Answer
$$\frac{1}{{108}}$$
Work Step by Step
$$\eqalign{
& \int_0^{ + \infty } {{x^2}{e^{ - 6x}}dx} \cr
& {\text{Using the definition of improper integrals see page 1053}} \cr
& \underbrace {\int_a^{ + \infty } {f\left( x \right)dx = \mathop {\lim }\limits_{M \to + \infty } } \int_a^M {f\left( x \right)} dx}_ \Downarrow \cr
& \int_0^{ + \infty } {{x^2}{e^{ - 6x}}dx} = \mathop {\lim }\limits_{M \to + \infty } \int_0^M {{x^2}{e^{ - 6x}}dx} \cr
& {\text{Integrating by tabulation}} \cr
& \to \left( + \right){x^2},\left( - \right)2x,\left( + \right){\text{2,0}} \cr
& \to {e^{ - 6x}}, - \frac{1}{6}{e^{ - 6x}},{\text{ }}\frac{1}{{36}}{e^{ - 6x}}, - \frac{1}{{216}}{e^{ - 6x}} \cr
& = {x^2}\left( { - \frac{1}{6}{e^{ - 6x}}} \right) - \left( {2x} \right)\left( {\frac{1}{{36}}{e^{ - 6x}}} \right) + 2\left( { - \frac{1}{{216}}{e^{ - 6x}}} \right) \cr
& = - \frac{1}{6}{x^2}{e^{ - 6x}} - \frac{1}{{18}}x{e^{ - 6x}} - \frac{1}{{108}}{e^{ - 6x}} + C \cr
& = \mathop {\lim }\limits_{M \to + \infty } \left[ { - \frac{1}{6}{x^2}{e^{ - 6x}} - \frac{1}{{18}}x{e^{ - 6x}} - \frac{1}{{108}}{e^{ - 6x}}} \right]_0^M \cr
& = \mathop {\lim }\limits_{M \to + \infty } \left( { - \frac{1}{6}{M^2}{e^{ - 6M}} - \frac{1}{{18}}M{e^{ - 6M}} - \frac{1}{{108}}{e^{ - 6M}}} \right) \cr
& {\text{ }} - \mathop {\lim }\limits_{M \to + \infty } \left( { - \frac{1}{6}\left( 0 \right) - \frac{1}{{18}}\left( 0 \right) - \frac{1}{{108}}{e^{ - 0}}} \right) \cr
& = \mathop {\lim }\limits_{M \to + \infty } \left[ {\left( { - \frac{1}{6}{M^2}{e^{ - 6M}} - \frac{1}{{18}}M{e^{ - 6M}} - \frac{1}{{108}}{e^{ - 6M}}} \right) + \frac{1}{{108}}} \right] \cr
& {\text{Evaluate the limit when }}M \to + \infty \cr
& = \left( { - \frac{1}{6}{{\left( { + \infty } \right)}^2}{e^{ - \infty }} - \frac{1}{{18}}M{e^{ - \infty }} - \frac{1}{{108}}{e^{ - \infty }}} \right) + \frac{1}{{108}} \cr
& = 0 + \frac{1}{{108}} \cr
& = \frac{1}{{108}} \cr} $$
