| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Trapezium Rule Approximation with Area |
| Difficulty | Moderate -0.3 This is a structured multi-part question requiring standard C2 techniques: basic area of triangle, routine integration of √x, inequality reasoning from a diagram, and trapezium rule application. While it has 5 parts and requires understanding of why trapezium rule underestimates, each individual step is straightforward with no novel problem-solving required. Slightly easier than average due to heavy scaffolding and standard methods. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{1}{2}\) | B1 | Total: 1 mark |
| (ii) \(\int_0^1 x^2 \, dx = \left[\frac{2}{3}x^{\frac{3}{2}}\right]_0^1\) | M1, A1 | |
| \(= \frac{2}{3}\) | E1 | Total: 3 marks |
| (iii) \(\frac{1}{2} < A < \frac{2}{3}\) | B1 | Total: 1 mark |
| (iv) Values of \(y_0, y_1, y_2, y_3, y_4\) are \(0, 0.39528, 0.61237, 0.81009, 1\) | B1 | |
| Area \(= \frac{1}{2} \times 0.25 \times \left(0 + 2(0.39528 + 0.61237 + 0.81009) + 1\right)\) | M1, A1, A1 | |
| \(= 0.57874\) | Total: 4 marks | |
| (v) Diagram showing: The region lies between the triangle and the upper curve. The region is below the middle curve. | B1, B1, B1 | Total: 3 marks |
**(i)** $\frac{1}{2}$ | B1 | **Total: 1 mark**
**(ii)** $\int_0^1 x^2 \, dx = \left[\frac{2}{3}x^{\frac{3}{2}}\right]_0^1$ | M1, A1 |
$= \frac{2}{3}$ | E1 | **Total: 3 marks**
**(iii)** $\frac{1}{2} < A < \frac{2}{3}$ | B1 | **Total: 1 mark**
**(iv)** Values of $y_0, y_1, y_2, y_3, y_4$ are $0, 0.39528, 0.61237, 0.81009, 1$ | B1 |
Area $= \frac{1}{2} \times 0.25 \times \left(0 + 2(0.39528 + 0.61237 + 0.81009) + 1\right)$ | M1, A1, A1 |
$= 0.57874$ | | **Total: 4 marks**
**(v)** Diagram showing: The region lies between the triangle and the upper curve. The region is below the middle curve. | B1, B1, B1 | **Total: 3 marks**9 Fig. 9 shows\\
$P \quad$ The line $y = x$\\
$Q$ The curve $y = \sqrt { \frac { 1 } { 2 } \left( x + x ^ { 2 } \right) }$\\
$R \quad$ The curve $\quad y = \sqrt { x }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1c52d6b5-84b4-455a-9620-c377ae457069-4_471_1103_762_374}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{figure}
(i) Write down the area of the triangle formed by the line $y = x$, the line $x = 1$ and the $x$-axis.\\
(ii) Show that the area of the region formed by the curve $y = \sqrt { x }$, the line $x = 1$ and the $x$-axis is $\frac { 2 } { 3 }$.
An estimate is required of the Area, $A$, of the region formed by the curve $y = \sqrt { \frac { 1 } { 2 } \left( x + x ^ { 2 } \right) }$, the line $x = 1$ and the $x$-axis.\\
(iii) Use results to parts (i) and (ii) to complete the statement
$$\ldots \ldots \ldots \ldots . . < A < \ldots \ldots \ldots \ldots \ldots . .$$
(iv) Use the Trapezium Rule with 4 strips to find an estimate for $A$.\\
(v) Draw a sketch of Fig. 9. Use it to illustrate the area found as the trapezium rule estimate for $A$.\\
Explain how your diagram shows that the trapezium rule estimate must be:\\
consistent with the answer to part (iv);\\
an under-estimate for A .
\hfill \mbox{\textit{OCR MEI C2 Q9 [12]}}