Answer:
[tex]V_{total} = \displaystyle\int_0^{60} A e^{-k t}~dt[/tex]
Step-by-step explanation:
We are given the following in the question:
Oil leaks out of a tanker at a rate of r = f(t) liters per minute, where t is in minutes.
[tex]f(t) = A e^{-k t}[/tex]
Let V be the volume, then we are given that rate of leakage is:
[tex]\displaystyle\frac{dV}{dt} = f(t) = A e^{-k t}[/tex]
Thus, we can write:
[tex]dV = f(t).dt = A e^{-k t}~dt[/tex]
We have to find the a definite integral expressing the total quantity of oil which leaks out of the tanker in the first hour.
Thus, total amount of oil leaked will be the definite integral from 0 minutes to 60 minutes.
[tex]dV =A e^{-k t}~dt\\V_{total} = \displaystyle\int_a^b f(t)~dt\\\\a = 0\text{ minutes}\\b = 60\text{ minutes}\\\\V_{total} = \displaystyle\int_0^{60} A e^{-k t}~dt[/tex]
The units of integral will be liters.
The definite integral must be:
[tex]\int\limits^b_a {A*e^{-k*t}} \, dt = -\frac{A}{k}*(e^{-k*b } - e^{-k*a}})[/tex]
And the units of that integral are volume units.
We know that the rate at which the tank leaks is:
f(t) = A*e^{-k*t}
We want to write a definite integral between the times t = a and t = b, this will be:
[tex]\int\limits^b_a {A*e^{-k*t}} \, dt[/tex]
Solving that integral, we get:
[tex]\int\limits^b_a {A*e^{-k*t}} \, dt = -\frac{A}{k}*(e^{-k*b } - e^{-k*a}})[/tex]
Now, which units has this integral?
f(t) defines the rate at which the oil tank leaks, then the integral over the interval [a, b] will give the volume of oil leaked on that interval of time. So the integral must-have volume units.
If you want to learn more about integrals, you can read:
https://brainly.com/question/21023287
