MATHEMATICS 1 CONTENTS Areas under a curve Consumer and producer - PowerPoint PPT Presentation

Applications of integrals BUSINESS MATHEMATICS 1

CONTENTS Areas under a curve Consumer and producer surplus From marginal to total Old exam question Further study 2

AREAS UNDER A CURVE How do we compute the area 𝐵 under the graph of a non- negative function 𝑔 over the interval 𝑏, 𝑐 ? Take a point 𝑦 in the interval 𝑏, 𝑐 Move to a point slightly further, at 𝑦 + Δ𝑦 Form a small rectangle enclosed by these two points, the horizontal axis, and the curve ▪ width Δ𝑦 ▪ height 𝑔 𝑦 ▪ area Δ𝐵 = 𝑔 𝑦 Δ𝑦 𝑏 𝑦𝑦 + Δ𝑦 𝑐 3

AREAS UNDER A CURVE Let 𝑦 run from 𝑏 to 𝑐 in 𝑜 steps 𝑐−𝑏 ▪ clearly Δ𝑦 = 𝑜 ▪ each rectangle starts at 𝑦 𝑗 = 𝑏 + 𝑗 − 1 Δ𝑦 𝑜 𝑜 Then 𝐵 = σ 𝑗=1 Δ𝐵 𝑗 = σ 𝑗=1 𝑔 𝑦 𝑗 Δ𝑦 ▪ Riemann sum 𝑏 𝑐 4

AREAS UNDER A CURVE Now take Δ𝑦 very small (so 𝑜 very large) 𝑐 𝑜 𝐵 = lim Δ𝑦→0 ෍ 𝑔 𝑦 𝑗 Δ𝑦 = න 𝑔 𝑦 𝑒𝑦 𝑗=1 𝑏 The Riemann sum provides one way of proving that the area under a non-negative function 𝑔 𝑦 within in an interval 𝑐 𝑔 𝑦 𝑒𝑦 𝑏, 𝑐 (or 𝑏, 𝑐 ) is ׬ 𝑏 5

AREAS UNDER A CURVE Another way of proving this: ▪ suppose 𝐵 𝑦 is the area between 𝑏 and 𝑦 . ▪ take 𝑦 + Δ𝑦 , the area is 𝐵 𝑦 + Δ𝑦 ≈ 𝐵 𝑦 + 𝑔 𝑦 Δ𝑦 Δ𝐵 𝑦 ▪ so Δ𝐵 𝑦 ≈ 𝑔 𝑦 Δ𝑦 and ≈ 𝑔 𝑦 Δ𝑦 Δ𝐵 𝑦 𝑒𝐵 𝑦 ▪ so in the limit lim = = 𝑔 𝑦 Δ𝑦 𝑒𝑦 Δ𝑦→0 In words: ▪ the derivative of the area function 𝐵 is the function 𝑔 ▪ the area function 𝐵 is a primitive function of the function 𝑔 6

AREAS UNDER A CURVE Example: let 𝑔 𝑦 = 𝑦 2 + 5 , with integration boundaries 1,2 ▪ check that 𝑔 𝑦 ≥ 0 within 1,2 The area enclosed by: ▪ the 𝑦 -axis ( 𝑧 = 0 ) ▪ the function ( 𝑧 = 𝑔 𝑦 ) ▪ the lower boundary ( 𝑦 = 1 ) ▪ the upper boundary ( 𝑦 = 2 ) 2 𝑔 𝑦 𝑒𝑦 ... is 𝐵 = ׬ 1 7

AREAS UNDER A CURVE Example (continued): ▪ ׬ 𝑔 𝑦 𝑒𝑦 = ׬ 𝑦 2 + 5 𝑒𝑦 = 1 3 𝑦 3 + 5𝑦 + 𝐷 2 2 𝑔 𝑦 𝑒𝑦 = 3 𝑦 3 + 5𝑦 1 ▪ so 𝐵 = ׬ = 1 1 1 1 1 3 2 3 + 5 ⋅ 2 − 3 1 3 + 5 ⋅ 1 = 7 ▪ = 3 8

EXERCISE 1 Given 𝑧 = 𝑓 3𝑦 , compute the area enclosed by the function graph, 𝑦 = 0 , 𝑦 = 1 and 𝑧 = 0 . 9

EXERCISE 2 Given 𝑧 = 𝑓 3𝑦 , compute the area enclosed by the function graph, 𝑦 = 0 , 𝑦 = 1 and 𝑧 = −2 . 11

AREAS UNDER A CURVE 1 1 𝑦𝑒𝑦 = 1 1 1 Let’s compute ׬ 2 𝑦 2 ቃ = 2 − 2 = 0 −1 −1 Note that 0 is not the area If 𝑔 𝑦 takes a negative value, area and definite integral are not equal In order to compute the area one proceeds as follows: 0 1 0 −𝑦 𝑒𝑦 + ׬ 1 𝑦𝑒𝑦 = 1 1 2 𝑦 2 2 𝑦 2 ▪ 𝐵 = ׬ ቃ − ቃ = 1 −1 0 −1 0 13

EXERCISE 3 Given 𝑧 = 𝑦 3 − 4𝑦 2 + 3𝑦 , compute the area enclosed by the function graph, 𝑦 = −1 , 𝑦 = 4 and 𝑧 = 0 . 14

CONSUMER AND PRODUCER SURPLUS Let 𝑔 𝑅 denote the demand curve ▪ consumers will buy quantity 𝑅 if the price is 𝑔 𝑅 Let 𝑕 𝑅 denote the supply curve ▪ producers will produce quantity 𝑅 if the price is 𝑕 𝑅 Let 𝐹 = 𝑅 ∗ , 𝑄 ∗ be the equilibrium point ▪ where 𝑔 𝑅 ∗ = 𝑄 ∗ = 𝑕 𝑅 ∗ 16

CONSUMER AND PRODUCER SURPLUS Define the consumer surplus 𝑅 ∗ 𝑔 𝑅 − 𝑄 ∗ 𝑒𝑅 𝐷𝑇 = න 0 and the producer surplus 𝑅 ∗ 𝑄 ∗ − 𝑕 𝑅 𝑄𝑇 = න 𝑒𝑅 0 17

CONSUMER AND PRODUCER SURPLUS Example: 12 ▪ 𝑔 𝑅 = 1 + 𝑅+1 ▪ 𝑕 𝑅 = 𝑅 2 + 1 Equilibrium point: 𝑅 ∗ +1 = 𝑅 ∗2 + 1 ⇒ 𝑅 ∗2 𝑅 ∗ + 1 = 12 ⇒ 12 ▪ 1 + 𝑅 ∗ , 𝑄 ∗ = 2,5 ▪ Surplus: 2 2 = 12 ln 3 − 12 ▪ 𝐷𝑇 = ׬ ሿ 𝑅+1 − 4 𝑒𝑅 = 12 ln 𝑅 + 1 − 4𝑅 0 0 8 ≈ 5.18 2 2 4 − 𝑅 2 𝑒𝑅 = 𝑅 3 16 ▪ 𝑄𝑇 = ׬ 4𝑅 − ቃ = 3 ≈ 5.33 0 3 0 18

FROM MARGINAL TO TOTAL Given a cost function, we can calculate the marginal cost with a derivative ▪ example: 𝐷 𝑟 = 20 + 4𝑟 0.6 ▪ marginal costs: 𝐷 ′ 𝑟 = 2.4𝑟 −0.4 But suppose we know the marginal costs and need the cost function ▪ example: 𝐷 ′ 𝑟 = 2.4𝑟 −0.4 ▪ cost function: 𝐷 𝑟 = ׬ 𝐷 ′ 𝑟 𝑒𝑟 = 4𝑟 0.6 + 𝐷 ▪ unique up to the integration constant 𝐷 19

EXERCISE 4 An oil tank has been leaking with a flow rate 𝜚 𝑢 = 4𝑓 −𝑢 (liters/s) starting at 𝑢 = 0 . Each liter leaked represents a damage of 50$. How much damage has occurred 𝑢 = 𝑈 ? 20

OLD EXAM QUESTION 9 December 2015, Q2c 22

OLD EXAM QUESTION 27 March 2015, Q1i 23

FURTHER STUDY Sydsæter et al. 5/E 9.4 Tutorial exercises week 5 area under a curve area between two curves consumer surplus as an area surplus as an integral 24