AQA C3 2013 January — Question 2 9 marks

Exam BoardAQA
ModuleC3 (Core Mathematics 3)
Year2013
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndefinite & Definite Integrals
TypeShow definite integral equals value
DifficultyModerate -0.3 Part (a) is a routine Simpson's rule application with straightforward function evaluation. Part (b) requires recognizing the standard integral form ∫f'(x)/f(x)dx = ln|f(x)| with u = x² + 2, which is a common C3 technique. The 'show that' format and finding k adds minimal challenge beyond standard integration by substitution.
Spec1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.09f Trapezium rule: numerical integration

2
  1. Use Simpson's rule, with five ordinates (four strips), to calculate an estimate for $$\int _ { 0 } ^ { 4 } \frac { x } { x ^ { 2 } + 2 } \mathrm {~d} x$$ Give your answer to four significant figures.
  2. Show that the exact value of \(\int _ { 0 } ^ { 4 } \frac { x } { x ^ { 2 } + 2 } \mathrm {~d} x\) is \(\ln k\), where \(k\) is an integer. (5 marks)

Question 2:
Part (a)
AnswerMarks Guidance
AnswerMarks Guidance
\(h = 1\), ordinates at \(x = 0, 1, 2, 3, 4\)B1 Correct strip width
\(y_0 = 0,\ y_1 = \frac{1}{3},\ y_2 = \frac{2}{6} = \frac{1}{3},\ y_3 = \frac{3}{11},\ y_4 = \frac{4}{18} = \frac{2}{9}\)B1 All ordinates correct
\(I \approx \frac{1}{3}[y_0 + 4y_1 + 2y_2 + 4y_3 + y_4]\)M1 Correct Simpson's rule structure
\(= \frac{1}{3}\left[0 + \frac{4}{3} + \frac{2}{6} + \frac{12}{11} + \frac{2}{9}\right]\)
\(\approx 0.6870\)A1 Answer to 4 significant figures
Part (b)
AnswerMarks Guidance
AnswerMarks Guidance
\(\int_0^4 \frac{x}{x^2+2}\,dx = \left[\frac{1}{2}\ln(x^2+2)\right]_0^4\)M1 A1 Recognise form; correct integration
\(= \frac{1}{2}\ln 18 - \frac{1}{2}\ln 2\)M1 Apply limits
\(= \frac{1}{2}\ln 9 = \ln 3\)A1 A1 Simplify to \(\ln k\); \(k = 3\)
# Question 2:

## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $h = 1$, ordinates at $x = 0, 1, 2, 3, 4$ | B1 | Correct strip width |
| $y_0 = 0,\ y_1 = \frac{1}{3},\ y_2 = \frac{2}{6} = \frac{1}{3},\ y_3 = \frac{3}{11},\ y_4 = \frac{4}{18} = \frac{2}{9}$ | B1 | All ordinates correct |
| $I \approx \frac{1}{3}[y_0 + 4y_1 + 2y_2 + 4y_3 + y_4]$ | M1 | Correct Simpson's rule structure |
| $= \frac{1}{3}\left[0 + \frac{4}{3} + \frac{2}{6} + \frac{12}{11} + \frac{2}{9}\right]$ | | |
| $\approx 0.6870$ | A1 | Answer to 4 significant figures |

## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^4 \frac{x}{x^2+2}\,dx = \left[\frac{1}{2}\ln(x^2+2)\right]_0^4$ | M1 A1 | Recognise form; correct integration |
| $= \frac{1}{2}\ln 18 - \frac{1}{2}\ln 2$ | M1 | Apply limits |
| $= \frac{1}{2}\ln 9 = \ln 3$ | A1 A1 | Simplify to $\ln k$; $k = 3$ |

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2
\begin{enumerate}[label=(\alph*)]
\item Use Simpson's rule, with five ordinates (four strips), to calculate an estimate for

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

Give your answer to four significant figures.
\item Show that the exact value of $\int _ { 0 } ^ { 4 } \frac { x } { x ^ { 2 } + 2 } \mathrm {~d} x$ is $\ln k$, where $k$ is an integer. (5 marks)
\end{enumerate}

\hfill \mbox{\textit{AQA C3 2013 Q2 [9]}}