| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume with numerical methods |
| Difficulty | Standard +0.3 Part (a) is a standard volume of revolution requiring recognition that y² = 1 + e^(2x) simplifies the integral to a routine exponential integration. Parts (b)(i) and (b)(ii) are straightforward trapezium rule applications with minimal calculation and conceptual reasoning about concavity. This is easier than average as it's mostly procedural with helpful simplification. |
| Spec | 1.09f Trapezium rule: numerical integration4.08d Volumes of revolution: about x and y axes |
| \(x\) | 0 | 0.5 | 1 | 1.5 | 2 |
| \(y\) | 1.9283 | 2.8964 | 4.5919 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(V = \int_0^2 \pi y^2 dx = \int_0^2 \pi(1+e^{2x})dx\) | M1 | \(\int_0^2 \pi(1+e^{2x})dx\) — limits must appear but may be later; condone omission of \(dx\) if intention clear |
| \(= \pi\left[x + \frac{1}{2}e^{2x}\right]_0^2\) | B1 | \(\left[x + \frac{1}{2}e^{2x}\right]\) independent of \(\pi\) and limits |
| \(= \pi(2 + \frac{1}{2}e^4 - \frac{1}{2})\) | DM1 | Dependent on first M1. Need both limits substituted in integral of form \(ax + be^{2x}\), where \(a,b\) non-zero constants. Accept answers including \(e^0\) for M1. Condone absence of \(\pi\) for M1 at this stage |
| \(= \frac{1}{2}\pi(3+e^4)\) | A1 | cao exact only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x=0,\ y=1.4142;\ x=2,\ y=7.4564\) | B1 | 1.414, 7.456 or better |
| \(A = \frac{0.5}{2}\{(1.4142+7.4564) + 2(1.9283+2.8964+4.5919)\} = 6.926\) | M1 | Correct formula seen (can be implied by correct intermediate step e.g. \(27.7038.../4\)) |
| \(= 6.926\) | A1 | 6.926 or 6.93 (do not allow more dp) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 8 strips: 6.823, 16 strips: 6.797. Trapezium rule overestimates this area, but the overestimate gets less as the number of strips increases. | B1 | oe |
## Question 5:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = \int_0^2 \pi y^2 dx = \int_0^2 \pi(1+e^{2x})dx$ | M1 | $\int_0^2 \pi(1+e^{2x})dx$ — limits must appear but may be later; condone omission of $dx$ if intention clear |
| $= \pi\left[x + \frac{1}{2}e^{2x}\right]_0^2$ | B1 | $\left[x + \frac{1}{2}e^{2x}\right]$ independent of $\pi$ and limits |
| $= \pi(2 + \frac{1}{2}e^4 - \frac{1}{2})$ | DM1 | Dependent on first M1. Need **both** limits substituted in integral of form $ax + be^{2x}$, where $a,b$ non-zero constants. Accept answers including $e^0$ for M1. Condone absence of $\pi$ for M1 at this stage |
| $= \frac{1}{2}\pi(3+e^4)$ | A1 | cao exact only |
**[4 marks]**
---
### Part (b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x=0,\ y=1.4142;\ x=2,\ y=7.4564$ | B1 | 1.414, 7.456 or better |
| $A = \frac{0.5}{2}\{(1.4142+7.4564) + 2(1.9283+2.8964+4.5919)\} = 6.926$ | M1 | Correct formula seen (can be implied by correct intermediate step e.g. $27.7038.../4$) |
| $= 6.926$ | A1 | 6.926 or 6.93 (do not allow more dp) |
**[3 marks]**
---
### Part (b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| 8 strips: 6.823, 16 strips: 6.797. Trapezium rule overestimates this area, but the overestimate gets less as the number of strips increases. | B1 | oe |
**[1 mark]**
---5 Fig. 4 shows the curve $y = \sqrt { 1 + \mathrm { e } ^ { 2 x } }$, and the region between the curve, the $x$-axis, the $y$-axis and the line $x = 2$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{252453c9-9afa-435c-b64b-5ea37ec69eed-5_657_732_375_667}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Find the exact volume of revolution when the shaded region is rotated through $360 ^ { \circ }$ about the $x$-axis.
\item \begin{enumerate}[label=(\roman*)]
\item Complete the table of values, and use the trapezium rule with 4 strips to estimate the area of the shaded region.
\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | }
\hline
$x$ & 0 & 0.5 & 1 & 1.5 & 2 \\
\hline
$y$ & & 1.9283 & 2.8964 & 4.5919 & \\
\hline
\end{tabular}
\end{center}
\item The trapezium rule for $\int _ { 0 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { 2 x } } \mathrm {~d} x$ with 8 and 16 strips gives 6.797 and 6.823, although not necessarily in that order. Without doing the calculations, say which result is which, explaining your reasoning.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI C4 Q5 [8]}}