## Thomas' Calculus 13th Edition

$$- \frac{1}{9}$$
\eqalign{ & \int_{ - \infty }^0 {x{e^{3x}}} dx \cr & {\text{By the definition of the improper integrals, we have }} \cr & \int_{ - \infty }^b {f\left( x \right)dx = \mathop {\lim }\limits_{a \to - \infty } } \int_a^b {f\left( x \right)} dx,{\text{ so}} \cr & {\text{then}}{\text{,}} \cr & \int_{ - \infty }^0 {x{e^{3x}}} dx = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {x{e^{3x}}} dx \cr & \cr & {\text{Integrating }}\int {x{e^{3x}}dx{\text{ by parts method}}} \cr & \,\,\,\,u = x,\,\,\,\,du = dx \cr & \,\,\,\,dv = {e^{3x}}dx,\,\,\,\,v = \frac{1}{3}{e^{3x}} \cr & \,\,\,\,\int {udv} = uv - \int {vdu} \to \int {x{e^{3x}}dx} = \frac{x}{3}{e^{3x}} - \int {\left( {\frac{1}{3}{e^{3x}}} \right)} \left( {dx} \right) \cr & \,\,\,\,\,\,\int {x{e^{3x}}dx} = \frac{x}{3}{e^{3x}} - \frac{1}{3}\int {{e^{3x}}} dx \cr & \,\,\,\,\,\,\int {x{e^{3x}}dx} = \frac{x}{3}{e^{3x}} - \frac{1}{9}{e^{3x}} + C \cr & \cr & \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {x{e^{3x}}} dx = \mathop {\lim }\limits_{a \to - \infty } \left( {\frac{x}{3}{e^{3x}} - \frac{1}{9}{e^{3x}}} \right)_a^0 \cr & = \mathop {\lim }\limits_{a \to - \infty } \left( {\frac{0}{3}{e^{3\left( 0 \right)}} - \frac{1}{9}{e^{3\left( 0 \right)}}} \right) - \mathop {\lim }\limits_{a \to - \infty } \left( {\frac{a}{3}{e^{3\left( a \right)}} - \frac{1}{9}{e^{3\left( a \right)}}} \right) \cr & = \mathop {\lim }\limits_{a \to - \infty } \left( { - \frac{1}{9}} \right) - \mathop {\lim }\limits_{a \to - \infty } \left( {\frac{a}{3}{e^{3a}} - \frac{1}{9}{e^{3a}}} \right) \cr & {\text{Evaluate the limit when }}a \to - \infty \cr & = \left( { - \frac{1}{9}} \right) - \left( {\frac{{ - \infty }}{3}{e^{3\left( { - \infty } \right)}} - \frac{1}{9}{e^{3\left( { - \infty } \right)}}} \right) \cr & = - \frac{1}{9} - \left( {0 - 0} \right) \cr & = - \frac{1}{9} \cr}