Answer
Use the improper integral:
$$
\int_{0}^{\infty} P e^{-k t} d t
$$
The total amount of the waste that will enter the atmosphere for $P=50 , \quad k=0.06$. is given by:
$$
\begin{aligned} \int_{0}^{\infty} 50 e^{-0.06 t} d t &=50 \lim _{b \rightarrow \infty} \int_{0}^{b} e^{-0.06 t} d t \\ &=\left.50 \lim _{b \rightarrow \infty} \frac{e^{-0.06 t}}{-0.06}\right|_{0} ^{b} \\ &=\frac{50}{-0.06} \lim _{b \rightarrow \infty}\left(e^{-0.06 b}-e^{0}\right) \\ &=-\frac{50}{0.06}(0-1)=\frac{50}{0.06} \\ & \approx 833.3 \end{aligned}
$$
Work Step by Step
Use the improper integral:
$$
\int_{0}^{\infty} P e^{-k t} d t
$$
The total amount of the waste that will enter the atmosphere for $P=50 , \quad k=0.06$. is given by:
$$
\begin{aligned} \int_{0}^{\infty} 50 e^{-0.06 t} d t &=50 \lim _{b \rightarrow \infty} \int_{0}^{b} e^{-0.06 t} d t \\ &=\left.50 \lim _{b \rightarrow \infty} \frac{e^{-0.06 t}}{-0.06}\right|_{0} ^{b} \\ &=\frac{50}{-0.06} \lim _{b \rightarrow \infty}\left(e^{-0.06 b}-e^{0}\right) \\ &=-\frac{50}{0.06}(0-1)=\frac{50}{0.06} \\ & \approx 833.3 \end{aligned}
$$
