(a)     Using l’Hopital’s Rule, show that \(\mathop {\lim }\limits_{x \to \infty } x{{\text{e}}^{ - x}} = 0\) .

(b)     Determine \(\int_0^a {x{{\text{e}}^{ - x}}{\text{d}}x} \) .

(c)     Show that the integral \(\int_0^\infty  {x{{\text{e}}^{ - x}}{\text{d}}x} \) is convergent and find its value.

(a)     \(\mathop {\lim }\limits_{x \to \infty } \frac{x}{{{{\text{e}}^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{1}{{{{\text{e}}^x}}}\)     M1A1

= 0     AG

[2 marks]

(b)     Using integration by parts     M1

\(\int_0^a {x{{\text{e}}^{ - x}}{\text{d}}x}  = \left[ { - x{{\text{e}}^{ - x}}} \right]_0^a + \int_0^a {{{\text{e}}^{ - x}}{\text{d}}x} \)     A1A1

\( = - a{{\text{e}}^{ - a}} - \left[ {{e^{ - x}}} \right]_0^a\)     A1

\( = 1 - a{{\text{e}}^{ - a}} - {{\text{e}}^{ - a}}\)     A1

[5 marks]

(c)     Since \({{\text{e}}^{ - a}}\) and \(a{{\text{e}}^{ - a}}\) are both convergent (to zero), the integral is convergent.     R1

Its value is 1.     A1

[2 marks]

Total [9 marks]

Most candidates made a reasonable attempt at (a). In (b), however, it was disappointing to note that some candidates were unable to use integration by parts to perform the integration. In (c), while many candidates obtained the correct value of the integral, proof of its convergence was often unconvincing.