| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Rectangle bounds for definite integral |
| Difficulty | Moderate -0.8 This is a straightforward application question requiring standard techniques: finding upper bounds using rectangles (mirror of given lower bound method), applying the trapezium rule formula with given ordinates, and multiplying by length for volume. All steps are routine with no conceptual challenges beyond basic understanding of numerical integration. |
| Spec | 1.08d Evaluate definite integrals: between limits1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Finding at least 2 distinct values for \(h\) (table of values not required): \(x = 0, \pm0.5, \pm1, \pm1.5, 2\) giving \(h = \frac{1}{5}, \frac{4}{25}, \frac{1}{10}, \frac{4}{65}, \frac{1}{25}\), i.e. \(h = 0.2, 0.16, 0.1, 0.061538, 0.04\) | M1 | Finding at least 2 distinct values for \(h\). Need not be a table of values |
| \(= 2 \times 0.5(0.2 + 0.16 + 0.1 + 0.061538)\) | M1 | Using rectangles forming UB for area with width 0.5. Need not be drawn. Allow both M marks if only half the area considered |
| \(= 0.522\) to 3sf | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area \(\approx \frac{0.5}{2}(0.2 + 0.04 + 2(0.16 + 0.1 + 0.061538))\) | M1 | Using trapezium rule with 5 \(h\) values. FT incorrect \(h\) values for the M mark. Allow awrt 0.22. Using the definite integral functionality of calculator gives 0.2214297… so method must be seen |
| \(= 0.221\) (to 3 sf) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Volume \(= 2 \times (0.2208 \times 10)\) | M1 | Condone missing factor of 2 for method mark. FT part (b) |
| \(= 4.42\) m\(^3\) | A1 [2] |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Finding at least 2 distinct values for $h$ (table of values not required): $x = 0, \pm0.5, \pm1, \pm1.5, 2$ giving $h = \frac{1}{5}, \frac{4}{25}, \frac{1}{10}, \frac{4}{65}, \frac{1}{25}$, i.e. $h = 0.2, 0.16, 0.1, 0.061538, 0.04$ | M1 | Finding at least 2 distinct values for $h$. Need not be a table of values |
| $= 2 \times 0.5(0.2 + 0.16 + 0.1 + 0.061538)$ | M1 | Using rectangles forming UB for area with width 0.5. Need not be drawn. Allow both M marks if only half the area considered |
| $= 0.522$ to 3sf | A1 [3] | |
---
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area $\approx \frac{0.5}{2}(0.2 + 0.04 + 2(0.16 + 0.1 + 0.061538))$ | M1 | Using trapezium rule with 5 $h$ values. FT incorrect $h$ values for the M mark. Allow awrt 0.22. Using the definite integral functionality of calculator gives 0.2214297… so method must be seen |
| $= 0.221$ (to 3 sf) | A1 [2] | |
---
## Question 8(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Volume $= 2 \times (0.2208 \times 10)$ | M1 | Condone missing factor of 2 for method mark. FT part (b) |
| $= 4.42$ m$^3$ | A1 [2] | |
---8 Fig. 8.1 shows the cross-section of a straight driveway 4 m wide made from tarmac.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7de77679-59c0-4431-a9cb-6ab11d2f9062-06_139_1135_1027_248}
\captionsetup{labelformat=empty}
\caption{Fig. 8.1}
\end{center}
\end{figure}
The height $h \mathrm {~m}$ of the cross-section at a displacement $x \mathrm {~m}$ from the middle is modelled by $\mathrm { h } = \frac { 0.2 } { 1 + \mathrm { x } ^ { 2 } }$ for $- 2 \leqslant x \leqslant 2$.
A lower bound of $0.3615 \mathrm {~m} ^ { 2 }$ is found for the area of the cross-section using rectangles as shown in Fig. 8.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7de77679-59c0-4431-a9cb-6ab11d2f9062-06_266_1276_1594_248}
\captionsetup{labelformat=empty}
\caption{Fig. 8.2}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Use a similar method to find an upper bound for the area of the cross-section.
\item Use the trapezium rule with 4 strips to estimate $\int _ { 0 } ^ { 2 } \frac { 0.2 } { 1 + x ^ { 2 } } d x$.
\item The driveway is 10 m long. Use your answer in part (b) to find an estimate of the volume of tarmac needed to make the driveway.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2020 Q8 [7]}}