| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Trapezium rule for applications |
| Difficulty | Moderate -0.8 This is a straightforward application of two standard C2 techniques: trapezium rule (routine calculation with given values) and polynomial integration (direct application of power rule). Both parts are multi-step but require no problem-solving insight—just methodical execution of learned procedures. The context is realistic but doesn't add conceptual difficulty. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(= (-)0.136\) seen [m²] www | 4 | M3 for \(0.1/2 \times (0.14 + 0.16 + 2[0.22 + 0.31 + 0.36 + 0.32])\) M2 for one slip; M1 for two slips must be positive |
| Volume \(= 0.34\) [m³] or f.t from their area \(\times 2.5\) | 1 | |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(2x^4 - x^3 - 0.25x^2 - 0.15x\) o.e. | M2 | M1 for 2 terms correct dep on integral attempted must have neg sign |
| value at 0.5 [– value at 0] \(= –0.1375\) | M1 | |
| area of cross section (of trough) or area between curve and \(x\)-axis \(0.34375\) r.o.t. to 3 or more sf [m³] m² seen in (i) or (ii) | A1 | |
| E1 | ||
| B1 | ||
| U1 | ||
| 7 |
### Part i
Area $= (-)0.136$ seen [m²] www | 4 | M3 for $0.1/2 \times (0.14 + 0.16 + 2[0.22 + 0.31 + 0.36 + 0.32])$ M2 for one slip; M1 for two slips must be positive
Volume $= 0.34$ [m³] or f.t from their area $\times 2.5$ | 1 |
| 5 |
### Part ii
$2x^4 - x^3 - 0.25x^2 - 0.15x$ o.e. | M2 | M1 for 2 terms correct dep on integral attempted must have neg sign
value at 0.5 [– value at 0] $= –0.1375$ | M1 |
area of cross section (of trough) or area between curve and $x$-axis $0.34375$ r.o.t. to 3 or more sf [m³] m² seen in (i) or (ii) | A1 |
| E1 |
| B1 |
| U1 |
| 7 |\includegraphics{figure_12}
A water trough is a prism 2.5 m long. Fig. 12 shows the cross-section of the trough, with the depths in metres at 0.1 m intervals across the trough. The trough is full of water.
\begin{enumerate}[label=(\roman*)]
\item Use the trapezium rule with 5 strips to calculate an estimate of the area of cross-section of the trough.
Hence estimate the volume of water in the trough. [5]
\item A computer program models the curve of the base of the trough, with axes as shown and units in metres, using the equation $y = 8x^3 - 3x^2 - 0.5x - 0.15$, for $0 \leq x \leq 0.5$.
Calculate $\int_0^{0.5} (8x^3 - 3x^2 - 0.5x - 0.15) \, \text{d}x$ and state what this represents.
Hence find the volume of water in the trough as given by this model. [7]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 2008 Q12 [12]}}