Chapter 8: Techniques of Integration - Practice Exercises - Page 519: 59

$$2$$

$$\eqalign{ & \int_0^\infty {{x^2}{e^{ - x}}} dx \cr & {\text{By the definition of the improper integrals, we have }} \cr & \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx \cr & {\text{then}}{\text{,}} \cr & \int_0^\infty {{x^2}{e^{ - x}}} dx = \mathop {\lim }\limits_{b \to \infty } \int_0^b {{x^2}{e^{ - x}}} dx \cr & \cr & {\text{Integrating }}\int {{x^2}{e^{ - x}}dx{\text{ by parts method}}} \cr & \,\,\,\,u = {x^2},\,\,\,\,du = 2xdx \cr & \,\,\,\,dv = {e^{ - x}}dx,\,\,\,\,v = - {e^{ - x}} \cr & \,\,\,\,\int {udv} = uv - \int {vdu} \to \int {{x^2}{e^{ - x}}dx} = - {x^2}{e^{ - x}} - \int {\left( { - {e^{ - x}}} \right)} \left( {2xdx} \right) \cr & \,\,\,\,\int {{x^2}{e^{ - x}}dx} = - {x^2}{e^{ - x}} + \int {2x} {e^{ - x}}dx \cr & \,\,\,{\text{Integrating by parts again}} \cr & \,\,\,\,u = 2x,\,\,\,\,du = 2dx \cr & \,\,\,\,dv = {e^{ - x}}dx,\,\,\,\,v = - {e^{ - x}} \cr & \,\,\,\,\int {{x^2}{e^{ - x}}dx} = - {x^2}{e^{ - x}} + \left( { - 2x{e^{ - x}} - \int {\left( { - {e^{ - x}}} \right)\left( {2dx} \right)} } \right) \cr & \,\,\,\,\int {{x^2}{e^{ - x}}dx} = - {x^2}{e^{ - x}} + \left( { - 2x{e^{ - x}} + \int {2{e^{ - x}}dx} } \right) \cr & \,\,\,\,\int {{x^2}{e^{ - x}}dx} = - {x^2}{e^{ - x}} - 2x{e^{ - x}} + \int {2{e^{ - x}}dx} \cr & \,\,\,\,\int {{x^2}{e^{ - x}}dx} = - {x^2}{e^{ - x}} - 2x{e^{ - x}} - 2{e^{ - x}} + C \cr & \cr & \mathop {\lim }\limits_{b \to \infty } \int_0^b {{x^2}{e^{ - x}}} dx = \mathop {\lim }\limits_{b \to \infty } \left( { - {x^2}{e^{ - x}} - 2x{e^{ - x}} - 2{e^{ - x}}} \right)_0^b \cr & = \mathop {\lim }\limits_{b \to \infty } \left( { - {b^2}{e^{ - b}} - 2b{e^{ - b}} - 2{e^{ - b}}} \right) - \mathop {\lim }\limits_{b \to \infty } \left( { - {{\left( 0 \right)}^2}{e^{ - 0}} - 2\left( 0 \right){e^{ - 0}} - 2{e^{ - 0}}} \right) \cr & = \mathop {\lim }\limits_{b \to \infty } \left( { - {b^2}{e^{ - b}} - 2b{e^{ - b}} - 2{e^{ - b}}} \right) - \mathop {\lim }\limits_{b \to \infty } \left( { - 2} \right) \cr & {\text{Evaluate the limit when }}b \to \infty \cr & = - {\left( \infty \right)^2}{e^{ - \infty }} - 2\left( \infty \right){e^{ - \left( \infty \right)}} - 2{e^{ - \left( \infty \right)}} - \left( { - 2} \right) \cr & = - 0 - 0 - 0 + 2 \cr & = 2 \cr} $$