Mathematics 9709/32 — October/November 2017

Cambridge A-Level · Pure Mathematics 3 · worked solutions for every part, with the mark scheme

questions

marks

105

minutes

Topics Integration · Differentiation · Logarithmic and Exponential Functions · Trigonometry · Differential Equations · Complex Numbers · +3 more

Q1IntegrationFree sample

Fig. 1

The diagram shows a sketch of the curve $y = \frac{3}{\sqrt{9 - x^3}}$ for values of $x$ from $-1.2$ to $1.2$ .

(i)

Use the trapezium rule, with two intervals, to estimate the value of

\int_{-1.2}^{1.2} \frac{3}{\sqrt{9 - x^3}}\,dx,

giving your answer correct to 2 decimal places.

DifficultyMedium-Easy

Worked solution

Approach

Apply the trapezium rule formula $\frac{h}{2}[y_0 + 2y_1 + y_2]$ with $n=2$ intervals. Calculate the width $h$ and the ordinates at $x = -1.2$ , $x = 0$ , and $x = 1.2$ .

Working

The interval is $[-1.2, 1.2]$ with 2 intervals, so the width of each interval is:

h = \frac{1.2 - (-1.2)}{2} = 1.2

The ordinates are evaluated at $x_0 = -1.2$ , $x_1 = 0$ , and $x_2 = 1.2$ :

y_0 = \frac{3}{\sqrt{9 - (-1.2)^3}} = \frac{3}{\sqrt{9 + 1.728}} = \frac{3}{\sqrt{10.728}} \approx 0.915929

y_1 = \frac{3}{\sqrt{9 - 0^3}} = \frac{3}{3} = 1

y_2 = \frac{3}{\sqrt{9 - 1.2^3}} = \frac{3}{\sqrt{9 - 1.728}} = \frac{3}{\sqrt{7.272}} \approx 1.112485

Apply the trapezium rule formula:

\int_{-1.2}^{1.2} \frac{3}{\sqrt{9 - x^3}}\,dx \approx \frac{h}{2}[y_0 + 2y_1 + y_2]

Substitute the values:

\approx \frac{1.2}{2}[0.915929 + 2(1) + 1.112485]

\approx 0.6[0.915929 + 2 + 1.112485]

\approx 0.6[4.028414]

\approx 2.4170484

Rounding to 2 decimal places:

\approx 2.42

Answer

2.42

Final answer

2.42

Detailed explanation

Walkthrough

First, determine the width of each sub-interval. The total range is from $-1.2$ to $1.2$ , which is $2.4$ . With two intervals, the width $h$ is $1.2$ . Next, calculate the $y$ -values (ordinates) at the three $x$ -values that define these intervals: $x = -1.2$ , $x = 0$ , and $x = 1.2$ . Substitute these into the function $y = \frac{3}{\sqrt{9 - x^3}}$ to get $y_0 \approx 0.915929$ , $y_1 = 1$ , and $y_2 \approx 1.112485$ . Finally, plug these into the trapezium rule formula $\frac{h}{2}[y_0 + 2y_1 + y_2]$ to find the approximate area, and round to 2 decimal places.

Key Takeaways

The trapezium rule approximates the area under a curve by dividing it into trapezia.
The formula for $n$ intervals is $\frac{h}{2}[y_0 + 2y_1 + 2y_2 + \dots + y_n]$ , where $h$ is the interval width.
Always keep extra decimal places during intermediate calculations to avoid rounding errors.

Common Mistakes

Using the wrong interval width $h$ . Remember $h = \frac{\text{upper} - \text{lower}}{\text{number of intervals}}$ , not the number of ordinates.
Forgetting to multiply the middle ordinates by 2 in the formula.
Rounding intermediate values too early, which can lead to an incorrect final answer.

Things to Be Careful About

Ensure you evaluate the function correctly, especially with negative values and powers (e.g., $(-1.2)^3 = -1.728$ ).
The question asks for the answer correct to 2 decimal places, so do not round until the very last step.

Techniques used

calculate interval width for trapezium ruleevaluate function at interval endpoints and midpointapply trapezium rule formularound to required decimal places

(ii)

Explain, with reference to the diagram, why the trapezium rule may be expected to give a good approximation to the true value of the integral in this case.

DifficultyEasy

Worked solution

Approach

Examine the shape of the curve in the diagram to determine why the trapezium rule will yield a good approximation.

Working

The trapezium rule works by approximating the curve with straight line segments. If the curve is nearly straight (or has very little curvature) over the intervals, the trapezia will closely match the actual area under the curve.

From the diagram, the curve is slightly increasing and slightly concave upwards. Because the curvature is small and the function is nearly linear over the given interval, the straight edges of the trapezia will lie very close to the actual curve.

Therefore, the trapezium rule is expected to give a good approximation.

Answer

The curve is nearly straight (or has very small curvature) over the interval, so the trapezia closely approximate the area under the curve.

Final answer

The curve is nearly straight (or has very small curvature) over the interval.

Detailed explanation

Walkthrough

The trapezium rule replaces the curved area under the graph with straight-edged trapezia. The accuracy of this method depends on how much the curve deviates from a straight line. By observing the provided diagram, we can see that the curve $y = \frac{3}{\sqrt{9 - x^3}}$ is only slightly increasing and slightly concave upwards between $x = -1.2$ and $x = 1.2$ . This means the curvature is very small, and the curve is almost linear. Because the curve is nearly straight, the straight-line tops of the trapezia will closely follow the actual curve, resulting in a very good approximation of the true integral.

Key Takeaways

The trapezium rule is most accurate when the function being integrated is nearly linear or has very low curvature over the interval.
Visual inspection of the graph can provide insight into the expected accuracy of numerical integration methods.

Common Mistakes

Stating that the curve is 'flat' or 'constant' when it is actually slightly increasing.
Not referencing the diagram or the shape of the curve in the explanation.

Things to Be Careful About

The mark scheme awards a mark for justifying the statement, so the explanation must clearly link the curve's shape (small curvature / nearly straight) to the accuracy of the approximation. Avoid vague statements like 'the graph is nice'.

Techniques used

analyze curve shape from diagramrelate curvature to trapezium rule accuracy

The rest of this paper

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Q7Complex Numbers8M
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Q10Vectors11M

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