Question 13
The sketch shows sections of the functions $f(x)=-0.5x^2+7.5x-18$ and $g(x)=4\,\text{cosec}\!\left(\dfrac{\pi x}{24}\right)-4$.
Two points of intersection at $(12,0)$ and point A are shown.
Determine the coordinates of point A.
[1 mark]Consider the shaded bounded region between the functions.
Determine an approximate area of this region using Simpson’s rule with four intervals. Show evidence of the values substituted into this rule in your solution.
[3 marks] c)State a definite integral that represents the area of the shaded bounded region.
[1 mark] d)Determine the value of your result from Question 13c). Give your answer to two decimal places.
[1 mark] e)Other than working to more decimal places, state a strategy involving Simpson’s rule that could be used to improve the accuracy of your result from Question 13b).
[1 mark]Simpson’s rule only needs five heights
Watch the region form between the curves. Each orange line is a gap between the curves, and those five gaps are all you substitute.
The heights at both ends are zero because the curves meet there. Those two terms drop out of the sum.
Add intervals and watch the estimate close in
The black curve is the gap $f(x)-g(x)$. Simpson’s rule fits a parabola through each set of three points, drawn dashed in orange. Move the slider to change the number of intervals.
This is the rule in part b), with four intervals. The two parabolas already sit very close to the curve.
With two intervals, one parabola has to cover the whole region, so the estimate is furthest from the exact area.
More intervals make $w$ smaller, so each parabola covers less of the curve and fits it more closely. This is the strategy part e) is asking for.
The slider moves in twos because Simpson’s rule needs an even number of intervals.
Work through the solution one mark at a time
Try each step yourself before you reveal it. Part b) follows the QCAA’s first method, which works with the gap between the curves.
Prefer to watch? Watch me work through it on YouTube (opens in a new tab)
Solve $f(x)=g(x)$ on your calculator, or find the intersection of the two graphs. Take the solution that isn’t $x=12$.
Some graphics calculators have no cosec function. Since $\text{cosec}\,\theta=\dfrac{1}{\sin\theta}$, enter $g$ as $\dfrac{4}{\sin\left(\frac{\pi x}{24}\right)}-4$ instead, with your calculator in radians.
You can check it: $f(4)=-8+30-18=4$ and $g(4)=\dfrac{4}{\sin\frac{\pi}{6}}-4=4$.
Marker: correctly determines A
Four intervals from $x=4$ to $x=12$ gives $w=\frac{12-4}{4}=2$. The region is between the curves, with $f$ on top, so tabulate $f(x)-g(x)$.
$$\begin{array}{c|ccccc}x&4&6&8&10&12\\\hline f(x)-g(x)&0&7.34&9.38&6.86&0\end{array}$$
This mark can be implied by your substitution in the next step.
Marker: correctly recognises the requirement to determine values of $f(x)-g(x)$
The question asks for evidence of the values substituted, so write the rule out with your numbers in it.
$$\begin{aligned}\text{Area}&\approx\frac{w}{3}\Big[y_0+4y_1+2y_2+4y_3+y_4\Big]\\&\approx\frac{2}{3}\Big[0+4(7.34)+2(9.38)+4(6.86)+0\Big]\end{aligned}$$
Marker: shows evidence of use of Simpson’s rule
Evaluate.
$$\frac{2}{3}\times75.56\approx50.37$$
With unrounded heights you get 50.38, which is also fine. The QCAA’s second method uses Simpson’s rule on $f$ and $g$ separately, then subtracts: $58.67-8.29\approx50.37$. The area must be positive.
Marker: determines required approximate area
Top curve minus bottom curve, from A to $(12,0)$. Include the $dx$.
Writing $\displaystyle\int_4^{12}\big(f(x)-g(x)\big)\,dx$ is also accepted.
Marker: correctly determines a definite integral, including the $dx$ term, that represents the exact area
Evaluate the integral on your calculator and round to two decimal places, as the question asks. If your calculator has no cosec function, type the cosec term as $\dfrac{4}{\sin\left(\frac{\pi x}{24}\right)}$ here too.
Marker: determines rounded value of exact area
The question rules out more decimal places, so the strategy has to change the rule itself.
Marker: correctly states a suitable strategy
Each phrase of the question gave you something
Simpson’s rule with a given number of intervals is a routine you have practised, and the calculator does the intersection and the exact integral for you. Each part points you to the next.
If I had to guess where you might lose marks, it would be using the values of $f$ alone in part b). That finds the area under the parabola, about 58.67, rather than the area between the curves. In part c), a missing $dx$ costs the mark. In part e), “use a smaller interval” only works if you say the number of intervals stays even.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2025 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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