AQA C3 2006 January — Question 2 4 marks

Exam BoardAQA
ModuleC3 (Core Mathematics 3)
Year2006
SessionJanuary
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNumerical integration
TypeSimpson's rule application
DifficultyModerate -0.5 This is a straightforward application of Simpson's rule with clearly specified parameters (5 ordinates, 4 strips). The function evaluation requires only calculator work with no algebraic manipulation. While it tests understanding of the Simpson's rule formula, it's a routine procedural question slightly easier than average due to its mechanical nature and lack of problem-solving elements.
Spec1.09f Trapezium rule: numerical integration
2 Use Simpson's rule with 5 ordinates (4 strips) to find an approximation to $$\int _ { 1 } ^ { 3 } \frac { 1 } { \sqrt { 1 + x ^ { 3 } } } \mathrm {~d} x$$ giving your answer to three significant figures.
Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(x=1\): \(y=0.707(1)\); \(x=1.5\): \(y=0.478(1)\); \(x=2\): \(y=0.333(3)\); \(x=2.5\): \(y=0.245(3)\); \(x=3\): \(y=0.189(0)\)B1 3 correct — SC B1 for all correct expressions but wrongly evaluated
All correctB1
\(A = \frac{1}{3} \times 0.5 \left[ y(1)+y(3) + 4(y(1.5)+y(2.5))+2(y(2)) \right]\)M1 Use of Simpson's rule
\(= 0.743\)A1 
## Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $x=1$: $y=0.707(1)$; $x=1.5$: $y=0.478(1)$; $x=2$: $y=0.333(3)$; $x=2.5$: $y=0.245(3)$; $x=3$: $y=0.189(0)$ | B1 | 3 correct — SC B1 for all correct expressions but wrongly evaluated |
| All correct | B1 | |
| $A = \frac{1}{3} \times 0.5 \left[ y(1)+y(3) + 4(y(1.5)+y(2.5))+2(y(2)) \right]$ | M1 | Use of Simpson's rule |
| $= 0.743$ | A1 | |

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2 Use Simpson's rule with 5 ordinates (4 strips) to find an approximation to

$$\int _ { 1 } ^ { 3 } \frac { 1 } { \sqrt { 1 + x ^ { 3 } } } \mathrm {~d} x$$

giving your answer to three significant figures.

\hfill \mbox{\textit{AQA C3 2006 Q2 [4]}}
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Q1 5 Q2 4 Q3 10 Q4 7 Q5 12 Q6 12 Q7 5 Q8 10 Q9 14