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Algebra Review & Applications, Mathematical Series, Differentiation and Integration

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This assignment is a comprehensive and concise resource that focuses on Algebra Review and Applications, Mathematical Series, Differentiation, and Integration. This document provides simple examples of algebraic concepts through their applications, including mathematical series on arithmetic and geometric progression. The resource also covers differentiation and integration techniques, through the example of supply and demand changes. It further explains the cost function, profit function, and profit maximization of a monopoly using differentiation. It then takes cigarette consumption as an example to calculate consumer surplus and producer surplus using integration methods. Students studying basic mathematics can benefit from this document as it provides a clear and concise explanation of mathematical concepts in their applications in various contexts.

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UNIVERSITY OF PERADENIYA




Algebra Review and Applications
Mathematical Series
Differentiation and Integration




Name Thevuni Kotigala

University University of Peradeniya, Sri Lanka

Faculty Faculty of Arts

Programme Masters in Development Practice (MDP)

Course Year 2020/21

Academic Year Pre-MDP

Lecturer Dr. S.J.S. de Mel

Department Department of Economics & Statistics

Subject Code DPR 402

Subject Name Basic Mathematics

Medium English
Algebra Review & Applications, Mathematical Series,
Topic
Differentiation, Integration
Word Count 438 words

, 1. (i) Find the intersection points of the two graphs, y = x 2 − 6 x + 8 and y = x + 2
through the solution of simultaneous equations.

𝑦 = 𝑥 2 − 6𝑥 + 8 → (1) 𝑦 = 𝑥 + 2 → (2)

(1) – (2) → 𝑦 − 𝑦 = 𝑥 2 − 6𝑥 + 8 − 𝑥 − 2

𝑥 2 − 7𝑥 + 6 = 0

𝑥 2 − 6𝑥 − 𝑥 + 6 = 0

𝑥(𝑥 − 6) − 1(𝑥 − 6) = 0

(𝑥 − 6)(𝑥 − 1) = 0
Therefore,

𝑥−6=0 OR 𝑥−1=0

𝑥=6 𝑥=1
AND AND

𝑦 = 6+2 𝑦 = 1+2

𝑦=8 𝑦=3

Intersection points are, (6, 8) and (1, 3)


(ii) Find the sum of the following series:

(a) 3, 7, 11, 15, 19, 23 up to the 20th term

Arithmetic Progression: 𝑎1 = 3, 𝑑 = 4, 𝑛 = 20
𝑛
𝑆𝑛 = [2𝑎1 + (𝑛 − 1)𝑑]
2
20
𝑆20 = [2 × 3 + (20 − 1)4]
2
𝑆20 = 820
(b) 1, -3, 9, -27, 81 up to the 10th term

Geometric Progression: 𝑎 = 1, 𝑛 = 10, 𝑟 = −3 (𝑟 < 1)
𝑎(1 − 𝑟 𝑛 )
𝑆𝑛 =
(1 − 𝑟)

1 (1 − (−3)10 )
𝑆10 =
(1 − (−3))
1 (1 − (59049))
𝑆10 =
(1 + 3)

𝑆10 = −14762
1

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