June 2024 Paper 3 Q11

AQACurrent spec10 marksIntegration

11 The curve \(C\) with equation

\[y = \left(x^2 - 8x\right)\ln x\]

is defined for \(x \gt 0\) and is shown in the diagram below.

The curve C rising from O to a small maximum, falling below the x-axis to a minimum, then rising steeply; the region R between the curve and the x-axis, below the x-axis, is shaded

The shaded region, \(R\), lies below the \(x\)-axis and is bounded by \(C\) and the \(x\)-axis.

Show that the area of \(R\) can be written as

\[p + q\ln 2\]

where \(p\) and \(q\) are rational numbers to be found. [10 marks]

Mark scheme

Scheme	Marks	AO
Equates \(\left(x^2 - 8x\right)\ln x\) to zero PI by 1 or 8 or ±108.(2…) May be seen on diagram or integral	M1	3.1a
Obtains at least one of \(x = 1\) or \(x = 8\) PI by ±108.(2…) May be seen on diagram or integral	A1	1.1b
Deduces the limits are 1 and 8 PI by ±108.(2…) May be seen on integral or substituted into their integrated expression	R1	2.2a
States \(u = \ln x\) and \(v^{\prime} = x^2 - 8x\) Condone \(v = \ln x\) and \(u^{\prime} = x^2 - 8x\)	M1	3.1a
Finds \(u^{\prime} = \dfrac{1}{x}\) and \(v = \dfrac{x^3}{3} - 4x^2\)	A1	3.1a
Applies integration by parts formula correctly by substituting their \(u\), \(u^{\prime}\) and \(v\) PI by \(\left(\dfrac{x^3}{3} - 4x^2\right)\ln x - \left(\dfrac{x^3}{9} - 2x^2\right)\) or \(-\left(\dfrac{x^3}{3} + 4x^2\right)\ln x + \left(\dfrac{x^3}{9} - 2x^2\right)\) Condone missing brackets	M1	1.1a
Obtains \(\left(\dfrac{x^3}{3} - 4x^2\right)\ln x - \left(\dfrac{x^3}{9} - 2x^2\right)\) OE or \(-\left(\dfrac{x^3}{3} + 4x^2\right)\ln x + \left(\dfrac{x^3}{9} - 2x^2\right)\) OE	A1	1.1b
Substitutes their non-zero limits correctly into their integrated expression (the subtraction does not need to be seen) or obtains exact values for their integrated expression using their non-zero limits eg \(-\dfrac{256}{3}\ln 8 + \dfrac{640}{9}\) and \(\dfrac{17}{9}\)	M1	1.1a
Obtains \(\dfrac{623}{9} - 256\ln 2\) or \(\dfrac{623}{9} - \dfrac{256}{3}\ln 8\) ACF must be exact form with two terms PI \(-\dfrac{623}{9} + 256\ln 2\)	A1	1.1b
Completes a reasoned argument to obtain \(-\dfrac{623}{9} + 256\ln 2\) To be awarded R1, all marks must be scored	R1	2.1
	(10 marks)

Typical solution

\[\left(x^2 - 8x\right)\ln x = 0\]\[x(x - 8)\ln x = 0\]\[x = 8,\ \ln x = 0 \Rightarrow x = 1\]\[\int_1^8 \left(x^2 - 8x\right)\ln x \,\mathrm{d}x\]\[u = \ln x \qquad u^{\prime} = \frac{1}{x}\]\[v^{\prime} = x^2 - 8x \qquad v = \frac{x^3}{3} - 4x^2\]\[\left(\frac{x^3}{3} - 4x^2\right)\ln x - \int \left(\frac{x^3}{3} - 4x^2\right)\frac{1}{x}\,\mathrm{d}x\]\[= \left(\frac{x^3}{3} - 4x^2\right)\ln x - \left(\frac{x^3}{9} - 2x^2\right)\]\[\int_1^8 \left(x^2 - 8x\right)\ln x \,\mathrm{d}x\]\[= \left[\left(\frac{8^3}{3} - 4 \times 8^2\right)\ln 8 - \left(\frac{8^3}{9} - 2 \times 8^2\right)\right]\]\[- \left[\left(\frac{1^3}{3} - 4 \times 1^2\right)\ln 1 - \left(\frac{1^3}{9} - 2 \times 1^2\right)\right]\]\[= -\frac{256}{3}\ln 8 + \frac{640}{9} - \frac{17}{9}\]\[= \frac{623}{9} - 256\ln 2\]\[\therefore \text{Area} = -\frac{623}{9} + 256\ln 2\]