Answer
An oil leak from an uncapped well is polluting a bay at a rate of
$ f(t)=125 e^{-0.025 t}$ gallons per year.
The total amount of oil that will enter the bay, assuming the well is never capped, is given by:
$$
\begin{aligned}
\int_{0}^{\infty} 125 e^{-0.025 t} d t &=\lim _{b \rightarrow \infty} \int_{0}^{b} 125 e^{-0.025 t} d t \\
&=5000 \text{ gallons}
\end{aligned}
$$
Work Step by Step
An oil leak from an uncapped well is polluting a bay at a rate of
$ f(t)=125 e^{-0.025 t}$ gallons per year.
The total amount of oil that will enter the bay, assuming the well is never capped, is given by:
$$
\begin{aligned}
\int_{0}^{\infty} 125 e^{-0.025 t} d t &=\lim _{b \rightarrow \infty} \int_{0}^{b} 125 e^{-0.025 t} d t \\
&=\left.\lim _{b \rightarrow \infty}\left(\frac{125}{-0.025} e^{-0.025 t}\right)\right|_{0} ^{b}\\
&=\lim _{b \rightarrow \infty}(-5000)\left(e^{-0.025 b}-e^{0}\right)\\
&=-5000\left(\lim _{b \rightarrow \infty} \frac{1}{e^{0.025 b}}-1\right)\\
&=-5000\left((0)-1\right)\\
&=5000.
\end{aligned}
$$
The total amount of oil that will enter the bay $5000$ gallons.