AQA C3 2007 January — Question 1 4 marks

Exam BoardAQA
ModuleC3 (Core Mathematics 3)
Year2007
SessionJanuary
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNumerical integration
TypeTrapezium rule with stated number of strips
DifficultyModerate -0.8 This is a straightforward application of the mid-ordinate rule with clear parameters (four strips, given interval). It requires only mechanical substitution into the formula and calculator work—no problem-solving, conceptual understanding of integration, or manipulation of the integrand is needed. Easier than average A-level questions which typically require some mathematical reasoning.
Spec1.09f Trapezium rule: numerical integration
1 Use the mid-ordinate rule with four strips of equal width to find an estimate for \(\int _ { 1 } ^ { 5 } \frac { 1 } { 1 + \ln x } \mathrm {~d} x\), giving your answer to three significant figures.
(4 marks)
Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(x = 1.5, 2.5, 3.5, 4.5\)M1 Method
A1\(x\) values
\(y_1 = 0.7115 \approx 0.712\), \(y_2 = 0.5218 \approx 0.522\), \(y_3 = 0.4439 \approx 0.444\), \(y_4 = 0.3993 \approx 0.399\)A1 (AWRT) 3 correct \(y\)'s
\(A = 1 \times (y_1 + y_2 + y_3 + y_4) = 2.08\)A1
Total: 4 marks
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = 1.5, 2.5, 3.5, 4.5$ | M1 | Method |
| | A1 | $x$ values |
| $y_1 = 0.7115 \approx 0.712$, $y_2 = 0.5218 \approx 0.522$, $y_3 = 0.4439 \approx 0.444$, $y_4 = 0.3993 \approx 0.399$ | A1 (AWRT) | 3 correct $y$'s |
| $A = 1 \times (y_1 + y_2 + y_3 + y_4) = 2.08$ | A1 | |

**Total: 4 marks**

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1 Use the mid-ordinate rule with four strips of equal width to find an estimate for $\int _ { 1 } ^ { 5 } \frac { 1 } { 1 + \ln x } \mathrm {~d} x$, giving your answer to three significant figures.\\
(4 marks)

\hfill \mbox{\textit{AQA C3 2007 Q1 [4]}}
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Q1 4 Q2 4 Q3 9 Q4 12 Q5 8 Q6 8 Q7 9 Q8 7 Q9 14