# Misc 14 - Chapter 8 Class 12 Application of Integrals

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: 2 + = 4, 3 2 =6 and 3 +5=0 Step 1: Make Figure 2 + = 4 3 2 = 6 3 + 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 = 1 4 Equation of AB 3 +5=0 +5=3 + 5 3 = = + 5 3 Therefore, Area ABED = 1 4 +5 3 = 1 3 1 4 +5 = 1 3 2 2 +5 1 4 = 1 3 4 2 2 +5.4 1 2 2 +5.1 = 1 3 8+20 1 2 5 = 1 3 45 2 = 15 2 Area ACD Area ACD = 1 2 Equation of line AC 2 + =4 =4 2 Area ACD = 1 2 4 2 = 4 2 2 2 1 2 = 4 2 1 2 = 4.2 2 2 4.1 1 2 = 8 4 4+1 = 1 Area CBE Area CBE = 2 4 Equation of line BC 3 +2 =6 3 6=2 3 6 2 = = 3 6 2 Therefore, Area CBE = 2 4 3 6 2 = 1 2 2 4 3 6 = 1 2 3 2 2 6 2 4 = 1 2 3.4 2 2 6.4 3.2 2 2 6.2 = 1 2 24 24 6+12 =3 Hence Area Required = Area ABED Area ACD Area CBE = 15 2 1 3 = 15 2 4 = 15 8 2 = 7 2 square units

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.