitute changed limits. A1: $\frac{64}{3}$ or equivalent |
| $I_5=\frac{64}{7}I_3$, $I_3=\frac{32}{5}I_1$ | M1, M1 | Applies reduction formula twice. First M1 for $I_5$ in terms of $I_3$, second M1 for $I_3$ in terms of $I_1$ (can be implied) |
| $I_5=\frac{131072}{105}$ | A1 | Any exact equivalent (depends on all previous marks) |
| | | (5) |
| | | **Total 11** |
---\begin{enumerate}
\item Given that
\end{enumerate}
$$I _ { n } = \int _ { 0 } ^ { 4 } x ^ { n } \sqrt { } \left( 16 - x ^ { 2 } \right) \mathrm { d } x , \quad n \geqslant 0$$
(a) prove that, for $n \geqslant 2$,
$$( n + 2 ) I _ { n } = 16 ( n - 1 ) I _ { n - 2 }$$
(b) Hence, showing each step of your working, find the exact value of $I _ { 5 }$\\
\hfill \mbox{\textit{Edexcel FP3 2013 Q6 [11]}}