Answer
\[2000\]
Work Step by Step
\[\begin{align}
& \text{Let }W\text{ the water drained, then} \\
& \text{The water drained is given by} \\
& W=\int_{0}^{\infty }{R\left( t \right)}dt \\
& W=\int_{0}^{\infty }{100{{e}^{-0.05t}}}dt \\
& \text{Use the definition of improper integrals} \\
& W=\underset{a\to \infty }{\mathop{\lim }}\,\int_{0}^{a}{100{{e}^{-0.05t}}}dt \\
& \text{Integrating} \\
& W=\underset{a\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{-0.05}\left( 100{{e}^{-0.05t}} \right) \right]_{0}^{a} \\
& W=\underset{a\to \infty }{\mathop{\lim }}\,\left[ -2000{{e}^{-0.05t}} \right]_{0}^{a} \\
& W=-2000\underset{a\to \infty }{\mathop{\lim }}\,\left[ {{e}^{-0.05t}} \right]_{0}^{a} \\
& W=-2000\underset{a\to \infty }{\mathop{\lim }}\,\left[ {{e}^{-0.05a}}-{{e}^{0}} \right] \\
& W=-2000\underset{a\to \infty }{\mathop{\lim }}\,\left[ {{e}^{-0.05a}}-1 \right] \\
& \text{Evaluate the limit} \\
& W=-2000\left[ {{e}^{-\infty }}-1 \right] \\
& W=-2000\left( 0-1 \right) \\
& W=2000 \\
\end{align}\]
