# Applications of Integrals in Economics

The concept of integration is widely used in business and economics. In this section, we consider the following applications of integrals in finance and economics:

• Marginal and total revenue, cost, and profit;
• Capital accumulation over a specified period of time;
• Consumer and producer surplus;
• Lorenz curve and Gini coefficient;

### Marginal and Total Revenue, Cost, and Profit

Marginal revenue $$\left({MR}\right)$$ is the additional revenue gained by producing one more unit of a product or service.

It can also be described as the change in total revenue $$\left({TR}\right)$$ divided by the change in number of units sold $$\left({Q}\right):$$

$MR = \frac{{dTR}}{{dQ}}.$

If a marginal revenue function $$MR\left( Q \right)$$ is known, the total revenue can be obtained by integrating the marginal revenue function:

$TR\left( Q \right) = \int {MR\left( Q \right)dQ} ,$

where integration is carried out over a certain interval of $$Q.$$

Marginal cost $$\left({MC}\right)$$ denotes the additional cost of producing one extra unit of output.

The similar relationship exists between the marginal cost $$MC$$ and the total cost $$TC:$$

$MC = \frac{{dTC}}{{dQ}},$

so

$TC\left( Q \right) = \int {MC\left( Q \right)dQ} .$

Since profit is defined as

$TP = TR – TC,$

we can write the following equation for marginal profit $$\left({MP}\right):$$

${MP = MR – MC,\;\;\text{or}\;\;}\kern0pt{\frac{{dTP}}{{dQ}} = \frac{{dTR}}{{dQ}} – \frac{{dTC}}{{dQ}}.}$

### Capital Accumulation Over a Period

Let $$I\left( t \right)$$ be the rate of investment. The total capital accumulation $$K$$ during the time interval $$\left[ {a,b} \right]$$ can be estimated by the formula

$K = \int\limits_a^b {I\left( t \right)dt} .$

### Consumer and Producer Surplus

The demand function or demand curve shows the relationship between the price of a certain product or service and the quantity demanded over a period of time.

The supply function or supply curve shows the quantity of a product or service that producers will supply over a period of time at any given price.

Both these price-quantity relationships are usually considered as functions of quantity $$\left( Q \right).$$

Generally, the demand function $$P = D\left( Q \right)$$ is decreasing, because consumers are likely to buy more of a product at lower prices. Unlike the law of demand, the supply function $$P = S\left( Q \right)$$ is increasing, because producers are willing to deliver a greater quantity of a product at higher prices.

The point $$\left( {{Q_0},{P_0}} \right)$$ where the demand and supply curves intersect is called the market equilibrium point.

The maximum price a consumer is willing and able to pay is defined by the demand curve $$P = D\left( Q \right).$$ For quantities $${Q \lt {Q_0}},$$ it is greater than the equilibrium price $${P_0}$$ in the market. Consumers gain by buying at the equilibrium price rather than at a higher price. This net gain is called consumer surplus.

Consumer surplus is represented by the area under the demand curve $$P = D\left( Q \right)$$ and above the horizontal line $$P = {P_0}$$ at the level of the market price.

Consumer surplus $$\left( {CS} \right)$$ is thus defined by the integration formula

${CS = \int\limits_0^{{Q_0}} {D\left( Q \right)dQ} – {P_0}{Q_0} }={ \int\limits_0^{{Q_0}} {\left[ {D\left( Q \right) – {P_0}} \right]dQ} .}$

A similar analysis shows that producers also gain if they trade their products at the market equilibrium price. Their gain is called producer surplus $$\left( {PS} \right)$$ and is given by the equation

${PS = {P_0}{Q_0} – \int\limits_0^{{Q_0}} {S\left( Q \right)dQ} }={ \int\limits_0^{{Q_0}} {\left[ {{P_0} – S\left( Q \right)} \right]dQ} .}$

### Lorenz Curve and Gini Coefficient

The Lorenz curve is a graphical representation of income or wealth distribution among a population.

The horizontal axis on a Lorenz curve typically shows the portion or percentage of total population, and the vertical axis shows the portion of total income or wealth. For instance, if a Lorenz curve has a point with coordinates $$\left( {0.4,0.2} \right),$$ this means that the first $$40\%$$ of population (ranked by income in increasing order) earned $$20\%$$ of total income.

The Lorenz Curve is represented by a convex curve. A more convex Lorenz curve implies more inequality in income distribution. The area between the $$45-$$degree line (the line of equality) and the Lorenz curve can be used as a measure of inequality.

The Gini coefficient $$G$$ is defined as the area between the line of equality and the Lorenz curve, divided by the total area under the line of equality:

${G = \frac{A}{{A + B}} }={ 2\int\limits_0^1 {\left[ {x – L\left( x \right)} \right]dx} .}$

The Gini coefficient is a relative measure of inequality. It ranges from $$0$$ (or $$0\%$$) to $$1$$ (or $$100\%$$), with $$0$$ representing perfect equality in a population and $$1$$ representing perfect inequality.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

The marginal revenue of a company is given by $$MR = 100 + 20Q + 3{Q^2},$$ where $$Q$$ is amount of units sold for a period. Find the total revenue function if at $$Q = 2$$ it is equal to $$260.$$

### Example 2

The rate of investment is given by $$I\left( t \right) = 6\sqrt t .$$ Calculate the capital growth between the $$4^{\text{th}}$$ and the $$9^{\text{th}}$$ years.

### Example 3

Assume the rate of investment is given by the function $$I\left( t \right) = \ln t .$$ Compute the total capital accumulation between the $$1^{\text{st}}$$ and the $$5^{\text{th}}$$ years.

### Example 4

For a certain product, the demand function is $$D\left( Q \right) = 1000 – 25Q,$$ and the supply function is $$S\left( Q \right) = 100 + {Q^2}.$$ Compute the consumer and producer surplus.

### Example 5

Assuming the demand function is $$D\left( Q \right) = 50 – Q$$ and the supply function is $$S\left( Q \right) = 20 + \sqrt Q,$$ compute the consumer and producer surplus.

### Example 6

Assume the demand and supply functions for a product are $$D\left( Q \right) = {\left( {Q – 2a} \right)^2}$$ and $$S\left( Q \right) = {Q^2},$$ where $$a \gt 0$$ is a parameter. Compute the consumer and producer surplus.

### Example 7

Calculate the Gini coefficient for the Lorenz function $$L\left( x \right) = {x^3}.$$

### Example 8

Calculate the Gini coefficient for the Lorenz function $$L\left( x \right) = {x^p},$$ where $$p \gt 1.$$

### Example 9

Suppose the Lorenz curve for a society is given by $$L\left( x \right) = \large{\frac{3}{5}}\normalsize {x^3} + \large{\frac{1}{5}}\normalsize {x^2} + \large{\frac{1}{5}}\normalsize x.$$ Find the Gini coefficient for this income distribution.

### Example 10

Suppose the Lorenz curve for a country is given by $$L\left( x \right) = 1 – \sqrt {1 – {x^2}}.$$ Determine the Gini coefficient for this income distribution.

### Example 1.

The marginal revenue of a company is given by $$MR = 100 + 20Q + 3{Q^2},$$ where $$Q$$ is amount of units sold for a period. Find the total revenue function if at $$Q = 2$$ it is equal to $$260.$$

Solution.

We find the total revenue function $$TR$$ by integrating the marginal revenue function $$MR:$$

${TR\left( Q \right) = \int {MR\left( Q \right)dQ} }={ \int {\left( {100 + 20Q + 3{Q^2}} \right)dQ} }={ 100Q + 10{Q^2} + {Q^3} + C.}$

The constant of integration $$C$$ can be determined using the initial condition $$TR\left( {Q = 2} \right) = 260.$$ Hence,

${200 + 40 + 8 + C = 260,}\;\; \Rightarrow {C = 12.}$

So, the total revenue function is given by

$TR\left( Q \right) = 100Q + 10{Q^2} + {Q^3} + 12.$

### Example 2.

The rate of investment is given by $$I\left( t \right) = 6\sqrt t .$$ Calculate the capital growth between the $$4^{\text{th}}$$ and the $$9^{\text{th}}$$ years.

Solution.

Using the integration formula

$K = \int\limits_a^b {I\left( t \right)dt} ,$

we have

${K = \int\limits_4^9 {6\sqrt t dt} }={ 6\int\limits_4^9 {{t^{\frac{1}{2}}}dt} }={ \left. {\frac{{12{t^{\frac{3}{2}}}}}{3}} \right|_4^9 }={ \left. {4{{\left( {\sqrt t } \right)}^3}} \right|_4^9 }={ 4\left( {{3^3} – {2^3}} \right) }={ 76.}$

### Example 3.

Assume the rate of investment is given by the function $$I\left( t \right) = \ln t .$$ Compute the total capital accumulation between the $$1^{\text{st}}$$ and the $$5^{\text{th}}$$ years.

Solution.

To calculate the capital accumulation, we use the formula

${K = \int\limits_a^b {I\left( t \right)dt} }={ \int\limits_1^5 {\ln tdt} .}$

Integrating by parts, we have

$\require{cancel}{\int {\ln tdt} = \left[ {\begin{array}{*{20}{l}} {u = \ln t}\\ {dv = dt}\\ {du = \frac{{dt}}{t}}\\ {v = t} \end{array}} \right] }={ t\ln t – \int {\cancel{t}\frac{{dt}}{\cancel{t}}} }={ t\ln t – \int {dt} }={ t\ln t – t.}$

Hence

${K = \left. {\left( {t\ln t – t} \right)} \right|_1^5 }={ \left( {5\ln 5 – 5} \right) – \left( {\ln 1 – 1} \right) }={ 5\ln 5 – 4 }\approx{ 4.05}$

### Example 4.

For a certain product, the demand function is $$D\left( Q \right) = 1000 – 25Q,$$ and the supply function is $$S\left( Q \right) = 100 + {Q^2}.$$ Compute the consumer and producer surplus.

Solution.

First we determine the equilibrium point by equating the demand and supply functions:

${D\left( Q \right) = S\left( Q \right),}\;\; \Rightarrow {1000 – 25Q = 100 + {Q^2},}\;\; \Rightarrow {{Q^2} + 25Q – 900 = 0.}$

The positive solution of the quadratic equation is $${Q_0} = 20.$$ The market equilibrium price is $${P_0} = 500.$$

The consumer surplus $$CS$$ is given by

${CS = \int\limits_0^{{Q_0}} {\left[ {D\left( Q \right) – {P_0}} \right]dQ} }={ \int\limits_0^{20} {\left( {1000 – 25Q – 500} \right)dQ} }={ \int\limits_0^{20} {\left( {500 – 25Q} \right)dQ} }={ \left. {\left( {500Q – \frac{{25{Q^2}}}{2}} \right)} \right|_0^{20} }={ 10000 – 5000 }={ 5000.}$

Similarly we find the producer surplus $$PS:$$

${PS = \int\limits_0^{{Q_0}} {\left[ {{P_0} – S\left( Q \right)} \right]dQ} }={ \int\limits_0^{20} {\left( {500 – 100 – {Q^2}} \right)dQ} }={ \int\limits_0^{20} {\left( {400 – {Q^2}} \right)dQ} }={ \left. {\left( {400Q – \frac{{{Q^3}}}{3}} \right)} \right|_0^{20} \approx 8000 – 2667 }={ 5333.}$

### Example 5.

Assuming the demand function is $$D\left( Q \right) = 50 – Q$$ and the supply function is $$S\left( Q \right) = 20 + \sqrt Q,$$ compute the consumer and producer surplus.

Solution.

First we determine the market equilibrium point.

${D\left( Q \right) = S\left( Q \right),}\;\; \Rightarrow {50 – Q = 20 + \sqrt Q ,}\;\; \Rightarrow {Q + \sqrt Q – 30 = 0.}$

Making the change $$z = \sqrt{Q},$$ we get the quadratic equation $${z^2} + z – 30 = 0,$$ which has the roots $$z = 5,$$ and $$z = -6.$$ The positive solution $$z = 5$$ gives the equilibrium point $${Q_0} = {z^2} = 25.$$ The price at this point is $${P_0} = 25.$$

Now we can calculate the consumer and producer surplus:

${CS = \int\limits_0^{{Q_0}} {\left[ {D\left( Q \right) – {P_0}} \right]dQ} }={ \int\limits_0^{25} {\left( {50 – Q – 25} \right)dQ} }={ \int\limits_0^{25} {\left( {25 – Q} \right)dQ} }={ \left. {\left( {25Q – \frac{{{Q^2}}}{2}} \right)} \right|_0^{25} }={ 625 – \frac{{625}}{2} }={ 312.5}$

${PS = \int\limits_0^{{Q_0}} {\left[ {{P_0} – S\left( Q \right)} \right]dQ} }={ \int\limits_0^{25} {\left( {25 – 20 – \sqrt Q } \right)dQ} }={ \int\limits_0^{25} {\left( {5 – {Q^{\frac{1}{2}}}} \right)dQ} }={ \left. {\left( {5Q – \frac{{2{Q^{\frac{3}{2}}}}}{3}} \right)} \right|_0^{25} }={ 125 – \frac{{250}}{3} }\approx{ 41.7}$

### Example 6.

Assume the demand and supply functions for a product are $$D\left( Q \right) = {\left( {Q – 2a} \right)^2}$$ and $$S\left( Q \right) = {Q^2},$$ where $$a \gt 0$$ is a parameter. Compute the consumer and producer surplus.

Solution.

First we find the point of market equilibrium:

${D\left( Q \right) = S\left( Q \right),}\;\; \Rightarrow {{\left( {Q – 2a} \right)^2} = {Q^2},}\;\; \Rightarrow {\left| {Q – 2a} \right| = \left| Q \right|.}$

This equation has the single solution $$Q = a.$$ So the market equilibrium is at the point $$\left( {{Q_0},{P_0}} \right) = \left( {a,{a^2}} \right).$$

Calculate the consumer surplus $$CS:$$

${CS = \int\limits_0^{{Q_0}} {\left[ {D\left( Q \right) – {P_0}} \right]dQ} }={ \int\limits_0^a {\left[ {{{\left( {Q – 2a} \right)}^2} – {a^2}} \right]dQ} }={ \int\limits_0^a {\left( {{Q^2} – 4aQ + 4{a^2} – {a^2}} \right)dQ} }={ \int\limits_0^a {\left( {{Q^2} – 4aQ + 3{a^2}} \right)dQ} }={ \left. {\left( {\frac{{{Q^3}}}{3} – 2a{Q^2} + 3{a^2}Q} \right)} \right|_0^a }={ \frac{{{a^3}}}{3} – 2{a^3} + 3{a^3} }={ \frac{{4{a^3}}}{3}.}$

Similarly we determine the producer surplus $$PS:$$

${PS = \int\limits_0^{{Q_0}} {\left[ {{P_0} – S\left( Q \right)} \right]dQ} }={ \int\limits_0^a {\left( {{a^2} – {Q^2}} \right)dQ} }={ \left. {\left( {{a^2}Q – \frac{{{Q^3}}}{3}} \right)} \right|_0^a }={ {a^3} – \frac{{{a^3}}}{3} }={ \frac{{2{a^3}}}{3}.}$

### Example 7.

Calculate the Gini coefficient for the Lorenz function $$L\left( x \right) = {x^3}.$$

Solution.

Substituting $$L\left( x \right) = {x^3}$$ and evaluating the integral, we find:

${G = 2\int\limits_0^1 {\left[ {x – L\left( x \right)} \right]dx} }={ 2\int\limits_0^1 {\left( {x – {x^3}} \right)dx} }={ 2\left. {\left( {\frac{{{x^2}}}{2} – \frac{{{x^4}}}{4}} \right)} \right|_0^1 }={ 2\left( {\frac{1}{2} – \frac{1}{4}} \right) }={ 0.50}$

### Example 8.

Calculate the Gini coefficient for the Lorenz function $$L\left( x \right) = {x^p},$$ where $$p \gt 1.$$

Solution.

Using the formula

$G = 2\int\limits_0^1 {\left[ {x – L\left( x \right)} \right]dx},$

we obtain

${G\left({p}\right) = 2\int\limits_0^1 {\left[ {x – L\left( x \right)} \right]dx} }={ 2\int\limits_0^1 {\left( {x – {x^p}} \right)dx} }={ 2\left. {\left( {\frac{{{x^2}}}{2} – \frac{{{x^{p + 1}}}}{{p + 1}}} \right)} \right|_0^1 }={ 2\left( {\frac{1}{2} – \frac{1}{{p + 1}}} \right) }={ 1 – \frac{2}{{p + 1}}.}$

In particular,

${G(p = 2) = 1 – \frac{2}{{2 + 1}} }={ \frac{1}{3} }\approx{ 0.33;}$

${G(p = 2) = 1 – \frac{2}{{3 + 1}} }={ \frac{1}{2} }={ 0.50;}$

${G(p = 4) = 1 – \frac{2}{{4 + 1}} }={ \frac{3}{5} }={ 0.60;}$

### Example 9.

Suppose the Lorenz curve for a society is given by $$L\left( x \right) = \large{\frac{3}{5}}\normalsize {x^3} + \large{\frac{1}{5}}\normalsize {x^2} + \large{\frac{1}{5}}\normalsize x.$$ Find the Gini coefficient for this income distribution.

Solution.

We use the integration formula

$G = 2\int\limits_0^1 {\left[ {x – L\left( x \right)} \right]dx}.$

The Gini index is then given by

${G \text{ = }}\kern0pt{2\int\limits_0^1 {\left[ {x – \left( {\frac{3}{5}{x^3} + \frac{1}{5}{x^2} + \frac{1}{5}x} \right)} \right]dx} }={ 2\int\limits_0^1 {\left( {\frac{4}{5}x – \frac{3}{5}{x^3} – \frac{1}{5}{x^2}} \right)dx} }={ \frac{2}{5}\int\limits_0^1 {\left( {4x – 3{x^3} – {x^2}} \right)dx} }={ \frac{2}{5}\left. {\left( {2{x^2} – \frac{{3{x^4}}}{4} – \frac{{{x^3}}}{3}} \right)} \right|_0^1 }={ \frac{2}{5}\left( {2 – \frac{3}{4} – \frac{1}{3}} \right) }={ \frac{{11}}{{30}} }\approx{ 0.37}$

### Example 10.

Suppose the Lorenz curve for a country is given by $$L\left( x \right) = 1 – \sqrt {1 – {x^2}}.$$ Determine the Gini coefficient for this income distribution.

Solution.

We compute the Gini coefficient using the formula

$G = 2\int\limits_0^1 {\left[ {x – L\left( x \right)} \right]dx} .$

This yields:

${G = 2\int\limits_0^1 {\left[ {x – \left( {1 – \sqrt {1 – {x^2}} } \right)} \right]dx} }={ 2\int\limits_0^1 {\left( {x – 1} \right)dx} + 2\int\limits_0^1 {\sqrt {1 – {x^2}} dx} }={ {I_1} + {I_2}.}$

Evaluate both integrals separately:

${{I_1} = 2\int\limits_0^1 {\left( {x – 1} \right)dx} }={ 2\left. {\left( {\frac{{{x^2}}}{2} – x} \right)} \right|_0^1 }={ 2\left( {\frac{1}{2} – 1} \right) }={ – 1.}$

To solve the second integral, we make the substitution:

${x = \sin t,\;\;}\kern0pt{dx = \cos tdt.}$

When $$x = 0,$$ $$t = 0,$$ and when $$x = 1,$$ $$t = \large{\frac{\pi }{2}}\normalsize.$$ So

${{I_2} = 2\int\limits_0^1 {\sqrt {1 – {x^2}} dx} }={ 2\int\limits_0^{\frac{\pi }{2}} {\sqrt {1 – {{\sin }^2}t} \cos tdt} }={ 2\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}tdt} }={ \int\limits_0^{\frac{\pi }{2}} {\left( {1 + \cos 2t} \right)dt} }={ \left. {\left( {t + \frac{{\sin 2t}}{2}} \right)} \right|_0^{\frac{\pi }{2}} }={ \frac{\pi }{2}.}$

Hence, the Gini coefficient is approximately equal to

$G = – 1 + \frac{\pi }{2} \approx 0.57$