Let \(R\left( t \right)\) represent the rate at which water flows into a tank. The rate can be measured in cubic meters per second or, for example, in gallons per minute.

Then the definite integral \(\int\limits_a^b {R\left( t \right)dt} \) expresses the total change in volume from time \(a\) to time \(b.\)

Instead of a water flow, we can consider any other quantity. In general case, if \(f\left( t \right)\) is the rate of change of some quantity, then the integral \(\int\limits_a^b {f\left( t \right)dt} \) represents the net change in that quantity over the time interval \(\left[ {a,b} \right].\)

This leads us to the Net Change Theorem, which states that if a quantity changes and is represented by a differentiable function, the final value equals the initial value plus the integral of the rate of change of that quantity:

The Net Change Theorem can be applied to various problems involving rate of change (such as finding volume, area, population, velocity, distance, cost, etc.)

Solved Problems

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Solution.

To estimate the fuel consumption, we use the net change theorem. This yields:

Solution.

The integral of the rate of change from \(t = 0\) to \(t = 4\) is the net change of the fish population:

Hence, the fish population will increase by \(F = 104\) thousands.

Solution.

Let \(T\) be the time required to fill the tank. Using integration, we have

To find the time \(T,\) we must solve the quadratic equation

The roots of the equation are \({T_{1,2}} = 100, 20\,\text{min}.\) The first root \({T_1} = 100\,\text{min}\) does not make sense as the rate of the water flow becomes negative for \(T \gt 60\,\text{min}.\)

Thus, the answer is \(T = 20\,\text{min}.\)

Solution.

By the net change theorem,

where \({G_0}\) and \(G\) are the initial and final populations, respectively.

By integrating from \(t = 0\) to \(T = 36\) months, we get the total increase in the number of people:

Thus, the final population in \(3\) years is expected to be