C4 June 2006 Q6

EdexcelOld spec13 marksIntegration

Figure 3: sketch of the curve y = (x - 1) ln x for x > 0, touching the x-axis at x = 1 — Figure 3

Figure 3 shows a sketch of the curve with equation \(y = (x - 1)\ln x\), \(x \gt 0\).

(a) Complete the table with the values of \(y\) corresponding to \(x = 1.5\) and \(x = 2.5\).

\(x\)	1	1.5	2	2.5	3
\(y\)	0		\(\ln 2\)		\(2\ln 3\)

(1)

Given that \(I = \displaystyle\int_1^3 (x - 1)\ln x\,\mathrm{d}x\),

(b) use the trapezium rule

(i) with values of \(y\) at \(x = 1\), 2 and 3 to find an approximate value for \(I\) to 4 significant figures,

(ii) with values of \(y\) at \(x = 1\), 1.5, 2, 2.5 and 3 to find another approximate value for \(I\) to 4 significant figures.

(5)

(c) Explain, with reference to Figure 3, why an increase in the number of values improves the accuracy of the approximation. (1)

(d) Show, by integration, that the exact value of \(\displaystyle\int_1^3 (x - 1)\ln x\,\mathrm{d}x\) is \(\frac{3}{2}\ln 3\). (6)

Mark scheme (a)

\(x\)	1	1.5	2	2.5	3
\(y\)	0	\(0.5\ln 1.5\)	\(\ln 2\)	\(1.5\ln 2.5\)	\(2\ln 3\)
or \(y\)	0	0.2027325541…	\(\ln 2\)	1.374436098…	\(2\ln 3\)

Scheme	Marks
Either \(0.5\ln 1.5\) and \(1.5\ln 2.5\) or awrt 0.20 and 1.37 (or mixture of decimals and ln’s)	B1
	(1)

Mark scheme (b)

Scheme	Marks
(i) \(I_1 \approx \dfrac{1}{2} \times 1 \times\) \(\underline{\left\{0 + 2(\ln 2) + 2\ln 3\right\}}\)	M1;
\(= \dfrac{1}{2} \times 3.583518938\ldots = 1.791759\ldots = 1.792\) (4sf)	A1 cao
(ii) \(I_2 \approx \dfrac{1}{2} \times 0.5\ ; \times \underline{\left\{0 + 2(0.5\ln 1.5 + \ln 2 + 1.5\ln 2.5) + 2\ln 3\right\}}\)	B1; M1ft
\(= \dfrac{1}{4} \times 6.737856242\ldots = 1.684464\ldots\)	A1
	(5)

Notes

M1 For structure of trapezium rule \(\{\ldots\ldots\ldots\}\);

A1 1.792

B1 Outside brackets \(\frac{1}{2} \times 0.5\)

M1ft For structure of trapezium rule \(\{\ldots\ldots\ldots\}\);

A1 awrt 1.684

Beware: In part (b) candidate can add up the individual trapezia:
(b)(i) \(I_1 \approx \frac{1}{2}(0 + \ln 2) + \frac{1}{2}(\ln 2 + \ln 3)\)
(ii) \(I_2 \approx \frac{1}{2}\cdot\frac{1}{2}(0 + 0.5\ln 1.5) + \frac{1}{2}\cdot\frac{1}{2}(0.5\ln 1.5 + \ln 2) + \frac{1}{2}\cdot\frac{1}{2}(\ln 2 + 1.5\ln 2.5) + \frac{1}{2}\cdot\frac{1}{2}(1.5\ln 2.5 + 2\ln 3)\)

Scheme	Marks
With increasing ordinates, the line segments at the top of the trapezia are closer to the curve.	B1
	(1)

Mark scheme (d)

Scheme	Marks
\(\left\{\begin{aligned} u &= \ln x \Rightarrow \tfrac{\mathrm{d}u}{\mathrm{d}x} = \tfrac{1}{x} \\ \tfrac{\mathrm{d}v}{\mathrm{d}x} &= x - 1 \Rightarrow v = \tfrac{x^2}{2} - x \end{aligned}\right\}\)
\(I = \left(\dfrac{x^2}{2} - x\right)\ln x - \displaystyle\int \frac{1}{x}\left(\frac{x^2}{2} - x\right)\mathrm{d}x\)	M1 A1
\(= \left(\dfrac{x^2}{2} - x\right)\ln x - \underline{\displaystyle\int \left(\frac{x}{2} - 1\right)\mathrm{d}x}\)
\(= \left(\dfrac{x^2}{2} - x\right)\ln x - \underline{\left(\dfrac{x^2}{4} - x\right)}\quad (+c)\)	M1; A1
\(\therefore I = \left[\left(\dfrac{x^2}{2} - x\right)\ln x - \dfrac{x^2}{4} + x\right]_1^3\)
\(= \left(\frac{3}{2}\ln 3 - \frac{9}{4} + 3\right) - \left(-\frac{1}{2}\ln 1 - \frac{1}{4} + 1\right)\)	ddM1
\(= \frac{3}{2}\ln 3 + \frac{3}{4} + 0 - \frac{3}{4} = \underline{\frac{3}{2}\ln 3}\) AG	A1 cso
	(6)
	(13 marks)

Notes

M1 Use of ‘integration by parts’ formula in the correct direction

A1 Correct expression

M1; An attempt to multiply at least one term through by \(\frac{1}{x}\) and an attempt to … … integrate;

A1 correct integration

ddM1 Substitutes limits of 3 and 1 and subtracts.

A1 cso \(\frac{3}{2}\ln 3\)

Aliter (d) Way 2

\(\displaystyle\int (x - 1)\ln x\,\mathrm{d}x = \int x\ln x\,\mathrm{d}x - \int \ln x\,\mathrm{d}x\)
\(\displaystyle\int x\ln x\,\mathrm{d}x = \frac{x^2}{2}\ln x - \int \frac{x^2}{2}\cdot\left(\frac{1}{x}\right)\mathrm{d}x\)	M1
\(= \dfrac{x^2}{2}\ln x - \dfrac{x^2}{4}\quad (+c)\)	A1
\(\displaystyle\int \ln x\,\mathrm{d}x = x\ln x - \int x\cdot\left(\frac{1}{x}\right)\mathrm{d}x\)	M1
\(= x\ln x - x\quad (+c)\)	A1
\(\therefore \displaystyle\int_1^3 (x - 1)\ln x\,\mathrm{d}x = \left(\tfrac{9}{2}\ln 3 - 2\right) - (3\ln 3 - 2) = \tfrac{3}{2}\ln 3\) AG	ddM1 A1 cso
	[6]

M1 Correct application of ‘by parts’

A1 Correct integration

M1 Correct application of ‘by parts’

A1 Correct integration

ddM1 Substitutes limits of 3 and 1 into both integrands and subtracts.

A1 cso \(\frac{3}{2}\ln 3\)

Aliter (d) Way 3

\(\left\{\begin{aligned} u &= \ln x \Rightarrow \tfrac{\mathrm{d}u}{\mathrm{d}x} = \tfrac{1}{x} \\ \tfrac{\mathrm{d}v}{\mathrm{d}x} &= (x - 1) \Rightarrow v = \tfrac{(x - 1)^2}{2} \end{aligned}\right\}\)
\(I = \dfrac{(x - 1)^2}{2}\ln x - \displaystyle\int \frac{(x - 1)^2}{2x}\,\mathrm{d}x\)	M1 A1
\(= \dfrac{(x - 1)^2}{2}\ln x - \displaystyle\int \frac{x^2 - 2x + 1}{2x}\,\mathrm{d}x\) \(= \dfrac{(x - 1)^2}{2}\ln x - \underline{\displaystyle\int \left(\frac{1}{2}x - 1 + \frac{1}{2x}\right)\mathrm{d}x}\)
\(= \dfrac{(x - 1)^2}{2}\ln x - \underline{\left(\dfrac{x^2}{4} - x + \dfrac{1}{2}\ln x\right)}\quad (+c)\)	M1; A1
\(\therefore I = \left[\dfrac{(x - 1)^2}{2}\ln x - \dfrac{x^2}{4} + x - \dfrac{1}{2}\ln x\right]_1^3\)
\(= \left(2\ln 3 - \frac{9}{4} + 3 - \frac{1}{2}\ln 3\right) - \left(0 - \frac{1}{4} + 1 - 0\right)\)	ddM1
\(= 2\ln 3 - \frac{1}{2}\ln 3 + \frac{3}{4} + \frac{1}{4} - 1 = \underline{\frac{3}{2}\ln 3}\) AG	A1 cso
	[6]

M1 Use of ‘integration by parts’ formula in the correct direction

A1 Correct expression

M1; Candidate multiplies out numerator to obtain three terms… … multiplies at least one term through by \(\frac{1}{x}\) and then attempts to … … integrate the result;

A1 correct integration

ddM1 Substitutes limits of 3 and 1 and subtracts.

A1 cso \(\frac{3}{2}\ln 3\)

Beware: \(\displaystyle\int \frac{1}{2x}\,\mathrm{d}x\) can also integrate to \(\dfrac{1}{2}\ln 2x\)

Beware: If you are marking using WAY 2 please make sure that you allocate the marks in the order they appear on the mark scheme. For example if a candidate only integrated \(\ln x\) correctly then they would be awarded M0A0M1A1M0A0 on ePEN.

Aliter (d) Way 4: By substitution

\(u = \ln x \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{x}\) \(I = \displaystyle\int (\mathrm{e}^u - 1).u\mathrm{e}^u\,\mathrm{d}u\)
\(= \displaystyle\int u(\mathrm{e}^{2u} - \mathrm{e}^u)\,\mathrm{d}u\)	M1
\(= u\left(\dfrac{1}{2}\mathrm{e}^{2u} - \mathrm{e}^u\right) - \underline{\displaystyle\int \left(\frac{1}{2}\mathrm{e}^{2u} - \mathrm{e}^u\right)\mathrm{d}x}\)	A1
\(= u\left(\dfrac{1}{2}\mathrm{e}^{2u} - \mathrm{e}^u\right) - \underline{\left(\dfrac{1}{4}\mathrm{e}^{2u} - \mathrm{e}^u\right)}\quad (+c)\)	M1; A1
\(\therefore I = \left[\dfrac{1}{2}u\mathrm{e}^{2u} - u\mathrm{e}^u - \dfrac{1}{4}\mathrm{e}^{2u} + \mathrm{e}^u\right]_{\ln 1}^{\ln 3}\)
\(= \left(\frac{9}{2}\ln 3 - 3\ln 3 - \frac{9}{4} + 3\right) - \left(0 - 0 - \frac{1}{4} + 1\right)\)	ddM1
\(= \frac{3}{2}\ln 3 + \frac{3}{4} + \frac{1}{4} - 1 = \underline{\frac{3}{2}\ln 3}\) AG	A1 cso
	[6]

Correct expression

M1 Use of ‘integration by parts’ formula in the correct direction

A1 Correct expression

M1; Attempt to integrate;

A1 correct integration

ddM1 Substitutes limits of \(\ln 3\) and \(\ln 1\) and subtracts.

A1 cso \(\frac{3}{2}\ln 3\)