A2 June 2025 Paper 2 Q16

AQACurrent spec5 marksIntegration

16 The diagram shows a design for a table top.

A shaded region symmetrical about the x-axis, bounded by two vertical lines to the right of the y-axis and by two curves, one above and one below the x-axis, that dip slightly towards the x-axis in the middle

The table top is modelled as being bounded by the lines \(x = 0.5\) and \(x = 1.5\), and the curves defined by \(y^2 = \dfrac{0.27}{2x - x^2}\), where \(x\) and \(y\) are measured in metres.

Find the area of the table top according to this model.

Give your answer in the form \(\dfrac{\pi\sqrt{p}}{q}\) square metres, where \(p\) and \(q\) are integers.

Fully justify your answer. [5 marks]

Mark scheme

Scheme	Marks	AO
Obtains a correct expression for the area. Condone omission of “2 ×”	B1	3.1a
Completes the square for the denominator of the integrand to obtain \(1 - (x \pm 1)^2\)	M1	3.1a
Integrates to obtain an inverse sine function.	M1	1.1a
Obtains \(k\sin^{-1}(x - 1)\) and substitutes in the limits.	M1	1.1a
Completes a reasoned argument to obtain \(\dfrac{\pi\sqrt{3}}{5}\) Condone missing units.	R1	2.1
	(5 marks)

Typical solution

\[A = 2 \times \int_{0.5}^{1.5} \sqrt{\frac{0.27}{2x - x^2}}\,\mathrm{d}x\]\[A = (2 \times \sqrt{0.27})\int_{0.5}^{1.5} \frac{1}{\sqrt{1 - (x - 1)^2}}\,\mathrm{d}x\]\[\begin{aligned} &= \left(\frac{3\sqrt{3}}{5}\right)\left[\sin^{-1}(x - 1)\right]_{0.5}^{1.5} \\ &= \left(\frac{3\sqrt{3}}{5}\right)\left(\sin^{-1}(0.5) - \sin^{-1}(-0.5)\right) \\ &= \left(\frac{3\sqrt{3}}{5}\right)\left(\frac{\pi}{6} - \left(-\frac{\pi}{6}\right)\right) \\ &= \left(\frac{3\sqrt{3}}{5}\right)\left(\frac{\pi}{3}\right) = \frac{\pi\sqrt{3}}{5} \text{ m}^2 \end{aligned}\]