| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2024 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Volume of revolution with substitution |
| Difficulty | Challenging +1.2 Part (a) requires executing a guided substitution with careful differentiation of u = 4x + 2sin(2x) to get du = (4 + 4cos(2x))dx = 8cos²x dx, then evaluating limits—this is a standard P4/FP2 technique with moderate algebraic manipulation. Part (b) is straightforward application of the volume formula V = π∫y² dx once part (a) is complete. The question is harder than basic C3/C4 integration but represents a typical Further Maths pure question with clear guidance, placing it slightly above average difficulty. |
| Spec | 1.08h Integration by substitution4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(u = 4x + 2\sin 2x \Rightarrow \frac{du}{dx} = 4 + 4\cos 2x\) | M1 | Attempts derivative of \(u\); look for \(\sin 2x \rightarrow \ldots\cos 2x\) |
| \(\frac{du}{dx} = 4 + 4\cos 2x = 8\cos^2 x \Rightarrow \int_{0}^{(\pi/2)} e^{4x+2\sin 2x}\cos^2 x\, dx = \int_{0}^{(2\pi)} \frac{e^u}{8}\, du\) | M1 | Applies double angle formula and carries out full substitution including \(dx\) replacement |
| \(= \left[\frac{1}{8}e^u\right]_0^{2\pi}\) or \(\left[\frac{1}{8}e^{4x+2\sin 2x}\right]_0^{\pi/2}\) | A1ft | Correct form with correct limits assigned |
| Alt by inspection: \(\int e^{4x+2\sin 2x}\cos^2 x\, dx = \int e^{4x+2\sin 2x} \times \frac{1}{2}(1 \pm \cos 2x)\, dx\) | (M1 A1) | Must apply double angle formula |
| \(= ke^{4x+2\sin 2x} = \left[\frac{1}{8}e^{4x+2\sin 2x}\right]_0^{\pi/2}\) | (M1; A1ft) | Achieves correct form |
| \(= \frac{1}{8}(e^{2\pi}-1)\) | A1*cso | Correct answer from fully correct working; must have identified correct limits |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(V = \pi\int_0^{\pi/2} y^2\, dx = \pi\int_0^{\pi/2}\left(6e^{x+\sin x}\cos x\right)^2 dx = 36\pi\int_0^{\pi/2} e^{4x+2\sin 2x}\cos^2 x\, dx\) | B1 | Correct volume formula with \(\pi\); squares correctly to achieve correct \(y^2\) |
| \(K\int_0^{\pi/2} e^{4x+2\sin 2x}\cos^2 x\, dx = \frac{K}{8}(e^{2\pi}-1)\) | M1 | Makes connection with part (a); multiplies answer to (a) by constant from \(y^2\) |
| \(= \frac{9\pi}{2}(e^{2\pi}-1)\) | A1 | cao; also accept \(\frac{9\pi}{2}e^{2\pi} - \frac{9\pi}{2}\) |
# Question 7:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $u = 4x + 2\sin 2x \Rightarrow \frac{du}{dx} = 4 + 4\cos 2x$ | M1 | Attempts derivative of $u$; look for $\sin 2x \rightarrow \ldots\cos 2x$ |
| $\frac{du}{dx} = 4 + 4\cos 2x = 8\cos^2 x \Rightarrow \int_{0}^{(\pi/2)} e^{4x+2\sin 2x}\cos^2 x\, dx = \int_{0}^{(2\pi)} \frac{e^u}{8}\, du$ | M1 | Applies double angle formula and carries out full substitution including $dx$ replacement |
| $= \left[\frac{1}{8}e^u\right]_0^{2\pi}$ or $\left[\frac{1}{8}e^{4x+2\sin 2x}\right]_0^{\pi/2}$ | A1ft | Correct form with correct limits assigned |
| **Alt by inspection:** $\int e^{4x+2\sin 2x}\cos^2 x\, dx = \int e^{4x+2\sin 2x} \times \frac{1}{2}(1 \pm \cos 2x)\, dx$ | (M1 A1) | Must apply double angle formula |
| $= ke^{4x+2\sin 2x} = \left[\frac{1}{8}e^{4x+2\sin 2x}\right]_0^{\pi/2}$ | (M1; A1ft) | Achieves correct form |
| $= \frac{1}{8}(e^{2\pi}-1)$ | A1*cso | Correct answer from fully correct working; must have identified correct limits |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $V = \pi\int_0^{\pi/2} y^2\, dx = \pi\int_0^{\pi/2}\left(6e^{x+\sin x}\cos x\right)^2 dx = 36\pi\int_0^{\pi/2} e^{4x+2\sin 2x}\cos^2 x\, dx$ | B1 | Correct volume formula with $\pi$; squares correctly to achieve correct $y^2$ |
| $K\int_0^{\pi/2} e^{4x+2\sin 2x}\cos^2 x\, dx = \frac{K}{8}(e^{2\pi}-1)$ | M1 | Makes connection with part (a); multiplies answer to (a) by constant from $y^2$ |
| $= \frac{9\pi}{2}(e^{2\pi}-1)$ | A1 | cao; also accept $\frac{9\pi}{2}e^{2\pi} - \frac{9\pi}{2}$ |
---\begin{enumerate}
\item (a) Using the substitution $u = 4 x + 2 \sin 2 x$, or otherwise, show that
\end{enumerate}
$$\int _ { 0 } ^ { \frac { \pi } { 2 } } \mathrm { e } ^ { 4 x + 2 \sin 2 x } \cos ^ { 2 } x \mathrm {~d} x = \frac { 1 } { 8 } \left( \mathrm { e } ^ { 2 \pi } - 1 \right)$$
Figure 3
The curve shown in Figure 3, has equation
$$y = 6 \mathrm { e } ^ { 2 x + \sin 2 x } \cos x$$
The region $R$, shown shaded in Figure 3, is bounded by the positive $x$-axis, the positive $y$-axis and the curve.
The region $R$ is rotated through $2 \pi$ radians about the $x$-axis to form a solid.\\
(b) Use the answer to part (a) to find the volume of the solid formed, giving the answer in simplest form.
\hfill \mbox{\textit{Edexcel P4 2024 Q7 [8]}}