A2 June 2019 Paper 2 Q13

AQACurrent spec10 marksIntegration

(a) Explain why \(\int_3^{\infty} x^2\mathrm{e}^{-2x}\,\mathrm{d}x\) is an improper integral. [1 mark]

(b) Evaluate \(\int_3^{\infty} x^2\mathrm{e}^{-2x}\,\mathrm{d}x\)

Show the limiting process. [9 marks]

Mark scheme (a)

Scheme	Marks	AO
Explains that one of the limits is infinity (or that the interval of integration is infinite)	R1	2.4

Typical solution

The upper limit is infinity, so it is an improper integral.

Mark scheme (b)

Scheme	Marks	AO
Uses integration by parts twice	M1	3.1a
Obtains the correct expressions for \(u^{\prime}\) and \(v\) when integrating the first time	B1	1.1b
Correctly applies integration by parts formula the first time	B1	1.1b
Correctly applies integration by parts formula to an integral of the form \(k\int x\mathrm{e}^{-2x}\,\mathrm{d}x\)	M1	1.1a
Finds complete correct expression for integral with or without \(c\). No limits needed at this stage. PI by later work	A1	1.1b
Defines the improper integral as a limit	E1	2.4
Applies the limiting process correctly, using \(\lim_{n \to \infty}(n^2\mathrm{e}^{-n}) = 0\) \(\lim_{n \to \infty}(n\mathrm{e}^{-n}) = 0\) and \(\lim_{n \to \infty}(\mathrm{e}^{-n}) = 0\)	M1	2.2a
Substitutes correct lower limit correctly into their three-term expression	M1	1.1a
Obtains correct exact value or awrt 0.0155	A1	1.1b
	(10 marks)

Typical solution

\[u = x^2 \qquad v^{\prime} = \mathrm{e}^{-2x}\]\[u^{\prime} = 2x \qquad v = \frac{-\mathrm{e}^{-2x}}{2}\]\[\int x^2\mathrm{e}^{-2x}\,\mathrm{d}x = -\frac{x^2\mathrm{e}^{-2x}}{2} + \int x\mathrm{e}^{-2x}\,\mathrm{d}x\]\[u = x \qquad v^{\prime} = \mathrm{e}^{-2x}\]\[u^{\prime} = 1 \qquad v = \frac{-\mathrm{e}^{-2x}}{2}\]\[\begin{aligned}\int x\mathrm{e}^{-2x}\,\mathrm{d}x &= \left(-\frac{x\mathrm{e}^{-2x}}{2}\right) + \int \frac{\mathrm{e}^{-2x}}{2}\,\mathrm{d}x \\ &= -\frac{x\mathrm{e}^{-2x}}{2} - \frac{\mathrm{e}^{-2x}}{4}\end{aligned}\]\[\therefore \int x^2\mathrm{e}^{-2x}\,\mathrm{d}x = -\frac{x^2\mathrm{e}^{-2x}}{2} - \frac{x\mathrm{e}^{-2x}}{2} - \frac{\mathrm{e}^{-2x}}{4}\]\[\int_3^{\infty} x^2\mathrm{e}^{-2x}\,\mathrm{d}x = \lim_{n \to \infty}\int_3^{n} x^2\mathrm{e}^{-2x}\,\mathrm{d}x\]\[= \lim_{n \to \infty}\left\{\left[-\frac{x^2\mathrm{e}^{-2x}}{2} - \frac{x\mathrm{e}^{-2x}}{2} - \frac{\mathrm{e}^{-2x}}{4}\right]_3^n\right\}\]\[\therefore \int_3^{\infty} x^2\mathrm{e}^{-2x}\,\mathrm{d}x = \frac{9\mathrm{e}^{-6}}{2} + \frac{3\mathrm{e}^{-6}}{2} + \frac{\mathrm{e}^{-6}}{4}\]\[= \frac{25\mathrm{e}^{-6}}{4}\]