| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area with exponential functions |
| Difficulty | Standard +0.3 This is a straightforward integration question requiring factorization of the exponential, finding the root, then integrating by parts or using a standard result. The 'show that' format and clear structure make it slightly easier than average, though integration by parts adds some technical demand. |
| Spec | 1.08d Evaluate definite integrals: between limits1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u=x,\ \frac{du}{dx}=1;\ \frac{dv}{dx}=e^{2x} \Rightarrow v=\frac{1}{2}e^{2x}\); attempts IBP on \(xe^{2x}\): \(\frac{1}{2}xe^{2x} - \int \frac{1}{2}e^{2x}\,dx\) | M1 | AO 3.1a |
| \(\int(2e^{2x} - xe^{2x})\,dx = e^{2x} - \left(\frac{1}{2}xe^{2x} - \int\frac{1}{2}e^{2x}\,dx\right)\) | M1 | AO 1.1b |
| \(= e^{2x} - \left(\frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x}\right)\) | A1 | AO 1.1b |
| \(\text{Area}(R) = \int_0^2 2e^{2x} - xe^{2x}\,dx = \left[\frac{5}{4}e^{2x} - \frac{1}{2}xe^{2x}\right]_0^2\) | M1 | AO 2.2a |
| \(= \left(\frac{5}{4}e^4 - e^4\right) - \left(\frac{5}{4}e^0 - 0\right) = \frac{1}{4}e^4 - \frac{5}{4}\) | A1 | AO 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = 2-x \Rightarrow \frac{du}{dx}=-1;\ \frac{dv}{dx}=e^{2x} \Rightarrow v=\frac{1}{2}e^{2x}\) | M1 | AO 3.1a |
| \(= \frac{1}{2}(2-x)e^{2x} - \int -\frac{1}{2}e^{2x}\,dx\) | M1 | AO 1.1b |
| \(= \frac{1}{2}(2-x)e^{2x} + \frac{1}{4}e^{2x}\) | A1 | AO 1.1b |
| \(\text{Area}(R) = \int_0^2(2-x)e^{2x}\,dx = \left[\frac{1}{2}(2-x)e^{2x} + \frac{1}{4}e^{2x}\right]_0^2\) | M1 | AO 2.2a |
| \(= \left(0 + \frac{1}{4}e^4\right) - \left(\frac{1}{2}(2)e^0 + \frac{1}{4}e^0\right) = \frac{1}{4}e^4 - \frac{5}{4}\) | A1 | AO 2.1 |
## Question 7:
### Main Method:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u=x,\ \frac{du}{dx}=1;\ \frac{dv}{dx}=e^{2x} \Rightarrow v=\frac{1}{2}e^{2x}$; attempts IBP on $xe^{2x}$: $\frac{1}{2}xe^{2x} - \int \frac{1}{2}e^{2x}\,dx$ | M1 | AO 3.1a |
| $\int(2e^{2x} - xe^{2x})\,dx = e^{2x} - \left(\frac{1}{2}xe^{2x} - \int\frac{1}{2}e^{2x}\,dx\right)$ | M1 | AO 1.1b |
| $= e^{2x} - \left(\frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x}\right)$ | A1 | AO 1.1b |
| $\text{Area}(R) = \int_0^2 2e^{2x} - xe^{2x}\,dx = \left[\frac{5}{4}e^{2x} - \frac{1}{2}xe^{2x}\right]_0^2$ | M1 | AO 2.2a |
| $= \left(\frac{5}{4}e^4 - e^4\right) - \left(\frac{5}{4}e^0 - 0\right) = \frac{1}{4}e^4 - \frac{5}{4}$ | A1 | AO 2.1 |
### Alt 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = 2-x \Rightarrow \frac{du}{dx}=-1;\ \frac{dv}{dx}=e^{2x} \Rightarrow v=\frac{1}{2}e^{2x}$ | M1 | AO 3.1a |
| $= \frac{1}{2}(2-x)e^{2x} - \int -\frac{1}{2}e^{2x}\,dx$ | M1 | AO 1.1b |
| $= \frac{1}{2}(2-x)e^{2x} + \frac{1}{4}e^{2x}$ | A1 | AO 1.1b |
| $\text{Area}(R) = \int_0^2(2-x)e^{2x}\,dx = \left[\frac{1}{2}(2-x)e^{2x} + \frac{1}{4}e^{2x}\right]_0^2$ | M1 | AO 2.2a |
| $= \left(0 + \frac{1}{4}e^4\right) - \left(\frac{1}{2}(2)e^0 + \frac{1}{4}e^0\right) = \frac{1}{4}e^4 - \frac{5}{4}$ | A1 | AO 2.1 |
**Notes:**
- M1: Recognises need for IBP on $xe^{2x}$ or $(2-x)e^{2x}$
- Second M1: Correct application giving $e^{2x}$ term plus remaining integral
- Final A1: Correct proof with $p = \frac{1}{4},\ q = -\frac{5}{4}$
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{96e004d9-c6b6-474b-9b67-06e1771c609e-14_604_1063_251_502}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of part of the curve with equation
$$y = 2 \mathrm { e } ^ { 2 x } - x \mathrm { e } ^ { 2 x } , \quad x \in \mathbb { R }$$
The finite region $R$, shown shaded in Figure 4, is bounded by the curve, the $x$-axis and the $y$-axis.
Use calculus to show that the exact area of $R$ can be written in the form $p \mathrm { e } ^ { 4 } + q$, where $p$ and $q$ are rational constants to be found.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\hfill \mbox{\textit{Edexcel Paper 1 Q7 [5]}}