# Calculus

## Applications of Integrals # Net Change Theorem

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:

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

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

Click or tap a problem to see the solution.

### 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?

### 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?

### 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.

### 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?

### 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?

Solution.

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?

Solution.

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.

Solution.

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?

Solution.

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.