| 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 requiring routine techniques: calculating a function value (simple substitution), applying the trapezium rule formula (standard procedure), error analysis (conceptual understanding of concavity), and volume of revolution (standard integration). While part (iii) requires recognizing that the integral simplifies nicely, all parts are textbook-standard with no novel problem-solving required. |
| 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 |
| \(A \approx 0.25/2\{1 + 1.4142 + 2(1.0308 + 1.1180 + 1.25)\}\) | M1 | \((0.125 \times 9.2118)\) |
| \(= 0.25 \times 4.6059 = 1.151475\) | ||
| \(= 1.151\) (3 d.p.) | E1 | Need evidence |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area is an over-estimate. 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 | Exact |
| [3] |
## Question 4(i):
| Answer | 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 | $(0.125 \times 9.2118)$ |
| $= 0.25 \times 4.6059 = 1.151475$ | | |
| $= 1.151$ (3 d.p.) | E1 | Need evidence |
| **[3]** | | |
---
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area is an over-estimate. 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]** | | |
---
## Question 4(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 | Exact |
| **[3]** | | |4 Fig. 2 shows the curve $y = \sqrt { 1 + x ^ { 2 } }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{252453c9-9afa-435c-b64b-5ea37ec69eed-4_572_939_551_638}
\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 Q4 [8]}}