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

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BUSINESS MATHEMATICS

Applications of integrals

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

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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 𝑐 𝑦𝑦 + Δ𝑦

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AREAS UNDER A CURVE Let 𝑦 run from 𝑏 to 𝑐 in 𝑜 steps ▪ clearly Δ𝑦 =

𝑐−𝑏 𝑜

▪ each rectangle starts at 𝑦𝑗 = 𝑏 + 𝑗 − 1 Δ𝑦 Then 𝐵 = σ𝑗=1

𝑜

Δ𝐵𝑗 = σ𝑗=1

𝑜

𝑔 𝑦𝑗 Δ𝑦 ▪ Riemann sum

𝑏 𝑐

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AREAS UNDER A CURVE Now take Δ𝑦 very small (so 𝑜 very large) 𝐵 = lim

Δ𝑦→0 ෍ 𝑗=1 𝑜

𝑔 𝑦𝑗 Δ𝑦 = න

𝑏 𝑐

𝑔 𝑦 𝑒𝑦 The Riemann sum provides one way of proving that the area under a non-negative function 𝑔 𝑦 within in an interval 𝑏, 𝑐 (or 𝑏, 𝑐 ) is ׬

𝑏 𝑐 𝑔 𝑦 𝑒𝑦

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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 ▪ so Δ𝐵 𝑦 ≈ 𝑔 𝑦 Δ𝑦 and

Δ𝐵 𝑦 Δ𝑦

≈ 𝑔 𝑦 ▪ so in the limit lim

Δ𝑦→0 Δ𝐵 𝑦 Δ𝑦

=

𝑒𝐵 𝑦 𝑒𝑦

= 𝑔 𝑦 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

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AREAS UNDER A CURVE Example: let 𝑔 𝑦 = 𝑦2 + 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) ... is 𝐵 = ׬

1 2 𝑔 𝑦 𝑒𝑦

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AREAS UNDER A CURVE Example (continued): ▪ ׬ 𝑔 𝑦 𝑒𝑦 = ׬ 𝑦2 + 5 𝑒𝑦 =

1 3 𝑦3 + 5𝑦 + 𝐷

▪ so 𝐵 = ׬

1 2 𝑔 𝑦 𝑒𝑦 = 1 3 𝑦3 + 5𝑦 1 2

= ▪ =

1 3 23 + 5 ⋅ 2 − 1 3 13 + 5 ⋅ 1 = 7 1 3

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EXERCISE 1 Given 𝑧 = 𝑓3𝑦, compute the area enclosed by the function graph, 𝑦 = 0, 𝑦 = 1 and 𝑧 = 0.

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EXERCISE 2 Given 𝑧 = 𝑓3𝑦, compute the area enclosed by the function graph, 𝑦 = 0, 𝑦 = 1 and 𝑧 = −2.

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AREAS UNDER A CURVE Let’s compute ׬

−1 1 𝑦𝑒𝑦 =

ቃ

1 2 𝑦2 −1 1

=

1 2 − 1 2 = 0

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: ▪ 𝐵 = ׬

−1 0 −𝑦 𝑒𝑦 + ׬ 1 𝑦𝑒𝑦 =

ቃ

1 2 𝑦2 −1

− ቃ

1 2 𝑦2 1

= 1

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EXERCISE 3 Given 𝑧 = 𝑦3 − 4𝑦2 + 3𝑦, compute the area enclosed by the function graph, 𝑦 = −1, 𝑦 = 4 and 𝑧 = 0.

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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∗

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CONSUMER AND PRODUCER SURPLUS Define the consumer surplus 𝐷𝑇 = න

𝑅∗

𝑔 𝑅 − 𝑄∗ 𝑒𝑅 and the producer surplus 𝑄𝑇 = න

𝑅∗

𝑄∗ − 𝑕 𝑅 𝑒𝑅

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CONSUMER AND PRODUCER SURPLUS

Example: ▪ 𝑔 𝑅 = 1 +

12 𝑅+1

▪ 𝑕 𝑅 = 𝑅2 + 1 Equilibrium point: ▪ 1 +

12 𝑅∗+1 = 𝑅∗2 + 1 ⇒ 𝑅∗2 𝑅∗ + 1 = 12 ⇒

▪ 𝑅∗, 𝑄∗ = 2,5 Surplus: ▪ 𝐷𝑇 = ׬

2 12 𝑅+1 − 4 𝑒𝑅 =

ሿ 12 ln 𝑅 + 1 − 4𝑅

2 = 12 ln 3 −

8 ≈ 5.18 ▪ 𝑄𝑇 = ׬

2 4 − 𝑅2 𝑒𝑅 =

ቃ 4𝑅 −

𝑅3 3 2

=

16 3 ≈ 5.33

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FROM MARGINAL TO TOTAL Given a cost function, we can calculate the marginal cost with a derivative ▪ example: 𝐷 𝑟 = 20 + 4𝑟0.6 ▪ 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 = 4𝑟0.6 + 𝐷 ▪ unique up to the integration constant 𝐷

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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?

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OLD EXAM QUESTION 9 December 2015, Q2c

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OLD EXAM QUESTION 27 March 2015, Q1i

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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