Question 1 - A-Level Maths

AQA C3 2014 June — Question 1 4 marks

Exam Board	AQA
Module	C3 (Core Mathematics 3)
Year	2014
Session	June
Marks	4
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Numerical integration
Type	Simpson's rule application
Difficulty	Moderate -0.8 This is a straightforward application of Simpson's rule with clearly specified parameters (five ordinates, four strips). Students only need to recall the formula, calculate ordinates at x = 0, π/4, π/2, 3π/4, π, and substitute into the standard Simpson's rule expression. No problem-solving or conceptual insight required beyond routine procedural execution.
Spec	1.09f Trapezium rule: numerical integration

1 Use Simpson's rule, with five ordinates (four strips), to calculate an estimate for $$\int _ { 0 } ^ { \pi } x ^ { \frac { 1 } { 2 } } \sin x d x$$ Give your answer to four significant figures.
[0pt] [4 marks]

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Question 1:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$h = \frac{\pi}{4}$	B1	Correct strip width stated or implied
Five ordinates: $x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi$
$y_0 = 0^{\frac{1}{2}}\sin 0 = 0$
$y_1 = \left(\frac{\pi}{4}\right)^{\frac{1}{2}}\sin\frac{\pi}{4} = 0.62573...$
$y_2 = \left(\frac{\pi}{2}\right)^{\frac{1}{2}}\sin\frac{\pi}{2} = 1.25331...$
$y_3 = \left(\frac{3\pi}{4}\right)^{\frac{1}{2}}\sin\frac{3\pi}{4} = 1.08948...$
$y_4 = \pi^{\frac{1}{2}}\sin\pi = 0$
Correct use of Simpson's Rule: $\frac{h}{3}[y_0 + y_4 + 4(y_1 + y_3) + 2(y_2)]$	M1	Correct Simpson's rule structure with their values
$= \frac{\pi}{12}[0 + 0 + 4(0.62573 + 1.08948) + 2(1.25331)]$	A1	Correct values used
$= \frac{\pi}{12}[6.90084 + 2.50663]$
$= \frac{\pi}{12}[9.40747...]$
$= 2.463$ (4 s.f.)	A1	Answer to 4 significant figures

# Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $h = \frac{\pi}{4}$ | B1 | Correct strip width stated or implied |
| Five ordinates: $x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi$ | | |
| $y_0 = 0^{\frac{1}{2}}\sin 0 = 0$ | | |
| $y_1 = \left(\frac{\pi}{4}\right)^{\frac{1}{2}}\sin\frac{\pi}{4} = 0.62573...$ | | |
| $y_2 = \left(\frac{\pi}{2}\right)^{\frac{1}{2}}\sin\frac{\pi}{2} = 1.25331...$ | | |
| $y_3 = \left(\frac{3\pi}{4}\right)^{\frac{1}{2}}\sin\frac{3\pi}{4} = 1.08948...$ | | |
| $y_4 = \pi^{\frac{1}{2}}\sin\pi = 0$ | | |
| Correct use of Simpson's Rule: $\frac{h}{3}[y_0 + y_4 + 4(y_1 + y_3) + 2(y_2)]$ | M1 | Correct Simpson's rule structure with their values |
| $= \frac{\pi}{12}[0 + 0 + 4(0.62573 + 1.08948) + 2(1.25331)]$ | A1 | Correct values used |
| $= \frac{\pi}{12}[6.90084 + 2.50663]$ | | |
| $= \frac{\pi}{12}[9.40747...]$ | | |
| $= 2.463$ (4 s.f.) | A1 | Answer to 4 significant figures |

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1 Use Simpson's rule, with five ordinates (four strips), to calculate an estimate for

$$\int _ { 0 } ^ { \pi } x ^ { \frac { 1 } { 2 } } \sin x d x$$

Give your answer to four significant figures.\\[0pt]
[4 marks]

\hfill \mbox{\textit{AQA C3 2014 Q1 [4]}}

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Q1 4 Q2 12 Q3 10 Q4 11 Q5 11 Q6 9 Q7 6 Q8 12