Calculus

Applications of Integrals

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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 point of equilibrium is the intersection of demand and supply curves.
Figure 1.

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 and producer surplus.
Figure 2.

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.

Lorenz curve and Gini coefficient
Figure 3.

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

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