# Misc 14

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Misc 14 Using the method of integration find the area of the region bounded by lines: 2𝑥 + 𝑦 = 4, 3𝑥–2𝑦=6 and 𝑥–3𝑦+5=0 Step 1: Make Figure 2𝑥 + 𝑦 = 4 3𝑥 – 2𝑦 = 6 𝑥 – 3𝑦 + 5 = 0 Step 2: Find intersecting Points A & B Point A Point A is intersection of lines x – 3y + 5 = 0 & 2x + y = 4 Now, x – 3y + 5 = 0 x = 3y – 5 Putting x = 3y – 5 in 2x + y = 4 2(3y – 5) + y = 4 6y – 10 + y = 4 7y = 14 y = 2 Putting y = 2 in x – 3y + 5 = 0 x – 3(2) + 5 = 0 x – 6 + 5 = 0 x = 1 So, point A (1, 2) Point B Point B is intersection of lines x – 3y + 5 = 0 & 3x – 2y = 6 Now, x – 3y + 5 = 0 x = 3y – 5 Putting x = 3y – 5 in 3x – 2y = 6 3(3y – 5) – 2y = 6 9y – 15 – 2y = 6 7y = 21 y = 3 Putting y = 3 in x – 3y + 5 = 0 x – 3(3) + 5 = 0 x – 9 + 5 = 0 x = 4 So, point A (4, 3) Finding area Area ABED Area ABED = 14𝑦 𝑑𝑥 𝑦→ Equation of AB 𝑥 – 3𝑦+5=0 𝑥+5=3𝑦 𝑥 + 53=𝑦 𝑦= 𝑥 + 53 Therefore, Area ABED = 14 𝑥+53 𝑑𝑥 = 13 14 𝑥+5 𝑑𝑥 = 13 𝑥22+5𝑥14 = 13 422+5.4− 122+5.1 = 13 8+20− 12−5 = 13 452 = 152 Area ACD Area ACD = 12𝑦 𝑑𝑥 𝑦→ Equation of line AC 2𝑥+𝑦=4 𝑦=4−2𝑥 Area ACD = 12 4−2𝑥 𝑑𝑥 = 4𝑥− 2 𝑥2212 = 4𝑥− 𝑥212 = 4.2− 22− 4.1− 12 = 8−4−4+1 = 1 Area CBE Area CBE = 24𝑦 𝑑𝑥 𝑦→ Equation of line BC 3𝑥+2𝑦=6 3𝑥−6=2𝑦 3𝑥 − 62=𝑦 𝑦= 3𝑥 − 62 Therefore, Area CBE = 24 3𝑥 − 62 𝑑𝑥 = 12 24 3𝑥−6 𝑑𝑥 = 12 3 𝑥22−6𝑥24 = 12 3.422−6.4− 3.222−6.2 = 12 24−24−6+12 =3 Hence Area Required = Area ABED – Area ACD – Area CBE = 152−1−3 = 152−4 = 15 − 82 = 72 square units

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .