Applications of Integrals in Economics
Solved Problems
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.
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 get:
\[G = 2\int\limits_0^1 {\left[ {x - L\left( x \right)} \right]dx} = 2\int\limits_0^1 {\left( {x - {x^3}} \right)dx} = 2\left. {\left( {\frac{{{x^2}}}{2} - \frac{{{x^4}}}{4}} \right)} \right|_0^1 = 2\left( {\frac{1}{2} - \frac{1}{4}} \right) = 0.50\]
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,
\[G(p = 2) = 1 - \frac{2}{{2 + 1}} = \frac{1}{3} \approx 0.33;\]
\[G(p = 3) = 1 - \frac{2}{{3 + 1}} = \frac{1}{2} = 0.50;\]
\[G(p = 4) = 1 - \frac{2}{{4 + 1}} = \frac{3}{5} = 0.60;\]
Example 9.
Suppose the Lorenz curve for a society is given by \[L\left( x \right) = \frac{3}{5} {x^3} + \frac{1}{5} {x^2} + \frac{1}{5} x.\] 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
\[G = 2\int\limits_0^1 {\left[ {x - \left( {\frac{3}{5}{x^3} + \frac{1}{5}{x^2} + \frac{1}{5}x} \right)} \right]dx} = 2\int\limits_0^1 {\left( {\frac{4}{5}x - \frac{3}{5}{x^3} - \frac{1}{5}{x^2}} \right)dx} = \frac{2}{5}\int\limits_0^1 {\left( {4x - 3{x^3} - {x^2}} \right)dx} = \frac{2}{5}\left. {\left( {2{x^2} - \frac{{3{x^4}}}{4} - \frac{{{x^3}}}{3}} \right)} \right|_0^1 = \frac{2}{5}\left( {2 - \frac{3}{4} - \frac{1}{3}} \right) = \frac{{11}}{{30}} \approx 0.37\]
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:
\[x = \sin t,\;\; dx = \cos tdt.\]
When \(x = 0,\) \(t = 0,\) and when \(x = 1,\) \(t = \frac{\pi }{2}.\) 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
\[G = - 1 + \frac{\pi }{2} \approx 0.57\]
