Answer
Divergent
Work Step by Step
\[\begin{align}
& \int_{0}^{\infty }{x{{e}^{4x}}}dx \\
& \text{By the definition of improper integrals} \\
& \int_{a}^{\infty }{f\left( x \right)}dx=\underset{b\to \infty }{\mathop{\lim }}\,\int_{a}^{b}{f\left( x \right)}dx \\
& \text{We obtain,} \\
& \int_{0}^{\infty }{x{{e}^{4x}}}dx=\underset{b\to \infty }{\mathop{\lim }}\,\int_{0}^{b}{x{{e}^{4x}}}dx \\
& \text{Integrating }\int{x{{e}^{4x}}dx}\text{ by parts we obtain }\frac{1}{4}x{{e}^{4x}}-\frac{1}{16}{{e}^{4x}}+C \\
& \underset{b\to \infty }{\mathop{\lim }}\,\int_{0}^{b}{x{{e}^{4x}}}dx=\underset{b\to \infty }{\mathop{\lim }}\,\left[ \text{ }\frac{1}{4}x{{e}^{4x}}-\frac{1}{16}{{e}^{4x}} \right]_{0}^{b} \\
& =\underset{b\to \infty }{\mathop{\lim }}\,\left[ \left( \text{ }\frac{1}{4}b{{e}^{bx}}-\frac{1}{16}{{e}^{bx}} \right)-\left( \text{ }\frac{1}{4}0{{e}^{0}}-\frac{1}{16}{{e}^{0}} \right) \right] \\
& =\underset{b\to \infty }{\mathop{\lim }}\,\left[ \left( \text{ }\frac{1}{4}b{{e}^{bx}}-\frac{1}{16}{{e}^{bx}} \right)+\frac{1}{16} \right] \\
& \text{Evaluate when }b\to \infty \\
& =\frac{1}{4}e^{\infty}\left( \text{ } \infty-\frac{1}{4}{} \right)+\frac{1}{16} \\
& =\infty \\
& \text{The improper integral diverges.} \\
& \text{Divergent} \\
\end{align}\]
