| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Trigonometric power reduction |
| Difficulty | Challenging +1.2 Part (a) is a standard reduction formula derivation using integration by parts—a core FP2 technique that follows a well-established method. Part (b) applies the formula to evaluate a definite integral with a substitution (u=4x), requiring careful bookkeeping but no novel insight. While this is Further Maths content and involves multiple steps, it's a textbook application of reduction formulae rather than a problem requiring creative problem-solving. |
| Spec | 1.08d Evaluate definite integrals: between limits8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(I_n = \int_0^{\frac{\pi}{2}} \sin x \sin^{n-1} x \, dx\) | M1 | Splits integrand into product shown and begins integration by parts (sign errors allowed) |
| \(= \left[-\cos x \sin^{n-1} x\right]_0^{\frac{\pi}{2}} - (-)\int_0^{\frac{\pi}{2}} \cos^2 x (n-1)\sin^{n-2} x \, dx\) | A1 | Correct work |
| \(= 0 - (-)\int_0^{\frac{\pi}{2}} (1-\sin^2 x)(n-1)\sin^{n-2} x \, dx\) | M1 | Uses limits on first term and expresses \(\cos^2\) in terms of \(\sin^2\) |
| \(I_n = (n-1)I_{n-2} - (n-1)I_n\), hence \(nI_n = (n-1)I_{n-2}\) | A1* | Completes proof collecting \(I_n\) terms correctly with all stages shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Uses \(I_n = \dfrac{n-1}{n}I_{n-2}\) to give \(I_{10} = \dfrac{9}{10}I_8\) or \(I_2 = \dfrac{1}{2}I_0\) | M1 | Attempts to find \(I_{10}\) and/or \(I_2\) |
| \(I_{10} = \dfrac{9 \times 7 \times 5 \times 3 \times 1}{10 \times 8 \times 6 \times 4 \times 2} I_0\) | M1 | Finds \(I_{10}\) in terms of \(I_0\) |
| \(I_0 = \dfrac{\pi}{2}\) | B1 | Finds \(I_0\) correctly |
| Required area is \(2(I_2 - I_{10})\) or \(8 \times \dfrac{1}{4}(I_2 - I_{10})\) | M1 | States the expression needed to find the required area |
| \(= 2\left(\dfrac{\pi}{4} - \dfrac{63\pi}{512}\right) = \dfrac{65\pi}{256} \text{ m}^2\) | A1 | Completes the calculation to give this exact answer |
# Question 7:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $I_n = \int_0^{\frac{\pi}{2}} \sin x \sin^{n-1} x \, dx$ | M1 | Splits integrand into product shown and begins integration by parts (sign errors allowed) |
| $= \left[-\cos x \sin^{n-1} x\right]_0^{\frac{\pi}{2}} - (-)\int_0^{\frac{\pi}{2}} \cos^2 x (n-1)\sin^{n-2} x \, dx$ | A1 | Correct work |
| $= 0 - (-)\int_0^{\frac{\pi}{2}} (1-\sin^2 x)(n-1)\sin^{n-2} x \, dx$ | M1 | Uses limits on first term and expresses $\cos^2$ in terms of $\sin^2$ |
| $I_n = (n-1)I_{n-2} - (n-1)I_n$, hence $nI_n = (n-1)I_{n-2}$ | A1* | Completes proof collecting $I_n$ terms correctly with all stages shown |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Uses $I_n = \dfrac{n-1}{n}I_{n-2}$ to give $I_{10} = \dfrac{9}{10}I_8$ or $I_2 = \dfrac{1}{2}I_0$ | M1 | Attempts to find $I_{10}$ and/or $I_2$ |
| $I_{10} = \dfrac{9 \times 7 \times 5 \times 3 \times 1}{10 \times 8 \times 6 \times 4 \times 2} I_0$ | M1 | Finds $I_{10}$ in terms of $I_0$ |
| $I_0 = \dfrac{\pi}{2}$ | B1 | Finds $I_0$ correctly |
| Required area is $2(I_2 - I_{10})$ or $8 \times \dfrac{1}{4}(I_2 - I_{10})$ | M1 | States the expression needed to find the required area |
| $= 2\left(\dfrac{\pi}{4} - \dfrac{63\pi}{512}\right) = \dfrac{65\pi}{256} \text{ m}^2$ | A1 | Completes the calculation to give this exact answer |
---7.
$$I _ { n } = \int _ { 0 } ^ { \frac { \pi } { 2 } } \sin ^ { n } x \mathrm {~d} x , \quad n \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Prove that, for $n \geqslant 2$,
$$n I _ { n } = ( n - 1 ) I _ { n - 2 }$$
\item \begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1c262813-4160-4eda-9a36-e4ba38182c8a-22_588_1018_630_520}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A designer is asked to produce a poster to completely cover the curved surface area of a solid cylinder which has diameter 1 m and height 0.7 m .
He uses a large sheet of paper with height 0.7 m and width of $\pi \mathrm { m }$.\\
Figure 2 shows the first stage of the design, where the poster is divided into two sections by a curve.
The curve is given by the equation
$$y = \sin ^ { 2 } ( 4 x ) - \sin ^ { 10 } ( 4 x )$$
relative to axes taken along the bottom and left hand edge of the paper.\\
The region of the poster below the curve is shaded and the region above the curve remains unshaded, as shown in Figure 2.
Find the exact area of the poster which is shaded.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 Q7 [9]}}