| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Complete table then apply trapezium rule |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard C4 techniques: calculating a function value, applying trapezium rule (routine procedure), error analysis (conceptual but basic), and volume of revolution (standard integration). All parts are textbook exercises requiring no novel insight, making it slightly easier than average. |
| Spec | 1.09f Trapezium rule: numerical integration4.08d Volumes of revolution: about x and y axes |
| \(x\) | 0 | 0.25 | 0.5 | 0.75 | 1 |
| \(y\) | 1 | 1.0308 | 1.25 | 1.4142 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(x = 0.5\), \(y = 1.1180\) | B1 | 4dp |
| \(A \approx 0.25/2\{1 + 1.4142 + 2(1.0308 + 1.1180 + 1.25)\}\) | M1 | |
| \(= 0.25 \times 4.6059 = 1.151475\) | \((0.125 \times 9.2118)\); need evidence | |
| \(= 1.151\) (3 d.p.) | E1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| The curve is below the trapezia, so the area is an over-estimate | B1 | or use a diagram to show why |
| This becomes less with more strips / greater number of strips improves accuracy so becomes less | B1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(V = \int_0^1 \pi y^2 \, dx = \int_0^1 \pi(1+x^2) \, dx\) | M1 | allow limits later |
| \(= \pi\left[(x + x^3/3)\right]_0^1\) | B1 | \(x + x^3/3\) |
| \(= 1\frac{1}{3}\pi\) | A1 [3] | exact |
# Question 2:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $x = 0.5$, $y = 1.1180$ | B1 | 4dp |
| $A \approx 0.25/2\{1 + 1.4142 + 2(1.0308 + 1.1180 + 1.25)\}$ | M1 | |
| $= 0.25 \times 4.6059 = 1.151475$ | | $(0.125 \times 9.2118)$; need evidence |
| $= 1.151$ (3 d.p.) | E1 [3] | |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| The curve is below the trapezia, so the area is an over-estimate | B1 | or use a diagram to show why |
| This becomes less with more strips / greater number of strips improves accuracy so becomes less | B1 [2] | |
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $V = \int_0^1 \pi y^2 \, dx = \int_0^1 \pi(1+x^2) \, dx$ | M1 | allow limits later |
| $= \pi\left[(x + x^3/3)\right]_0^1$ | B1 | $x + x^3/3$ |
| $= 1\frac{1}{3}\pi$ | A1 [3] | exact |
---2 Fig. 2 shows the curve $y = \sqrt { 1 + x ^ { 2 } }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5c149cb5-7392-4219-b285-486f4694aa6f-2_574_944_612_598}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
(i) The following table gives some values of $x$ and $y$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 0 & 0.25 & 0.5 & 0.75 & 1 \\
\hline
$y$ & 1 & 1.0308 & & 1.25 & 1.4142 \\
\hline
\end{tabular}
\end{center}
Find the missing value of $y$, giving your answer correct to 4 decimal places.\\
Hence show that, using the trapezium rule with four strips, the shaded area is approximately 1.151 square units.\\
(ii) Jenny uses a trapezium rule with 8 strips, and obtains a value of 1.158 square units. Explain why she must have made a mistake.\\
(iii) The shaded area is rotated through $360 ^ { \circ }$ about the $x$-axis. Find the exact volume of the solid of revolution formed.
\hfill \mbox{\textit{OCR MEI C4 2010 Q2 [8]}}