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 $P=50 , \quad k=0.04$. is given by:
$$
\begin{aligned} \int_{0}^{\infty} 50 e^{-0.04 t} d t &=50 \lim _{b \rightarrow \infty} \int_{0}^{b} e^{-0.04 t} d t \\ &=\left.50 \lim _{b \rightarrow \infty} \frac{e^{-0.04 t}}{-0.04}\right|_{0} ^{b} \\ &=\frac{50}{-0.04} \lim _{b \rightarrow \infty}\left(e^{-0.04 b}-e^{0}\right) \\ &=-\frac{50}{0.04}(0-1)=\frac{50}{0.04} \\ & \approx 1250 \end{aligned}
$$
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 $P=50 , \quad k=0.04$. is given by:
$$
\begin{aligned} \int_{0}^{\infty} 50 e^{-0.04 t} d t &=50 \lim _{b \rightarrow \infty} \int_{0}^{b} e^{-0.04 t} d t \\ &=\left.50 \lim _{b \rightarrow \infty} \frac{e^{-0.04 t}}{-0.04}\right|_{0} ^{b} \\ &=\frac{50}{-0.04} \lim _{b \rightarrow \infty}\left(e^{-0.04 b}-e^{0}\right) \\ &=-\frac{50}{0.04}(0-1)=\frac{50}{0.04} \\ & \approx 1250 \end{aligned}
$$
