| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2009 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Trapezium rule with stated number of strips |
| Difficulty | Moderate -0.3 This is a straightforward application of the trapezium rule with clearly specified ordinates and strip width. Part (a) requires careful arithmetic with a moderately complex function but follows a standard algorithm. Part (b) is routine recall (use more strips/ordinates). Slightly easier than average due to being purely procedural with no conceptual challenges. |
| Spec | 1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Integral} = h/2 \{\ldots\}\) | B1 | PI |
| \(\{\ldots\} = f(1.5) + 2[f(3) + f(4.5)] + f(6)\) | M1 | For the M1 covered range must be 1.5 to 6 OE summing of areas of the three traps. |
| \(\{\ldots\} = 2.51(5\ldots) + 2[25.4(5\ldots) + 88.8(4\ldots)] + 212(9\ldots)\) | A1 | Check at least 3sf values, rounded or truncated, or award if a combined value WRT 444 is seen or final answer is 333 or rounds to 333. Condone one numerical slip. |
| \(\text{Integral} = 0.75 \times 444.1 = 333 \text{ to 3sf}\) | A1cao | Must have 333. 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Increase the number of ordinates | E1 | OE eg increase the number of strips. 1 mark total |
**2(a)**
$h = 1.5$
$f(x) = x^2\sqrt{x^2-1}$
$\text{Integral} = h/2 \{\ldots\}$ | B1 | PI
$\{\ldots\} = f(1.5) + 2[f(3) + f(4.5)] + f(6)$ | M1 | For the M1 covered range must be 1.5 to 6 OE summing of areas of the three traps.
$\{\ldots\} = 2.51(5\ldots) + 2[25.4(5\ldots) + 88.8(4\ldots)] + 212(9\ldots)$ | A1 | Check at least 3sf values, rounded or truncated, or award if a combined value WRT 444 is seen or final answer is 333 or rounds to 333. Condone one numerical slip.
$\text{Integral} = 0.75 \times 444.1 = 333 \text{ to 3sf}$ | A1cao | Must have 333. 4 marks total
Treat using 4 strips as a MR and mark with max of B0M1A1A1cao as follows:
$h = 1.125$ B0
$\{\ldots\} = f(1.5) + 2[f(2.625) + f(3.75) + f(4.875)] + f(6)$ M1
$= 2.51(5) + 2[16.7(2) + 50.8(2) + 113.3)] + 212(9)$ A1
or award if a combined value WRT 577 is seen or final answer is 325 or rounds to 325. Condone one numerical slip.
Answer = 325 A1cao Must have 325
**2(b)**
Increase the number of ordinates | E1 | OE eg increase the number of strips. 1 mark total
**Total for Q2: 5 marks**
---2
\begin{enumerate}[label=(\alph*)]
\item Use the trapezium rule with four ordinates (three strips) to find an approximate value for
$$\int _ { 1.5 } ^ { 6 } x ^ { 2 } \sqrt { x ^ { 2 } - 1 } \mathrm {~d} x$$
giving your answer to three significant figures.
\item State how you could obtain a better approximation to the value of the integral using the trapezium rule.\\
(1 mark)
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2009 Q2 [5]}}