| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| 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 calculation), recognizing trapezium rule underestimates for convex curves, and computing a volume of revolution (standard integral). All parts are textbook exercises requiring recall and careful arithmetic rather than problem-solving or insight. |
| 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 | Marks | Guidance |
| When \(x = 0.5\), \(y = 1.1180\) | B1 | 4dp |
| \(A \approx \frac{0.25}{2}\{1 + 1.4142 + 2(1.0308 + 1.1180 + 1.25)\}\) | M1 | \((0.125 \times 9.2118)\) |
| \(= 0.25 \times 4.6059 = 1.151475\) | need evidence | |
| \(= 1.151\) (3 d.p.) | E1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | 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, or: greater number of strips improves accuracy so becomes less | B1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | 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 | Marks | Guidance |
|--------|-------|----------|
| When $x = 0.5$, $y = 1.1180$ | B1 | 4dp |
| $A \approx \frac{0.25}{2}\{1 + 1.4142 + 2(1.0308 + 1.1180 + 1.25)\}$ | M1 | $(0.125 \times 9.2118)$ |
| $= 0.25 \times 4.6059 = 1.151475$ | | need evidence |
| $= 1.151$ (3 d.p.) | E1 [3] | |
## Part (ii)
| Answer | 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, or: greater number of strips improves accuracy so becomes less | B1 [2] | |
## Part (iii)
| Answer | 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 $y = \overline { 1 + x ^ { 2 } }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ce44db53-2ec8-497b-a1d5-a8adf85e3929-2_577_941_549_636}
\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 Q2 [8]}}