Answer
$-25$
Work Step by Step
\[\begin{align}
& \int_{-\infty }^{0}{x{{e}^{0.2x}}}dx \\
& \text{By the definition of improper integrals} \\
& \int_{-\infty }^{b}{f\left( x \right)}dx=\underset{a\to -\infty }{\mathop{\lim }}\,\int_{a}^{b}{f\left( x \right)}dx \\
& \text{We obtain,} \\
& \int_{-\infty }^{0}{x{{e}^{0.2x}}}dx=\underset{a\to -\infty }{\mathop{\lim }}\,\int_{a}^{0}{x{{e}^{0.2x}}}dx \\
& \text{Integrating }\int{x{{e}^{0.2x}}dx}\text{ by parts we obtain }5x{{e}^{0.2x}}-25{{e}^{0.2x}}+C \\
& \underset{a\to -\infty }{\mathop{\lim }}\,\int_{a}^{0}{x{{e}^{0.2x}}}dx=\underset{a\to -\infty }{\mathop{\lim }}\,\left[ 5x{{e}^{0.2x}}-25{{e}^{0.2x}} \right]_{a}^{0} \\
& =\underset{a\to -\infty }{\mathop{\lim }}\,\left[ \left( 5\left( 0 \right){{e}^{0}}-25{{e}^{0}} \right)-\left( 5a{{e}^{a}}-25{{e}^{a}} \right) \right] \\
& =\underset{a\to -\infty }{\mathop{\lim }}\,\left[ -25-5a{{e}^{a}}+25{{e}^{a}} \right] \\
& \text{Evaluate when }a\to -\infty \\
& =-25-5\left( -\infty \right){{e}^{-\infty }}+25{{e}^{-\infty }} \\
& =-25-0+0 \\
& =-25 \\
\end{align}\]