C4 June 2017 Q3

EdexcelOld spec12 marksIntegration Numerical Methods

Figure 1: decreasing curve crossing the y-axis; region R shaded between the curve, the axes and x = 1 — Figure 1

Figure 1 shows a sketch of part of the curve with equation \(y = \dfrac{6}{(\mathrm{e}^x + 2)},\ x \in \mathbb{R}\)

The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(y\)-axis, the \(x\)-axis and the line with equation \(x = 1\)

The table below shows corresponding values of \(x\) and \(y\) for \(y = \dfrac{6}{(\mathrm{e}^x + 2)}\)

\(x\)	0	0.2	0.4	0.6	0.8	1
\(y\)	2		1.71830	1.56981	1.41994	1.27165

(a) Complete the table above by giving the missing value of \(y\) to 5 decimal places. (1)

(b) Use the trapezium rule, with all the values of \(y\) in the completed table, to find an estimate for the area of \(R\), giving your answer to 4 decimal places. (3)

(c) Use the substitution \(u = \mathrm{e}^x\) to show that the area of \(R\) can be given by \[\int_a^b \frac{6}{u(u + 2)}\,\mathrm{d}u\] where \(a\) and \(b\) are constants to be determined. (2)

(d) Hence use calculus to find the exact area of \(R\).
[Solutions based entirely on graphical or numerical methods are not acceptable.] (6)

Mark scheme (a)

Scheme

Marks

\(x\)	0	0.2	0.4	0.6	0.8	1
\(y\)	2	*1.8625426…*	1.71830	1.56981	1.41994	1.27165

\(y = \dfrac{6}{(2 + \mathrm{e}^x)}\)

\(\{\text{At } x = 0.2,\}\ y = 1.86254\) (5 dp)
1.86254

B1 cao

Note: Look for this value on the given table or in their working.

(1)

Mark scheme (b)

Scheme	Marks
\(\dfrac{1}{2}(0.2)\Big[\underline{2 + 1.27165 + 2\big(\text{their } 1.86254 + 1.71830 + 1.56981 + 1.41994\big)}\Big]\) Outside brackets \(\dfrac{1}{2} \times (0.2)\) or \(\dfrac{1}{10}\) or \(\dfrac{1}{2} \times \dfrac{1}{5}\) For structure of \(\big[\ldots\ldots\big]\)	B1 o.e. M1
\(\left\{= \dfrac{1}{10}(16.41283)\right\} = 1.641283 = 1.6413\) (4 dp) anything that rounds to 1.6413	A1
	(3)

Notes

Note: M1: Do not allow an extra \(y\)-value or a repeated \(y\) value in their […]
Do not allow an omission of a \(y\)-ordinate in their […] for M1 unless they give the correct answer of awrt 1.6413, in which case both M1 and A1 can be scored.

Note: A1: Working must be seen to demonstrate the use of the trapezium rule. (Actual area is 1.64150274…)

Note: Full marks can be gained in part (b) for awrt 1.6413 even if B0 is given in part (a)

Note: Award B1M1A1 for \(\dfrac{1}{10}(2 + 1.27165) + \dfrac{1}{5}(\text{their } 1.86254 + 1.71830 + 1.56981 + 1.41994)\) = awrt 1.6413

Bracketing mistakes: Unless the final answer implies that the calculation has been done correctly

Award B1M0A0 for \(\dfrac{1}{2}(0.2) + 2 + 2(\text{their } 1.86254 + 1.71830 + 1.56981 + 1.41994) + 1.27165\) (=16.51283)

Award B1M0A0 for \(\dfrac{1}{2}(0.2)(2 + 1.27165) + 2(\text{their } 1.86254 + 1.71830 + 1.56981 + 1.41994)\) (=13.468345)

Award B1M0A0 for \(\dfrac{1}{2}(0.2)(2) + 2(\text{their } 1.86254 + 1.71830 + 1.56981 + 1.41994) + 1.27165\) (=14.61283)

Alternative method: Adding individual trapezia

Area \(\approx 0.2 \times \left[\dfrac{2 + \text{"}1.86254\text{"}}{2} + \dfrac{\text{"}1.86254\text{"} + 1.71830}{2} + \dfrac{1.71830 + 1.56981}{2} + \dfrac{1.56981 + 1.41994}{2} + \dfrac{1.41994 + 1.27165}{2}\right]\)
\(= 1.641283\)

B1: 0.2 and a divisor of 2 on all terms inside brackets

M1: First and last ordinates once and two of the middle ordinates inside brackets ignoring the 2

A1: anything that rounds to 1.6413

Mark scheme (c)

Scheme	Marks
\(\{u = \mathrm{e}^x \text{ or } x = \ln u \Rightarrow\}\)
\(\dfrac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{e}^x\) or \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = u\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = \dfrac{1}{u}\) or \(\mathrm{d}u = u\,\mathrm{d}x\) etc., and \(\displaystyle\int \dfrac{6}{(\mathrm{e}^x + 2)}\,\mathrm{d}x = \displaystyle\int \dfrac{6}{(u + 2)u}\,\mathrm{d}u\) See notes	B1 *
\(\{x = 0\} \Rightarrow a = \mathrm{e}^0 \Rightarrow \underline{a = 1}\) \(\{x = 1\} \Rightarrow b = \mathrm{e}^1 \Rightarrow \underline{b = \mathrm{e}}\) \(a = 1\) and \(b = \mathrm{e}\) or \(b = \mathrm{e}^1\) or evidence of \(0 \to 1\) and \(1 \to \mathrm{e}\)	B1
NOTE: 1^st B1 mark CANNOT be recovered for work in part (d) NOTE: 2^nd B1 mark CAN be recovered for work in part (d)
	(2)

Notes

1^st B1: Must start from either

\(\displaystyle\int y\,\mathrm{d}x\), with integral sign and \(\mathrm{d}x\)
\(\displaystyle\int \dfrac{6}{(\mathrm{e}^x + 2)}\,\mathrm{d}x\), with integral sign and \(\mathrm{d}x\)
\(\displaystyle\int \dfrac{6}{(\mathrm{e}^x + 2)}\dfrac{\mathrm{d}x}{\mathrm{d}u}\,\mathrm{d}u\), with integral sign and \(\dfrac{\mathrm{d}x}{\mathrm{d}u}\mathrm{d}u\)

and state either \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{e}^x\) or \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = u\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = \dfrac{1}{u}\) or \(\mathrm{d}u = u\,\mathrm{d}x\)
and end at \(\displaystyle\int \dfrac{6}{u(u + 2)}\,\mathrm{d}u\), with integral sign and \(\mathrm{d}u\), with no incorrect working.

Note: So, just writing \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{e}^x\) and \(\displaystyle\int \dfrac{6}{(\mathrm{e}^x + 2)}\,\mathrm{d}x = \displaystyle\int \dfrac{6}{u(u + 2)}\,\mathrm{d}u\) is sufficient for 1^st B1

Note: Give 2^nd B0 for \(b = 2.718\ldots\), without reference to \(a = 1\) and \(b = \mathrm{e}\) or \(b = \mathrm{e}^1\)

Note: You can also give the 1^st B1 mark for using a reverse process. i.e.
Proceeding from \(\displaystyle\int \dfrac{6}{u(u + 2)}\,\mathrm{d}u\) to \(\displaystyle\int \dfrac{6}{(\mathrm{e}^x + 2)}\,\mathrm{d}x\), with no incorrect working,
and stating either \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{e}^x\) or \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = u\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = \dfrac{1}{u}\) or \(\mathrm{d}u = u\,\mathrm{d}x\)

Mark scheme (d)

Scheme	Marks
Way 1
\(\dfrac{6}{u(u + 2)} \equiv \dfrac{A}{u} + \dfrac{B}{(u + 2)}\) \(\Rightarrow 6 \equiv A(u + 2) + Bu\) Writing \(\dfrac{6}{u(u + 2)} \equiv \dfrac{A}{u} + \dfrac{B}{(u + 2)}\), o.e. or \(\dfrac{1}{u(u + 2)} \equiv \dfrac{P}{u} + \dfrac{Q}{(u + 2)}\), o.e., and a complete method for finding the value of at least one of their \(A\) or their \(B\) (or their \(P\) or their \(Q\))	M1
\(u = 0 \Rightarrow A = 3\) \(u = -2 \Rightarrow B = -3\) Both their* \(A = 3\) and their \(B = -3\). (Or their \(P = \tfrac{1}{2}\) and their \(Q = -\tfrac{1}{2}\) with the factor of 6 in front of the integral sign)*	A1
\(\displaystyle\int \dfrac{6}{u(u + 2)}\,\mathrm{d}u = \displaystyle\int \left(\dfrac{3}{u} - \dfrac{3}{(u + 2)}\right)\mathrm{d}u\) \(= 3\ln u - 3\ln(u + 2)\) or \(= 3\ln 2u - 3\ln(2u + 4)\) Integrates \(\dfrac{M}{u} \pm \dfrac{N}{u \pm k}\), \(M, N, k \neq 0\); (i.e. *a two term partial fraction) to obtain either \(\pm\lambda\ln(\alpha u)\) or \(\pm\mu\ln(\beta(u \pm k))\); \(\lambda, \mu, \alpha, \beta \neq 0\) Integration of both terms is correctly followed through from their \(M\) and from their \(N\).*	M1 A1 ft
\(\left\{\text{So } \big[3\ln u - 3\ln(u + 2)\big]_1^{\mathrm{e}}\right\}\) \(= \big(3\ln(\mathrm{e}) - 3\ln(\mathrm{e} + 2)\big) - \big(3\ln 1 - 3\ln 3\big)\) [Note: A proper consideration of the limit of \(u = 1\) is required for this mark] dependent on the 2^nd M mark Applies limits of e and 1 (or their \(b\) and their \(a\), where \(b > 0, b \neq 1, a > 0\)) in \(u\) or applies limits of 1 and 0 in \(x\) and subtracts the correct way round.	dM1
\(= 3 - 3\ln(\mathrm{e} + 2) + 3\ln 3\) or \(3(1 - \ln(\mathrm{e} + 2) + \ln 3)\) or \(3 + 3\ln\left(\dfrac{3}{\mathrm{e} + 2}\right)\) or \(3\ln\left(\dfrac{\mathrm{e}}{\mathrm{e} + 2}\right) - 3\ln\left(\dfrac{1}{3}\right)\) or \(3 - 3\ln\left(\dfrac{\mathrm{e} + 2}{3}\right)\) or \(3\ln\left(\dfrac{3\mathrm{e}}{\mathrm{e} + 2}\right)\) or \(\ln\left(\dfrac{27\mathrm{e}^3}{(\mathrm{e} + 2)^3}\right)\) see notes	A1 cso
Note: Allow \(\mathrm{e}^1\) in place of e for the final A1 mark.
	(6)
	(12 marks)

Notes

Note: Give final A0 for \(3 - 3\ln\mathrm{e} + 2 + 3\ln 3\) (i.e. bracketing error) unless recovered.

Note: Give final A0 for \(3 - 3\ln(\mathrm{e} + 2) + 3\ln 3 - 3\ln 1\), where \(3\ln 1\) has not been simplified to 0

Note: Give final A0 for \(3\ln\mathrm{e} - 3\ln(\mathrm{e} + 2) + 3\ln 3\), where \(3\ln\mathrm{e}\) has not been simplified to 3

Note: Give final A0 for \(3 - 3\ln(\mathrm{e} + 2) + 3\ln 3\) simplifying to \(1 - \ln(\mathrm{e} + 2) + \ln 3\) (i.e. dividing their correct final answer by 3)
Otherwise, you can ignore incorrect working (isw) following on from a correct exact value.

Note: A decimal answer of 1.641502724… (without a correct exact answer) is final A0

Note: \(\big[-3\ln(u + 2) + 3\ln u\big]_1^{\mathrm{e}}\) followed by awrt 1.64 (without a correct exact answer) is final M1A0

Note: BE CAREFUL! Candidates will assign their own “\(A\)” and “\(B\)” for this question.

Note: Writing down \(\dfrac{6}{(u + 2)u}\) in the form \(\dfrac{A}{(u + 2)} + \dfrac{B}{u}\) with at least one of \(A\) or \(B\) correct is 1^st M1

Note: Writing down \(\dfrac{6}{(u + 2)u}\) as \(\dfrac{-3}{(u + 2)} + \dfrac{3}{u}\) is 1^st M1 1^st A1.

Note: Condone \(\displaystyle\int \left(\dfrac{3}{u} - \dfrac{3}{(u + 2)}\right)\mathrm{d}u\) to give \(3\ln u - 3\ln u + 2\) (poor bracketing) for 2^nd A1.

Note: Award M0A0M1A1ft for a candidate who writes down
e.g. \(\displaystyle\int \dfrac{6}{u(u + 2)}\,\mathrm{d}u = \displaystyle\int \left(\dfrac{6}{u} + \dfrac{6}{(u + 2)}\right)\mathrm{d}u = 6\ln u + 6\ln(u + 2)\)
AS EVIDENCE OF WRITING \(\dfrac{6}{u(u + 2)}\) AS PARTIAL FRACTIONS.

Note: Award M0A0M0A0 for a candidate who writes down
\(\displaystyle\int \dfrac{6}{u(u + 2)}\,\mathrm{d}u = 6\ln u + 6\ln(u + 2)\) or \(\displaystyle\int \dfrac{6}{u(u + 2)}\,\mathrm{d}u = \ln u + 6\ln(u + 2)\)
WITHOUT ANY EVIDENCE OF WRITING \(\dfrac{6}{u(u + 2)}\) as partial fractions.

Note: Award M1A1M1A1 for a candidate who writes down
\(\displaystyle\int \dfrac{6}{u(u + 2)}\,\mathrm{d}u = 3\ln u - 3\ln(u + 2)\)
WITHOUT ANY EVIDENCE OF WRITING \(\dfrac{6}{u(u + 2)}\) as partial fractions.

Note: If they lose the “6” and find \(\displaystyle\int_1^{\mathrm{e}} \dfrac{1}{u(u + 2)}\,\mathrm{d}u\) we can allow a maximum of M1A0M1A1ftM1A0

Way 2

Scheme	Marks
\(\left\{\displaystyle\int \dfrac{6}{u^2 + 2u}\,\mathrm{d}u = \displaystyle\int \dfrac{3(2u + 2)}{u^2 + 2u}\,\mathrm{d}u - \displaystyle\int \dfrac{6u}{u^2 + 2u}\,\mathrm{d}u\right\}\)
\(= \displaystyle\int \dfrac{3(2u + 2)}{u^2 + 2u}\,\mathrm{d}u - \displaystyle\int \dfrac{6}{u + 2}\,\mathrm{d}u\) \(\displaystyle\int \dfrac{\pm\alpha(2u + 2)}{u^2 + 2u}\{\mathrm{d}u\} \pm \displaystyle\int \dfrac{\delta}{u + 2}\{\mathrm{d}u\},\ \alpha, \beta, \delta \neq 0\) Correct expression	M1 A1
\(= 3\ln(u^2 + 2u) - 6\ln(u + 2)\) Integrates \(\dfrac{\pm M(2u + 2)}{u^2 + 2u} \pm \dfrac{N}{u \pm k}\), \(M, N, k \neq 0\), to obtain any one of \(\pm\lambda\ln(u^2 + 2u)\) or \(\pm\mu\ln(\beta(u \pm k))\); \(\lambda, \mu, \beta \neq 0\) Integration of both terms is correctly followed through* from their \(M\) and from their \(N\)*	M1 A1 ft
\(\left\{\text{So, } \big[3\ln(u^2 + 2u) - 6\ln(u + 2)\big]_1^{\mathrm{e}}\right\}\) \(= \big(3\ln(\mathrm{e}^2 + 2\mathrm{e}) - 6\ln(\mathrm{e} + 2)\big) - \big(3\ln 3 - 6\ln 3\big)\) dependent on the 2^nd M mark Applies limits of e and 1 (or their \(b\) and their \(a\), where \(b > 0, b \neq 1, a > 0\)) in \(u\) or applies limits of 1 and 0 in \(x\) and subtracts the correct way round.	dM1
\(= 3\ln(\mathrm{e}^2 + 2\mathrm{e}) - 6\ln(\mathrm{e} + 2) + 3\ln 3\) \(3\ln(\mathrm{e}^2 + 2\mathrm{e}) - 6\ln(\mathrm{e} + 2) + 3\ln 3\)	A1 o.e.
	(6)

Way 3

Scheme	Marks
Applying \(u = \theta - 1\)
\(\left\{\displaystyle\int_1^{\mathrm{e}} \dfrac{6}{u(u + 2)}\,\mathrm{d}u =\right\} \displaystyle\int_2^{1 + \mathrm{e}} \dfrac{6}{(\theta - 1)(\theta + 1)}\,\mathrm{d}\theta = \displaystyle\int_2^{1 + \mathrm{e}} \dfrac{6}{\theta^2 - 1}\,\mathrm{d}u = \left[3\ln\left(\dfrac{\theta - 1}{\theta + 1}\right)\right]_2^{1 + \mathrm{e}}\)	M1A1M1A1
\(= 3\ln\left(\dfrac{1 + \mathrm{e} - 1}{\mathrm{e} + 1 + 1}\right) - 3\ln\left(\dfrac{2 - 1}{2 + 1}\right) = 3\ln\left(\dfrac{\mathrm{e}}{\mathrm{e} + 2}\right) - 3\ln\left(\dfrac{1}{3}\right)\) 3^rd M mark is dependent on 2^nd M mark	dM1A1
	(6)