Applications of Integrals

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Net Change Theorem

Let R (t) 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 (t) 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 [a, b].

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:

\[F\left( b \right) = F\left( a \right) + \int\limits_a^b {f\left( t \right)dt} .\]

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

Example 1.

The engine on a boat starts at \(t = 0\) and consumes fuel at the rate of \[c\left( t \right) = 5 - {e^{ - t}}\] litres per hour. How much fuel does it consume for the first 2 hours?


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

\[C = \int\limits_0^2 {c\left( t \right)dt} = \int\limits_0^2 {\left( {5 - {e^{ - t}}} \right)dt} = \left. {\left( {5t + {e^{ - t}}} \right)} \right|_0^2 = 10 + {e^{ - 2}} - 1 = 9 + \frac{1}{{{e^2}}} \approx 9.14\]

Example 2.

Suppose a fish population in a lake is increasing with a rate of \[f\left( t \right) = 10 + 5{t^{\frac{3}{2}}}\] thousands of fish per year, where \(t\) is the number of years from now. How much will the fish population increase in \(4\) years?


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

\[F = \int\limits_0^4 {f\left( t \right)dt} = \int\limits_0^4 {\left( {10 + 5{t^{\frac{3}{2}}}} \right)dt} = \left. {\left( {10t + 2{t^{\frac{5}{2}}}} \right)} \right|_0^4 = 40 + 2 \cdot 32 = 104.\]

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

Example 3.

A tank has a capacity of \(V = 1000\,\text{litres}.\) Water is pumped into the tank at the rate of \[R\left( t \right) = 60 - t\] litres per minute, where time \(t\) is measured in minutes. How long will it take to fill the tank to its full capacity.


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

\[V = \int\limits_0^T {R\left( t \right)dt} = \int\limits_0^T {\left( {60 - t} \right)dt} = \left. {\left( {60t - \frac{{{t^2}}}{2}} \right)} \right|_0^T = 60T - \frac{{{T^2}}}{2}.\]

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

\[\frac{{{T^2}}}{2} - 60T + V = 0,\;\text{ or }\;{T^2} - 120T + 2000 = 0.\]

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 \(T = 20\,\text{min}.\)

Example 4.

Suppose that \(t\) months from now the population of a town will be growing at the rate of \[g\left( t \right) = 50 + 10\sqrt t\] people per month. The initial population is 5000. What will be the population in \(3\) years?


By the net change theorem,

\[G = {G_0} + \int\limits_0^T {g\left( t \right)dt} ,\]

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:

\[\int\limits_0^T {g\left( t \right)dt} = \int\limits_0^{36} {\left( {50 + 10\sqrt t } \right)dt} = \left. {\left( {50t + \frac{{20{t^{\frac{3}{2}}}}}{3}} \right)} \right|_0^{36} = 1800 + 1440 = 3240.\]

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

\[G = 5000 + 3240 = 8240.\]

See more problems on Page 2.

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