AQA C2 2012 January — Question 2 5 marks

Exam BoardAQA
ModuleC2 (Core Mathematics 2)
Year2012
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNumerical integration
TypeTrapezium rule with stated number of strips
DifficultyModerate -0.8 This is a straightforward application of the trapezium rule with clearly specified ordinates and strip width. Part (a) requires only substitution into the standard formula with basic calculator work, while part (b) tests recall of how to improve trapezium rule accuracy (use more strips). No problem-solving or conceptual insight needed—purely procedural execution of a standard numerical method.
Spec1.09f Trapezium rule: numerical integration

2
  1. Use the trapezium rule with five ordinates (four strips) to find an approximate value for $$\int _ { 0 } ^ { 4 } \frac { 2 ^ { x } } { x + 1 } \mathrm {~d} x$$ giving your answer to three significant figures.
  2. State how you could obtain a better approximation to the value of the integral using the trapezium rule.

Question 2:
Part (a):
AnswerMarks Guidance
WorkingMark Guidance
\(h = 1\)B1 \(h=1\) stated or used (PI by x-values 0,1,2,3,4 provided no contradiction)
\(f(x) = \frac{2^x}{x+1}\); \(I \approx \frac{h}{2}\{f(0)+f(4)+2[f(1)+f(2)+f(3)]\}\)M1 OE summing of areas of the 'trapezia'
\(= 1 + \frac{16}{5} + 2\left(\frac{2}{2}+\frac{4}{3}+\frac{8}{4}\right) = 1+3.2+2(1+1.33\ldots+2)\)A1 OE Accept 1dp evidence. Can be implied by later correct work provided >1 term or a single term which rounds to 6.43
\((I \approx) 0.5[4.2+2\times4.333..] = 6.43\) (to 3sf)A1 (Total: 4) CAO Must be 6.43
Part (b):
AnswerMarks Guidance
WorkingMark Guidance
Increase the number of ordinatesE1 (Total: 1) OE e.g. increase the number of strips
## Question 2:

**Part (a):**

| Working | Mark | Guidance |
|---------|------|----------|
| $h = 1$ | B1 | $h=1$ stated or used (PI by x-values 0,1,2,3,4 provided no contradiction) |
| $f(x) = \frac{2^x}{x+1}$; $I \approx \frac{h}{2}\{f(0)+f(4)+2[f(1)+f(2)+f(3)]\}$ | M1 | OE summing of areas of the 'trapezia' |
| $= 1 + \frac{16}{5} + 2\left(\frac{2}{2}+\frac{4}{3}+\frac{8}{4}\right) = 1+3.2+2(1+1.33\ldots+2)$ | A1 | OE Accept 1dp evidence. Can be implied by later correct work provided >1 term or a single term which rounds to 6.43 |
| $(I \approx) 0.5[4.2+2\times4.333..] = 6.43$ (to 3sf) | A1 (Total: 4) | CAO Must be 6.43 |

**Part (b):**

| Working | Mark | Guidance |
|---------|------|----------|
| Increase the number of ordinates | E1 (Total: 1) | OE e.g. increase the number of strips |

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2
\begin{enumerate}[label=(\alph*)]
\item Use the trapezium rule with five ordinates (four strips) to find an approximate value for

$$\int _ { 0 } ^ { 4 } \frac { 2 ^ { x } } { x + 1 } \mathrm {~d} x$$

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.
\end{enumerate}

\hfill \mbox{\textit{AQA C2 2012 Q2 [5]}}