Question

Oil leaks out of a tanker at a rate of r=f(t) liters per minute, where t is in minutes. If f(t) = A e^(-kt), write a definite integral expressing the total quantity of oil which leaks out of the tanker in the first hour.What is the total quantity of ?ab ? what does a=? what does b=?What are the units of this integral?

Answer 1

Answer:

a=0 , b=1 hour =60 minutes , [I] = liters

Step-by-step explanation:

for leak out rate

r(t)=A e^(-kt)

then for the first hour , that is between a=0 and b=1 hour =60 minutes , then

$I=\int\limits^{60}_{0} {r(t)} \, dt = \int\limits^{60}_{0} {A*e^{-kt}} \, dt$

since the units of I , [I]=[r(t)]*[t] = liters/minute * minute = liters

then the integral has units [I] = liters

Answer 2

A definite integral is the value of the integral over an interval [a, b]

The given function for the rate of the leak is r = f(t) = $A \cdot e^{-k \cdot t}$

The definite integral is presented as follows;

• $\underline{The \ total \ quantity \ =\displaystyle \int\limits^a_b {A \cdot e^{-k \cdot t}} \, dt = \frac{A \cdot e^{k \cdot a} - A \cdot e^{k \cdot b}}{k \cdot e^{k \cdot b} \cdot e^{k \cdot a}}}$

The definite integral for the total quantity of oil which leaks in the first one hour is therefore;

$\displaystyle \int\limits^{60}_0 {A \cdot e^{-k \cdot t}} \, dt = \frac{A \cdot e^{k \times 60} - A \cdot e^{k \times 0}}{k \cdot e^{k \times 0} \cdot e^{k \times 60}} = \frac{A \cdot e^{60\cdot k } - A }{k\cdot e^{ 60 \cdot k}}$

$\displaystyle \int\limits^{60}_0 {A \cdot e^{-k \cdot t}} \, dt = \frac{A \cdot e^{60\cdot k } - A }{k\cdot e^{ 60 \cdot k}}$

Where;

• a = 60, b = 0

The integral expresses the quantity of oil that leaks out, over a period of time, based on a leakage rate f(t), measured in liters per minute

Therefore;

The unit of the integral = Liter/min × (minutes over which the integral is found) = Liters

• The unit of the integral is in liters, L.

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