A2 June 2022 Paper 2 Q6

AQACurrent spec3 marksIntegration

6 The diagram below shows part of the graph of \(y = \mathrm{f}(x)\)

The line \(TPQ\) is a tangent to the graph of \(y = \mathrm{f}(x)\) at the point \(P\left(\dfrac{a + b}{2}, \mathrm{f}\left(\dfrac{a + b}{2}\right)\right)\)

The points \(S(a, 0)\) and \(T\) lie on the line \(x = a\)

The points \(Q\) and \(R(b, 0)\) lie on the line \(x = b\)

A curve y = f(x) rising to a maximum; the tangent at P meets the vertical line x = a at T and the vertical line x = b at Q, with S at (a, 0) and R at (b, 0) on the x-axis, and a vertical line from P to the x-axis

Sharon uses the mid-ordinate rule with one strip to estimate the value of the integral \(\displaystyle\int_a^b \mathrm{f}(x)\,\mathrm{d}x\)

By considering the area of the trapezium \(QRST\), state, giving reasons, whether you would expect Sharon’s estimate to be an under-estimate or an over-estimate. [3 marks]

Mark scheme

Scheme	Marks	AO
States that the area of the trapezium is greater than the integral Condone “the area of the trapezium is greater than the curve”.	B1	2.2a
Explains that the area of the trapezium equals Sharon’s estimate/ result of using mid-ordinate rule	E1	2.4
States that Sharon’s estimate is an over-estimate and completes a reasoned argument to explain the required result	R1	2.1
	(3 marks)

Typical solution

The area of the trapezium is greater than the integral.

Area of trapezium \(= (b - a)y_{\frac{1}{2}}\)

The area of the trapezium is the same as the area of the rectangle from use of the mid-ordinate rule.

This is equal to Sharon’s estimate.

The trapezium includes the area represented by the integral, so Sharon’s estimate is an over-estimate.