Chapter 7.1, Problem 42E

To determine

To estimate: It needs to be determined the amount of pollution removed from the lake between 2015 and 2025.

Answer to Problem 42E

The amount of pollution removed is around 39.73 which is close to option C, thus option (C) is the correct choice.

Explanation of Solution

Given information: It is given that pollution is removed from a lake at a rate modeled by function $y=20e^{-0.5t}\text{ tons/yr}$ where t is the number of years since 2015.

Formula used: When the antiderivative of $f\left(x\right)$ is $F\left(x\right)$ then the antiderivative of $f\left(kx\right)=\frac{F\left(kx\right)}{k}$ .

Calculation: The amount of pollution removed from the lake between 2015 and 2025 is given by ${\displaystyle \underset{0}{\overset{10}{\int }}20e^{-0.5t}dt}=20{\displaystyle \underset{0}{\overset{10}{\int }}{e}^{-0.5t}dt}$ .

Now further as the antiderivative of $e^t$ is $e^t$ so the antiderivative of $e^{-0.5t}$ is $\frac{e^{-0.5t}}{-0.5}$, thus amount of pollution removed is as:

  $\begin{array}{c}20{\displaystyle \underset{0}{\overset{10}{\int }}{e}^{-0.5t}dt}=20{\left[-2e^{-0.5t}\right]}_0^{10}\\ =-40e^{-5}+40e^0\\ =40\left(1-e^{-5}\right)\\ \approx 39.73\end{array}$

Hence the option (C) is the correct choice.