| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2021 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Integration of e^(ax)·trig(bx) |
| Difficulty | Standard +0.8 This is a standard Further Maths integration by parts question requiring the cyclic method for e^(ax)sin(bx), followed by applying limits from 0 to π. While it requires multiple steps and careful algebra, it's a well-known technique taught explicitly in P4/FP2. The 'show that' format in part (b) provides the target answer, reducing difficulty slightly. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int e^{2x} \sin x \, dx = \frac{1}{2}e^{2x} \sin x - \int \frac{1}{2}e^{2x} \cos x \, dx\) | M1 | Attempts IBP with \(u = \sin x\), \(v' = e^{2x}\) to obtain \(Ae^{2x}\sin x \pm B\int e^{2x}\cos x \, dx\), \(A>0\) |
| \(= \ldots - \frac{1}{4}e^{2x}\cos x - \int \frac{1}{4}e^{2x}\sin x \, dx\) | dM1 | Attempts IBP again with \(u = \cos x\), \(v' = e^{2x}\) on \(B\int e^{2x}\cos x \, dx\) |
| \(\int e^{2x}\sin x \, dx = \frac{1}{2}e^{2x}\sin x - \frac{1}{4}e^{2x}\cos x - \int \frac{1}{4}e^{2x}\sin x \, dx\) | A1 | Correct unsimplified expression |
| \(\frac{5}{4}\int e^{2x}\sin x \, dx = \frac{1}{2}e^{2x}\sin x - \frac{1}{4}e^{2x}\cos x\) | ddM1 | Collecting \(\int e^{2x}\sin x \, dx\) terms and making it subject. Dependent on both M marks |
| \(= \frac{2}{5}e^{2x}\sin x - \frac{1}{5}e^{2x}\cos x + c\) | A1 | Allow with or without \(+c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int e^{2x}\sin x \, dx = -e^{2x}\cos x + \int 2e^{2x}\cos x \, dx\) | M1 | Attempts IBP with \(u = e^{2x}\), \(v' = \sin x\) |
| \(= \ldots + 2e^{2x}\sin x - \int 4e^{2x}\sin x \, dx\) | dM1 | Attempts IBP again |
| \(\int e^{2x}\sin x \, dx = -e^{2x}\cos x + 2e^{2x}\sin x - \int 4e^{2x}\sin x \, dx\) | A1 | Correct unsimplified expression |
| \(5\int e^{2x}\sin x \, dx = -e^{2x}\cos x + 2e^{2x}\sin x\) | ddM1 | Collecting terms, dependent on both M marks |
| \(= \frac{2}{5}e^{2x}\sin x - \frac{1}{5}e^{2x}\cos x + c\) | A1 | Allow with or without \(+c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\frac{2}{5}e^{2\pi}\sin\pi - \frac{1}{5}e^{2\pi}\cos\pi\right) - \left(\frac{2}{5}e^0\sin 0 - \frac{1}{5}e^0\cos 0\right)\) | M1 | Applying limits 0 and \(\pi\) to expression with at least one term \(Ae^{2x}\sin x\) and one term \(Be^{2x}\cos x\). Both limits must be used |
| \(= \frac{1}{5}e^{2\pi} + \frac{1}{5} = \frac{e^{2\pi}+1}{5}\) | A1* | Found correctly from correct answer in (a) via at least one intermediate line e.g. \(\frac{e^{2\pi}}{5} + \frac{1}{5}\) |
# Question 7(a) Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int e^{2x} \sin x \, dx = \frac{1}{2}e^{2x} \sin x - \int \frac{1}{2}e^{2x} \cos x \, dx$ | M1 | Attempts IBP with $u = \sin x$, $v' = e^{2x}$ to obtain $Ae^{2x}\sin x \pm B\int e^{2x}\cos x \, dx$, $A>0$ |
| $= \ldots - \frac{1}{4}e^{2x}\cos x - \int \frac{1}{4}e^{2x}\sin x \, dx$ | dM1 | Attempts IBP again with $u = \cos x$, $v' = e^{2x}$ on $B\int e^{2x}\cos x \, dx$ |
| $\int e^{2x}\sin x \, dx = \frac{1}{2}e^{2x}\sin x - \frac{1}{4}e^{2x}\cos x - \int \frac{1}{4}e^{2x}\sin x \, dx$ | A1 | Correct unsimplified expression |
| $\frac{5}{4}\int e^{2x}\sin x \, dx = \frac{1}{2}e^{2x}\sin x - \frac{1}{4}e^{2x}\cos x$ | ddM1 | Collecting $\int e^{2x}\sin x \, dx$ terms and making it subject. Dependent on both M marks |
| $= \frac{2}{5}e^{2x}\sin x - \frac{1}{5}e^{2x}\cos x + c$ | A1 | Allow with or without $+c$ |
# Question 7(a) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int e^{2x}\sin x \, dx = -e^{2x}\cos x + \int 2e^{2x}\cos x \, dx$ | M1 | Attempts IBP with $u = e^{2x}$, $v' = \sin x$ |
| $= \ldots + 2e^{2x}\sin x - \int 4e^{2x}\sin x \, dx$ | dM1 | Attempts IBP again |
| $\int e^{2x}\sin x \, dx = -e^{2x}\cos x + 2e^{2x}\sin x - \int 4e^{2x}\sin x \, dx$ | A1 | Correct unsimplified expression |
| $5\int e^{2x}\sin x \, dx = -e^{2x}\cos x + 2e^{2x}\sin x$ | ddM1 | Collecting terms, dependent on both M marks |
| $= \frac{2}{5}e^{2x}\sin x - \frac{1}{5}e^{2x}\cos x + c$ | A1 | Allow with or without $+c$ |
# Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{2}{5}e^{2\pi}\sin\pi - \frac{1}{5}e^{2\pi}\cos\pi\right) - \left(\frac{2}{5}e^0\sin 0 - \frac{1}{5}e^0\cos 0\right)$ | M1 | Applying limits 0 and $\pi$ to expression with at least one term $Ae^{2x}\sin x$ and one term $Be^{2x}\cos x$. Both limits must be used |
| $= \frac{1}{5}e^{2\pi} + \frac{1}{5} = \frac{e^{2\pi}+1}{5}$ | A1* | Found correctly **from correct answer in (a)** via at least one intermediate line e.g. $\frac{e^{2\pi}}{5} + \frac{1}{5}$ |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{216f5735-a7ad-4d70-9da9-ae1f098a97d9-14_620_615_278_662}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Find $\int \mathrm { e } ^ { 2 x } \sin x \mathrm {~d} x$
Figure 2 shows a sketch of part of the curve with equation
$$y = \mathrm { e } ^ { 2 x } \sin x \quad x \geqslant 0$$
The finite region $R$ is bounded by the curve and the $x$-axis and is shown shaded in Figure 2.
\item Show that the exact area of $R$ is $\frac { \mathrm { e } ^ { 2 \pi } + 1 } { 5 }$\\
(Solutions relying on calculator technology are not acceptable.)\\
Question 7 continue
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2021 Q7 [7]}}