| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Over/underestimate justification with graph |
| Difficulty | Moderate -0.8 This is a straightforward application of the trapezium rule with clearly specified parameters (4 strips, width 0.5), requiring only substitution into the formula and simple arithmetic. Part (ii) tests basic understanding that the trapezium rule overestimates for concave functions. Both parts are routine C2-level exercises with no problem-solving or novel insight required. |
| Spec | 1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempt at least 4 correct \(y\)-coords, and no others | M1 | Allow rounded or truncated decimals. Allow error in rearrangement e.g. \(\sqrt{x} - \sqrt{3}\). |
| Attempt correct trapezium rule, any \(h\), to find area between \(x=3\) and \(x=5\) | M1 | Correct structure: \(0.5 \times (\text{any } h) \times (\text{first} + \text{last} + 2 \times \text{middles})\). Using just one strip is M0. |
| Use correct \(h\) (soi) for their \(y\)-values — must be at equal intervals | M1 | Must be in attempt at trapezium rule, not Simpson's rule. Allow if one \(y\)-value missing or extra. Using \(h=2\) with only one strip is M0. |
| Obtain \(1.82\), or better | A1 (4) | More accurate solution is \(1.819479\ldots\) Answer only is 0/4. Using integration is 0/4. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State underestimate | B1* | Ignore any reasons given. |
| Convincing reason referring to trapezia being below curve | B1d* (2) | Referring to gaps between curve and trapezia can get B1. Sketch with brief explanation (sketch alone is B0). Trapezia must show clear intention to have top vertices on the curve. Only referring to concave/convex is B0. Can get B1 for 'rate of change of gradient (or second derivative) is negative', but not for 'gradient is decreasing'. |
# Question 3:
## Part (i): $0.5 \times 0.5 \times \left[\sqrt{0} + 2(\sqrt{0.5} + \sqrt{1} + \sqrt{1.5}) + \sqrt{2}\right] = 1.82$
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt at least 4 correct $y$-coords, and no others | M1 | Allow rounded or truncated decimals. Allow error in rearrangement e.g. $\sqrt{x} - \sqrt{3}$. |
| Attempt correct trapezium rule, any $h$, to find area between $x=3$ and $x=5$ | M1 | Correct structure: $0.5 \times (\text{any } h) \times (\text{first} + \text{last} + 2 \times \text{middles})$. Using just one strip is M0. |
| Use correct $h$ (soi) for their $y$-values — must be at equal intervals | M1 | Must be in attempt at trapezium rule, not Simpson's rule. Allow if one $y$-value missing or extra. Using $h=2$ with only one strip is M0. |
| Obtain $1.82$, or better | A1 (4) | More accurate solution is $1.819479\ldots$ Answer only is 0/4. Using integration is 0/4. |
## Part (ii): Underestimate as tops of trapezia are below curve
| Answer/Working | Mark | Guidance |
|---|---|---|
| State underestimate | B1* | Ignore any reasons given. |
| Convincing reason referring to trapezia being below curve | B1d* (2) | Referring to gaps between curve and trapezia can get B1. Sketch with brief explanation (sketch alone is B0). Trapezia must show clear intention to have top vertices on the curve. Only referring to concave/convex is B0. Can get B1 for 'rate of change of gradient (or second derivative) is negative', but not for 'gradient is decreasing'. |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{c52fe7e9-0442-4b3e-b924-2e5e4b3e98f5-02_510_791_991_678}
The diagram shows the curve $y = \sqrt { x - 3 }$.\\
(i) Use the trapezium rule, with 4 strips each of width 0.5 , to find an approximate value for the area of the region bounded by the curve, the $x$-axis and the line $x = 5$. Give your answer correct to 3 significant figures.\\
(ii) State, with a reason, whether this approximation is an underestimate or an overestimate.
\hfill \mbox{\textit{OCR C2 2011 Q3 [6]}}