Question

Oil is leaking from a tanker at the rate of $R\left(t\right)=2000e^{(-0.2t)}\mathrm{gallons}\mathrm{per}\mathrm{hour}$, where $t$ is measured in hours. How much oil leaks out of the tanker from the time $t=0$ to $t=10$?

Answer 1

Find the oil leaks out of the tanker from the time $t=0$ to $t=10$:

The rate of leaking is $R\left(t\right)=2000e^{-0.2t}$.

The integration rule of exponential is $\int e^{px}dx=\frac{e^{px}}{p}+C$, where $C$ is the constant of integration and $p$ is any non-zero real number.

Integrate the expression of $R\left(t\right)$ between $t=0,10$ and simplify.

$\begin{array}{rcl}{\int }_0^{10}R\left(t\right)dt& =& {\int }_0^{10}2000e^{-0.2t}dt\\ & =& 2000{\left[\frac{e^{-0.2t}}{-0.2}\right]}_0^{10}\\ & =& \frac{2000}{-0.2}\left[e^{-2}-e^0\right]\\ & =& 10000\left(1-\frac{1}{e^2}\right)\\ & \approx & 8647\end{array}$

Hence, the volume of oil that leaks out of the tanker at the time $t=0$ to $t=10$ is $8,647g\mathrm{allons}.$