Chapter 8.1, Problem 36E

To determine

Tofind the amount of pollution removed from the lake between $1995$ and $2005$ .

Answer to Problem 36E

Approximately the value is found to be $40$ option (a).

Explanation of Solution

Given information:

Pollution is being removed from a lake at a rate modelled by the function

  $y=20e^{-0.5t}tons/yr$ where $t$ -number of years since $1995$ .

From given the period is totally about $10$ years between $1995$ and $2005$ .

Hence, in $1995$ the time is found to be $t=0$ and in $2005$ the time is found to be $t=10$

The totalpollution removed from the lake is given by solving the integral,

  ${\displaystyle \underset{0}{\overset{10}{\int }}20e^{-0.5t}}dt=20\cdot {\displaystyle {\int }_0^{10}{e}^{-0.5t}}dt$

Substitute $u=-0.5$ ,

when $\begin{array}{l}t=0\Rightarrow u=0\\ t=10\Rightarrow u=-5\end{array}$

1. ,

  $\begin{array}{l}=20\cdot {\displaystyle {\int }_0^{-5}-}\frac{1}{0.5}{e}^{u}du\\ =20\left(-{\displaystyle {\int }_{-5}^0-}\frac{1}{0.5}{e}^{u}du\right)\\ =20\left(-\left(-\frac{1}{0.5}\cdot {\displaystyle {\int }_{-5}^0{e}^u}du\right)\right)\\ =20\left(-\left(-2\cdot {\displaystyle {\int }_{-5}^0{e}^u}du\right)\right)\\ =20\left(-\left(-2{\left[e^u\right]}_{-5}^0\right)\right)\\ =40{\left[e^u\right]}_{-5}^0\\ =40\left(1-\frac{1}{e^5}\right)\approx 39.730\end{array}$

Therefore, approximately the value is found to be $40$ option (a).