| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2017 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume with exponential functions |
| Difficulty | Standard +0.3 This is a standard C4 volumes of revolution question requiring rotation about the x-axis with an exponential function. Students must square the function (giving e^(2x) + 4 + 4e^(-2x)), integrate term-by-term using standard exponential integrals, and evaluate between 0 and ln 4. While it involves multiple steps and careful algebraic manipulation, it follows a completely routine procedure with no novel insight required, making it slightly easier than average. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{V=\}\pi\int_0^{\ln 4}(e^x+2e^{-x})^2\,dx\) | B1 | For \(\pi\int(e^x+2e^{-x})^2\); ignore limits and \(dx\); can be implied |
| \(=\{\pi\}\int_0^{\ln 4}(e^{2x}+4e^{-2x}+4)\,dx\) | M1 | Expands \((e^x+2e^{-x})^2 \to \pm\alpha e^{2x}\pm\beta e^{-2x}\pm\delta\) where \(\alpha,\beta,\delta\neq 0\); ignore \(\pi\), integral sign, limits and \(dx\) |
| \(=\{\pi\}\left[\frac{1}{2}e^{2x}-2e^{-2x}+4x\right]_0^{\ln 4}\) | M1 (dep on 2nd M1) | Integrates at least one of \(\pm\alpha e^{2x}\) to give \(\pm\frac{\alpha}{2}e^{2x}\), or \(\pm\beta e^{-2x}\) to give \(\pm\frac{\beta}{2}e^{-2x}\), \(\alpha,\beta\neq 0\) |
| \(e^{2x}+4e^{-2x} \to \frac{1}{2}e^{2x}-2e^{-2x}\) | A1 | Can be simplified or un-simplified |
| \(4 \to 4x\) or \(4e^0x\) | B1 cao | |
| \(=\{\pi\}\left\{\left(\frac{1}{2}e^{2\ln 4}-2e^{-2\ln 4}+4\ln 4\right)-\left(\frac{1}{2}e^0-2e^0+4(0)\right)\right\}\) | dM1 | Dependent on previous method mark; some evidence of applying limits of \(\ln 4\) o.e. and 0; subtracts correct way round; proper consideration of limit 0 required |
| \(=\frac{75}{8}\pi+4\pi\ln 4\) or \(\frac{75}{8}\pi+8\pi\ln 2\) or \(\pi\left(\frac{75}{8}+4\ln 4\right)\) or \(\pi\left(\frac{75}{8}+8\ln 2\right)\) or \(\frac{75}{8}\pi+\ln 2^{8\pi}\) or \(\frac{75}{8}\pi+\pi\ln 256\) or \(\ln\left(2^{8\pi}e^{\frac{75}{8}\pi}\right)\) or \(\frac{1}{8}\pi(75+32\ln 4)\) etc | A1 isw |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{V=\}\pi\int_0^{\ln 4}(e^x+2e^{-x})^2\,dx\) | B1 | For \(\pi\int(e^x+2e^{-x})^2\); ignore limits and \(dx\) |
| \(u=e^x \Rightarrow \frac{du}{dx}=e^x=u\); \(x=\ln 4\Rightarrow u=4\), \(x=0\Rightarrow u=e^0=1\) | ||
| \(V=\{\pi\}\int_1^4\left(u+\frac{2}{u}\right)^2\frac{1}{u}\,du = \{\pi\}\int_1^4\left(u^2+\frac{4}{u^2}+4\right)\frac{1}{u}\,du\) | ||
| \((e^x+2e^{-x})^2 \to \pm\alpha u\pm\beta u^{-3}\pm\delta u^{-1}\) where \(u=e^x\), \(\alpha,\beta,\delta\neq 0\) | M1 | Ignore \(\pi\), integral sign, limits and \(du\) |
| \(=\{\pi\}\left[\frac{1}{2}u^2-\frac{2}{u^2}+4\ln u\right]_1^4\) | M1 (dep on 2nd M1) | Integrates at least one of \(\pm\alpha u\) to give \(\pm\frac{\alpha}{2}u^2\), or \(\pm\beta u^{-3}\) to give \(\pm\frac{\beta}{2}u^{-2}\), \(\alpha,\beta\neq 0\), where \(u=e^x\) |
| \(u+4u^{-3} \to \frac{1}{2}u^2-2u^{-2}\) | A1 | Simplified or un-simplified, where \(u=e^x\) |
| \(4u^{-1} \to 4\ln u\), where \(u=e^x\) | B1 cao | |
| Applies limits 4 and 1 to changed function in \(u\) and subtracts correct way round | dM1 | Dependent on previous method mark |
| \(=\frac{75}{8}\pi+4\pi\ln 4\) etc (same alternatives as Way 1) | A1 isw |
# Question 5:
## Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{V=\}\pi\int_0^{\ln 4}(e^x+2e^{-x})^2\,dx$ | B1 | For $\pi\int(e^x+2e^{-x})^2$; ignore limits and $dx$; can be implied |
| $=\{\pi\}\int_0^{\ln 4}(e^{2x}+4e^{-2x}+4)\,dx$ | M1 | Expands $(e^x+2e^{-x})^2 \to \pm\alpha e^{2x}\pm\beta e^{-2x}\pm\delta$ where $\alpha,\beta,\delta\neq 0$; ignore $\pi$, integral sign, limits and $dx$ |
| $=\{\pi\}\left[\frac{1}{2}e^{2x}-2e^{-2x}+4x\right]_0^{\ln 4}$ | M1 (dep on 2nd M1) | Integrates at least one of $\pm\alpha e^{2x}$ to give $\pm\frac{\alpha}{2}e^{2x}$, or $\pm\beta e^{-2x}$ to give $\pm\frac{\beta}{2}e^{-2x}$, $\alpha,\beta\neq 0$ |
| $e^{2x}+4e^{-2x} \to \frac{1}{2}e^{2x}-2e^{-2x}$ | A1 | Can be simplified or un-simplified |
| $4 \to 4x$ or $4e^0x$ | B1 cao | |
| $=\{\pi\}\left\{\left(\frac{1}{2}e^{2\ln 4}-2e^{-2\ln 4}+4\ln 4\right)-\left(\frac{1}{2}e^0-2e^0+4(0)\right)\right\}$ | dM1 | Dependent on previous method mark; some evidence of applying limits of $\ln 4$ o.e. and 0; subtracts correct way round; proper consideration of limit 0 required |
| $=\frac{75}{8}\pi+4\pi\ln 4$ or $\frac{75}{8}\pi+8\pi\ln 2$ or $\pi\left(\frac{75}{8}+4\ln 4\right)$ or $\pi\left(\frac{75}{8}+8\ln 2\right)$ or $\frac{75}{8}\pi+\ln 2^{8\pi}$ or $\frac{75}{8}\pi+\pi\ln 256$ or $\ln\left(2^{8\pi}e^{\frac{75}{8}\pi}\right)$ or $\frac{1}{8}\pi(75+32\ln 4)$ etc | A1 isw | |
## Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{V=\}\pi\int_0^{\ln 4}(e^x+2e^{-x})^2\,dx$ | B1 | For $\pi\int(e^x+2e^{-x})^2$; ignore limits and $dx$ |
| $u=e^x \Rightarrow \frac{du}{dx}=e^x=u$; $x=\ln 4\Rightarrow u=4$, $x=0\Rightarrow u=e^0=1$ | | |
| $V=\{\pi\}\int_1^4\left(u+\frac{2}{u}\right)^2\frac{1}{u}\,du = \{\pi\}\int_1^4\left(u^2+\frac{4}{u^2}+4\right)\frac{1}{u}\,du$ | | |
| $(e^x+2e^{-x})^2 \to \pm\alpha u\pm\beta u^{-3}\pm\delta u^{-1}$ where $u=e^x$, $\alpha,\beta,\delta\neq 0$ | M1 | Ignore $\pi$, integral sign, limits and $du$ |
| $=\{\pi\}\left[\frac{1}{2}u^2-\frac{2}{u^2}+4\ln u\right]_1^4$ | M1 (dep on 2nd M1) | Integrates at least one of $\pm\alpha u$ to give $\pm\frac{\alpha}{2}u^2$, or $\pm\beta u^{-3}$ to give $\pm\frac{\beta}{2}u^{-2}$, $\alpha,\beta\neq 0$, where $u=e^x$ |
| $u+4u^{-3} \to \frac{1}{2}u^2-2u^{-2}$ | A1 | Simplified or un-simplified, where $u=e^x$ |
| $4u^{-1} \to 4\ln u$, where $u=e^x$ | B1 cao | |
| Applies limits 4 and 1 to changed function in $u$ and subtracts correct way round | dM1 | Dependent on previous method mark |
| $=\frac{75}{8}\pi+4\pi\ln 4$ etc (same alternatives as Way 1) | A1 isw | |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cd958ff3-ed4e-4bd7-aa4b-339da6d618a6-16_589_540_248_705}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Diagram not drawn to scale
The finite region $S$, shown shaded in Figure 2, is bounded by the $y$-axis, the $x$-axis, the line with equation $x = \ln 4$ and the curve with equation
$$y = \mathrm { e } ^ { x } + 2 \mathrm { e } ^ { - x } , \quad x \geqslant 0$$
The region $S$ is rotated through $2 \pi$ radians about the $x$-axis.\\
Use integration to find the exact value of the volume of the solid generated. Give your answer in its simplest form.\\[0pt]
[Solutions based entirely on graphical or numerical methods are not acceptable.]
\hfill \mbox{\textit{Edexcel C4 2017 Q5 [7]}}