Essential Mathematics for Economics

Integration

Outline
 Several concepts help to quantify benefits from
consumption and costs from production
 Antiderivatives

 Integrals

 Fundamental Theorem of Calculus

1
Consumers’ Surplus
 We consider a measure of the total value to
consumers from consuming a good
 First, we determine the gross Consumers’ Surplus
 It equals the total value
 Then, we determine the net Consumers’ Surplus
 It equals the value after accounting for the
expenditure

100

60

40 100

2
Determining areas

ℎ
2

Determining areas

h2

h1

3
Notation
 We will see three related concepts in this
section:
‫׬‬ = antiderivative
𝑏
‫׬‬
𝑎
= definite integral
𝑥
‫׬‬
𝑎
= indefinite integral

Antiderivatives
 An antiderivative of 𝑓(𝑥) is a function 𝐹(𝑥)
satisfying the property 𝐹 ′ 𝑥 = 𝑓(𝑥)
 Antiderivatives of 𝑓(𝑥) differ by a constant
 If 𝑓 𝑥 = 2𝑥, its antiderivatives are of the form
𝐹 𝑥 = ‫ 𝑥 = 𝑥𝑑 𝑥 𝑓 ׬‬2 + 𝑐

4
 Finding an antiderivative

𝑛
𝑥 𝑛+1
1. Power Rule න𝑥 𝑑𝑥 = +𝑐
𝑛+1
𝑥 𝑛+1
2. Multiplicative constant ‫𝐴 = 𝑥𝑑 𝑛 𝑥𝐴 ׬‬ +𝑐
𝑛+1

𝑓(𝑥) 𝑛+1
3. Function of a function න 𝑓(𝑥) 𝑛 𝑓′(𝑥)𝑑𝑥 = +𝑐
𝑛+1

4. Exponential function න𝑒 𝑓(𝑥) 𝑓′(𝑥)𝑑𝑥 = 𝑒 𝑓(𝑥) + 𝑐

𝑓′(𝑥)
5. Logarithmic function න 𝑑𝑥 = ln(𝑓 𝑥 ) + 𝑐
𝑓(𝑥)

Examples
𝑥3
 Power Rule: ‫ 𝑥 ׬‬2 𝑑𝑥 = +𝑐
3
 Multiplicative constant: ‫ ׬‬20𝑥 3 𝑑𝑥 = 5𝑥 4 + 𝑐
 Function of a function: ‫ 𝑥(׬‬3 + 5)3𝑥 2 𝑑𝑥 =
(𝑥 3 +5)2
+𝑐
2

5
Examples
2 2
 Exponential function: ‫ ׬‬2𝑥𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 + 𝑐
2𝑥+3
 Logarithmic function: ‫ ׬‬2 𝑑𝑥 =
𝑥 +3𝑥+5
ln(𝑥 2 + 3𝑥 + 5) + 𝑐

The definite integral

 The definite integral of a function over the
𝑏
interval [a,b] is written ‫𝑥𝑑 𝑥 𝑓 𝑎׬‬
 It is the area under the graph of f, from a to b

6
The definite integral

𝑓(𝑥)

a b
𝑋

The definite integral

𝑓 𝑥𝑟 𝑥𝑟+1 − 𝑥𝑟
𝑓(𝑥𝑟 )

a 𝑥𝑟 b
𝑋

7
The definite integral
 The typical rectangle has area 𝑓 𝑥𝑟 𝑥𝑟+1 − 𝑥𝑟
 With n rectangles the combined area is:
σ𝑛𝑟=1 𝑓 𝑥𝑟 𝑥𝑟+1 − 𝑥𝑟
 As n goes to infinity, the sum approaches a
limiting value:
𝑏
lim σ𝑛𝑟=1 𝑓 𝑥𝑟 𝑥𝑟+1 − 𝑥𝑟 = ‫𝑥𝑑)𝑥(𝑓 𝑎׬‬
𝑛→∞

The indefinite integral

𝑥
 In the indefinite integral ‫𝑧𝑑 𝑧 𝑓 𝑎׬‬, a limit of
integration is a variable
 This indefinite integral equals the area under the
graph, from a to 𝑥

8
Fundamental Theorem of Calculus

 Let 𝑓(𝑥) be a continuous function on the interval

[a, b]
 The Fundamental Theorem of Calculus says that if
𝐹 ′ 𝑥 = 𝑓(𝑥), then:
𝑏
‫ )𝑏(𝐹 = 𝑥𝑑)𝑥(𝑓 𝑎׬‬− 𝐹(𝑎)

Fundamental Theorem of Calculus

 Consider the indefinite integral:

𝑥
‫ )𝑥(𝐹 = 𝑧𝑑)𝑧(𝑓 𝑎׬‬− 𝐹(𝑎)
 Now take the derivative of both sides with
respect to 𝑥:
𝑑 𝑥
𝑑𝑥
‫𝐹 = 𝑧𝑑)𝑧(𝑓 𝑎׬‬′(𝑥) = 𝑓(𝑥)

9
Integration

𝑓(𝑥)

a 𝑥

Total cost and total variable cost

 A firm incurs a total cost 𝑇𝐶(𝑞) = 𝑉(𝑞) + 𝑇𝐶0
 Then, 𝑀𝐶 𝑞 = 𝑉 ′ 𝑞
𝑏 𝑏
⇒ ‫𝑉 𝑎׬ = 𝑞𝑑)𝑞(𝐶𝑀 𝑎׬‬′(𝑞)𝑑𝑞
= 𝑉 𝑏 − 𝑉 𝑎 = 𝑇𝑉𝐶 𝑏 − 𝑇𝑉𝐶 𝑎
 The change in total variable cost equals the area
under the MC curve

10
Total revenue and marginal revenue

 Suppose that total revenue is 𝑇𝑅(𝑞) = 𝐹(𝑞)

 Then, 𝑀𝑅 𝑞 = 𝐹 ′ 𝑞 = 𝑓(𝑞)
 The change in total revenue equals the area
under the MR curve:
𝑏
‫𝑞𝑑)𝑞(𝑅𝑀 𝑎׬‬
𝑏
= ‫𝑞𝑑)𝑞(𝑓 𝑎׬‬
= 𝐹 𝑏 − 𝐹 𝑎 = 𝑇𝑅 𝑏 − 𝑇𝑅(𝑎)

Total revenue and marginal revenue

 A monopolist’s demand curve is 𝑝 = 100 − 𝑞

 Total revenue is:
𝑇𝑅(𝑞) = 𝑝𝑞 = 100 − 𝑞 𝑞 = 100𝑞 − 𝑞 2
 Marginal revenue is
𝑑𝑇𝑅 𝑞
𝑀𝑅 𝑞 = = 100 − 2𝑞
𝑑𝑞

11
Demand, total revenue and marginal
revenue
 Total revenue satisfies:
𝑇𝑅 𝑞 − 𝑇𝑅 0
𝑞
= ‫׬‬0 𝑀𝑅 𝑧 𝑑𝑧
𝑞
= ‫׬‬0 (100 − 2𝑧) 𝑑𝑧
𝑧=𝑞
= [100𝑧 − 𝑧 2 ]ȁ𝑧=0
= 100𝑞 − 𝑞 2
 We conclude that 𝑇𝑅 𝑞 = 100𝑞 − 𝑞 2

p,
MR
100

60

20

MR D

40 50 100

12
p,
MR
100

60

20

MR D

40 50 100

The area between curves

 The area between 𝑓(𝑥) and 𝑔(𝑥), from a to b, is:
𝑏
‫ 𝑞 𝑓 𝑎׬‬− 𝑔 𝑞 𝑑𝑞
𝑏 𝑏
= ‫ 𝑞𝑑)𝑞(𝑓 𝑎׬‬− ‫𝑞𝑑)𝑞(𝑔 𝑎׬‬

13
 y

𝑔(𝑥)

𝑓(𝑥)

The benefit to consumers

 The area under the demand curve 𝑝 = 100 − 𝑞
is:
𝑞
‫׬‬0 (100 − 𝑧) 𝑑𝑧
𝑧=𝑞
𝑧2
= [100𝑧 − ]ቚ
2 𝑧=0
𝑞2
= 100𝑞 −
2
 When q = 40, gross CS = 4,000 – 800 = 3,200

14
The benefit to consumers
 The area under the demand curve and above the
price (60) is:
𝑞
‫׬‬0 (100 − 𝑧 − 60) 𝑑𝑧
𝑧=𝑞
𝑧2
= [40𝑧 − ]ቚ
2 𝑧=0
𝑞2
= 40𝑞 −
2
 Given q = 40, net CS = 1,600 – 800 = 800

Consumers’ surplus
 Let the demand curve be 𝑝 = 𝑓(𝑞)
 If the price is p0 and the quantity is q0, the gross
𝑞
Consumers’ Surplus is ‫׬‬0 0 𝑓(𝑞)𝑑𝑞
 The net Consumers’ Surplus is
𝑞0 0 𝑞
‫׬‬0 [𝑓 𝑞 − 𝑝0 ]𝑑𝑞 = ‫׬‬0 𝑓(𝑞)𝑑𝑞 − 𝑝0 𝑞0

15
 p

A
p0
B
D
q0 q

Example
 The demand curve for Georgetown caps is
𝑝 = 1,000𝑞−0.5
 Find an antiderivative of 1,000𝑞 −0.5
 Find the value of the gross consumers’ surplus
when the quantity sold is q0

16
Example
 An antiderivative of 1,000𝑞 −0.5 is 2,000𝑞 0.5 + 𝑐
 The gross CS equals:
𝑞 𝑞0
‫׬‬0 1,000𝑞 −0.5 𝑑𝑞 = 2,000𝑞0.5 ห0 = 2,000𝑞00.5
0

 Find the value of the net CS when the quantity

sold is 2,500

Example
 When the quantity is q0, the gross CS equals:
2,000𝑞00.5
 When the quantity is 2,500 the gross CS equals:
2,000(2,500)0.5 = 2,000 50 = 100,000
 When q = 2,500 the price is:
1,000
𝑝 = 1,000𝑞−0.5 = = 20
50
 The value of the net CS is:
CS = 100,000 – 20(2,500) = 50,000

17
Example
 The net CS equals:
𝑞
‫׬‬0 [1,000𝑞 −0.5 − 20]𝑑𝑞
0

𝑞0
= [2,000𝑞0.5 − 20𝑞]ห0
= 2,000𝑞00.5 − 20𝑞0
 When q = 2,500 the net CS is:
CS = 100,000 – 20(2,500) = 50,000

Producers’ Surplus
 Consider a perfectly competitive market with
supply curve p = 2q + 8
 The area below the price and above the supply
curve is the Producers’ Surplus

18
Producers’ Surplus
S
p

p0

q0
q

Producers’ Surplus
 The Producers’ Surplus equals:
𝑞
0
‫׬‬0 [𝑝0 − (2𝑞 + 8)]𝑑𝑞
𝑞 𝑞
= ‫׬‬0 0 𝑝0 𝑑𝑞 − ‫׬‬0 0(2𝑞 + 8)𝑑𝑞
𝑞
= 𝑝0 𝑞0 − ‫׬‬0 0(2𝑞 + 8)𝑑𝑞

19
Present value of a series of payments

 Here is the value of receiving $𝑎 at the end of n

periods, with annual compounding:
𝑎 + 𝑎 1 + 𝑟 + ⋯ + 𝑎(1 + 𝑟)𝑛−1
𝑎 𝑛
= 1+𝑟 −1
𝑟
 Now consider continuous payments with
continuous compounding:
𝑛
‫׬‬0 𝑎 ∙ 𝑒 𝑟𝑥 𝑑𝑥

Present value of a series of payments

 The integral equals:

𝑛
‫׬‬0 𝑎 ∙ 𝑒 𝑟𝑥 𝑑𝑥
𝑎 𝑥=𝑛
= [ ∙ 𝑒 𝑟𝑥 ]ቚ
𝑟 𝑥=0
𝑎 𝑎
= ∙ 𝑒 𝑟𝑛 −
𝑟 𝑟
𝑎
= 𝑒 𝑟𝑛 − 1
𝑟

20
Tossing dice
 Consider rolling one die
 If you roll 𝑥, you will receive $𝑥
 Possible outcomes are 𝑥1 = 1, 𝑥2 = 2, 𝑥3 = 3,
𝑥4 = 4, 𝑥5 = 5, and 𝑥6 = 6
 The outcomes are equally likely, so outcome
1
𝑥𝑗 occurs with probability 𝑝𝑗 = , for all j
6

Tossing dice
 The probability that your roll is less than or
equal to 𝑟 is
𝑟
𝐹 𝑟 = σ𝑟𝑗=1 𝑝𝑗 =
6
 The expected value of a roll is:
σ6𝑗=1 𝑝𝑗 𝑥𝑗
= 𝑝1 𝑥1 + 𝑝2 𝑥2 + 𝑝3 𝑥3 + 𝑝4 𝑥4 + 𝑝5 𝑥5 +𝑝6 𝑥6
1
= 1 + 2 + 3 + 4 + 5 + 6 = 3.50
6

21
Expected value

Outcome Expected Sum

1 100 100
2 100 200
3 100 300
4 100 400
5 100 500
6 100 600
Total 600 2,100

Continuous distributions
 Suppose that you select a real number randomly
from the interval [a, b]
 We say you “draw a number” or “take a draw”
 The probability that your draw is in the interval
[𝑥, 𝑥 + ∆] is (approximately) equal to 𝑓 𝑥 ∙ ∆,
where 𝑓 𝑥 is the density function

22
Continuous distributions
 The probability that your draw is less than or
equal to 𝑥 is
𝑥
𝐹 𝑥 = ‫𝑧𝑑 𝑧 𝑓 𝑎׬‬
 The expected value of your draw is
𝑏
𝐸𝑉 = ‫𝑥𝑑 𝑥 𝑓𝑥 𝑎׬‬

Example
 Suppose that you take a draw from the interval
[0, 10]
3
 Let 𝑓 𝑧 = (10𝑧 − 𝑧 2 ) for z  [0, 10]
500
 The probability that a draw is in the interval [z, z+∆]
is equal to 𝑓 𝑧 ∙ ∆

23
Density

0.15

5 10

Example
 The probability that your draw is less than or
equal to 𝑥 is
𝑥 𝑥 3
𝐹 𝑥 = ‫׬‬0 𝑓 𝑧 𝑑𝑧 = ‫׬‬0 (10𝑧 − 𝑧 2 )𝑑𝑧
500
𝑥
3 𝑧3
= 5𝑧 2 − ቚ
500 3 0
3𝑥 2 𝑥3
= −
100 500

24
Example
 The expected value of the draw is:
10 100 3𝑥
𝐸𝑉 = ‫׬‬0 𝑥𝑓 𝑥 𝑑𝑥 = ‫׬‬0 (10𝑥 − 𝑥 2 )𝑑𝑥
500
10
10𝑥 3 3𝑥 4
= − ቚ = 20 − 15 = 5
500 2,000 0

Example
 Suppose that you take a draw from a uniform
distribution on [0, 100]
1
 Let 𝑓 𝑧 = for all z  [0, 100]
100
 The probability that a draw is in the interval [z, z+∆]
∆
is equal to 𝑓 𝑧 ∙ ∆ = 100

 Find the probability that your draw is less than

or equal to 𝑥, where 0 < 𝑥 < 100

25
 Example

Density

1
100

𝑥
100

Example
 The probability that your draw is less than or
equal to 𝑥 is
𝑥 𝑥 1
𝐹 𝑥 = ‫׬‬0 𝑓 𝑧 𝑑𝑧 = ‫׬‬0 𝑑𝑧
100
𝑧 𝑥 𝑥
= ቚ =
100 0 100

 Find the expected value of the draw

26
Example
 The expected value of the draw is:
100 100 𝑥
𝐸𝑉 = ‫׬‬0 𝑥𝑓 𝑥 𝑑𝑥 = ‫׬‬0 𝑑𝑥
100
100
𝑥2 1002 100
= ቚ = = = 50
200 0 200 2

The newsvendor problem

 A newsvendor sells copies of a daily newspaper
 The newsvendor buys the newspapers first (in
bulk), not knowing what the day’s demand will
be

27
The newsvendor problem
 The newsvendor sells about 50 papers on a
typical day
 The number could be as high as 90 or as low as 10
 The newsvendor purchases newspapers for
$0.50 and sells them for $2.50 apiece
 The newsvendor must decide how many
newspapers to buy at the beginning of the day

The newsvendor problem

 Consider buying one more newspaper
 The loss if the newspaper does not sell is the
price paid for it; i.e., $0.50
 The gain if the newspaper does sell is the profit
margin; i.e., $(2.50 – 0.50) = $2.00
 Let p denote the probability that the newspaper
will not be sold

28
The newsvendor problem
 The newsvendor will want to buy additional
newspapers if:
– 0.50 𝑝 + 2.00 1 – 𝑝 ≥ 0
2.00
⇒𝑝≤ = 0.8
2.00+0.50
 If the cost is c and the benefit is b, the inequality
becomes:
–𝑐 𝑝 + 𝑏 1 – 𝑝 ≥ 0
𝑏 1
⇒𝑝≤ = 𝑐
𝑏+𝑐 1+𝑏

Summary
 In this section, we have seen several concepts
 Antiderivatives

 Indefinite integrals
 Definite integrals
 Fundamental Theorem of Calculus
 We have seen several applications

29