C4 June 2015 Q3

EdexcelOld spec8 marksIntegration Logs & Exponentials

Figure 1: curve from O rising then falling to cut the x-axis at A, region R shaded between curve and x-axis — Figure 1

Figure 1 shows a sketch of part of the curve with equation \(y = 4x - x\mathrm{e}^{\frac{1}{2}x},\ x \geqslant 0\)

The curve meets the \(x\)-axis at the origin \(O\) and cuts the \(x\)-axis at the point \(A\).

(a) Find, in terms of \(\ln 2\), the \(x\) coordinate of the point \(A\). (2)

(b) Find \[\int x\mathrm{e}^{\frac{1}{2}x}\,\mathrm{d}x\] (3)

The finite region \(R\), shown shaded in Figure 1, is bounded by the \(x\)-axis and the curve with equation \[y = 4x - x\mathrm{e}^{\frac{1}{2}x},\ x \geqslant 0\]

Mark scheme (a)

Scheme	Marks
\(y = 4x - x\mathrm{e}^{\frac{1}{2}x},\ x \geqslant 0\)
\(\left\{y = 0 \Rightarrow 4x - x\mathrm{e}^{\frac{1}{2}x} = 0 \Rightarrow x(4 - \mathrm{e}^{\frac{1}{2}x}) = 0 \Rightarrow\right\}\)
\(\mathrm{e}^{\frac{1}{2}x} = 4 \Rightarrow x_A = 4\ln 2\) Attempts to solve \(\mathrm{e}^{\frac{1}{2}x} = 4\) giving \(x = \ldots\) in terms of \(\pm\lambda\ln\mu\) where \(\mu > 0\) \(4\ln 2\) cao (Ignore \(x = 0\))	M1 A1
	(2)

Notes

M1: Attempts to solve \(\mathrm{e}^{\frac{1}{2}x} = 4\) giving \(x = \ldots\) in terms of \(\pm\lambda\ln\mu\) where \(\mu > 0\)

A1: \(4\ln 2\) cao stated in part (a) only (Ignore \(x = 0\))

Mark scheme (b)

Scheme	Marks
\(\left\{\displaystyle\int x\mathrm{e}^{\frac{1}{2}x}\,\mathrm{d}x\right\} = 2x\mathrm{e}^{\frac{1}{2}x} - \displaystyle\int 2\mathrm{e}^{\frac{1}{2}x}\,\{\mathrm{d}x\}\) \(\alpha x\mathrm{e}^{\frac{1}{2}x} - \beta\displaystyle\int \mathrm{e}^{\frac{1}{2}x}\,\{\mathrm{d}x\},\ \alpha > 0, \beta > 0\) \(2x\mathrm{e}^{\frac{1}{2}x} - \displaystyle\int 2\mathrm{e}^{\frac{1}{2}x}\,\{\mathrm{d}x\}\), with or without \(\mathrm{d}x\)	M1 A1 (M1 on ePEN)
\(= 2x\mathrm{e}^{\frac{1}{2}x} - 4\mathrm{e}^{\frac{1}{2}x}\ \{+c\}\) \(2x\mathrm{e}^{\frac{1}{2}x} - 4\mathrm{e}^{\frac{1}{2}x}\) o.e. with or without \(+c\)	A1
	(3)

Notes

NOTE: Part (b) appears as M1M1A1 on ePEN, but is now marked as M1A1A1.

M1: Integration by parts is applied in the form \(\alpha x\mathrm{e}^{\frac{1}{2}x} - \beta\displaystyle\int \mathrm{e}^{\frac{1}{2}x}\,\{\mathrm{d}x\}\), where \(\alpha > 0, \beta > 0\). (must be in this form) with or without \(\mathrm{d}x\)

A1: \(2x\mathrm{e}^{\frac{1}{2}x} - \displaystyle\int 2\mathrm{e}^{\frac{1}{2}x}\,\{\mathrm{d}x\}\) or equivalent, with or without \(\mathrm{d}x\). Can be un-simplified.

A1: \(2x\mathrm{e}^{\frac{1}{2}x} - 4\mathrm{e}^{\frac{1}{2}x}\) or equivalent with or without \(+c\). Can be un-simplified.

Note: You can also allow \(2\mathrm{e}^{\frac{1}{2}x}(x - 2)\) or \(\mathrm{e}^{\frac{1}{2}x}(2x - 4)\) for the final A1.

isw: You can ignore subsequent working following on from a correct solution.

SC: SPECIAL CASE: A candidate who uses \(u = x\), \(\dfrac{\mathrm{d}v}{\mathrm{d}x} = \mathrm{e}^{\frac{1}{2}x}\), writes down the correct “by parts” formula, but makes only one error when applying it can be awarded Special Case M1. (Applying their \(v\) counts for one consistent error.)

Mark scheme (c)

Scheme	Marks
\(\left\{\displaystyle\int 4x\,\mathrm{d}x\right\} = 2x^2\) \(4x \to 2x^2\) or \(\dfrac{4x^2}{2}\) o.e.	B1
\(\left\{\displaystyle\int_0^{4\ln 2} (4x - x\mathrm{e}^{\frac{1}{2}x})\,\mathrm{d}x\right\} = \left[2x^2 - \left(2x\mathrm{e}^{\frac{1}{2}x} - 4\mathrm{e}^{\frac{1}{2}x}\right)\right]_0^{4\ln 2 \text{ or } \ln 16 \text{ or their limits}}\)
\(= \left(2(4\ln 2)^2 - 2(4\ln 2)\mathrm{e}^{\frac{1}{2}(4\ln 2)} + 4\mathrm{e}^{\frac{1}{2}(4\ln 2)}\right) - \left(2(0)^2 - 2(0)\mathrm{e}^{\frac{1}{2}(0)} + 4\mathrm{e}^{\frac{1}{2}(0)}\right)\) See notes	M1
\(= \left(32(\ln 2)^2 - 32(\ln 2) + 16\right) - (4)\)
\(= 32(\ln 2)^2 - 32(\ln 2) + 12\) \(32(\ln 2)^2 - 32(\ln 2) + 12\), see notes	A1
	(3)
	(8 marks)

Notes

B1: \(4x \to 2x^2\) or \(\dfrac{4x^2}{2}\) oe

M1: Complete method of applying limits of their \(x_A\) and 0 to all terms of an expression of the form \(\pm Ax^2 \pm Bx\mathrm{e}^{\frac{1}{2}x} \pm C\mathrm{e}^{\frac{1}{2}x}\) (where \(A \neq 0\), \(B \neq 0\) and \(C \neq 0\)) and subtracting the correct way round.

Note: Evidence of a proper consideration of the limit of 0 is needed for M1. So subtracting 0 is M0.

Note: \(\ln 16\) or \(2\ln 4\) or equivalent is fine as an upper limit.

A1: A correct three term exact quadratic expression in \(\ln 2\).
For example allow for A1

\(32(\ln 2)^2 - 32(\ln 2) + 12\)
\(8(2\ln 2)^2 - 8(4\ln 2) + 12\)
\(2(4\ln 2)^2 - 32(\ln 2) + 12\)
\(2(4\ln 2)^2 - 2(4\ln 2)\mathrm{e}^{\frac{1}{2}(4\ln 2)} + 12\)

Note: Note that the constant term of 12 needs to be combined from \(4\mathrm{e}^{\frac{1}{2}(4\ln 2)} - 4\mathrm{e}^{\frac{1}{2}(0)}\) o.e.

Note: Also allow \(32\ln 2(\ln 2 - 1) + 12\) or \(32\ln 2\left(\ln 2 - 1 + \dfrac{12}{32\ln 2}\right)\) for A1.

Note: Do not apply “ignore subsequent working” for incorrect simplification.
Eg: \(32(\ln 2)^2 - 32(\ln 2) + 12 \to 64(\ln 2) - 32(\ln 2) + 12\) or \(32(\ln 4) - 32(\ln 2) + 12\)

Note: Bracketing error: \(32\ln 2^2 - 32(\ln 2) + 12\), unless recovered is final A0.

Note: Notation: Allow \(32(\ln^2 2) - 32(\ln 2) + 12\) for the final A1.

Note: 5.19378… without seeing \(32(\ln 2)^2 - 32(\ln 2) + 12\) is A0.

Note: 5.19378… following from a correct \(2x^2 - \left(2x\mathrm{e}^{\frac{1}{2}x} - 4\mathrm{e}^{\frac{1}{2}x}\right)\) is M1A0.

Note: 5.19378… from no working is M0A0.