# Misc 14 - Chapter 8 Class 12 Application of Integrals

Last updated at Dec. 12, 2019 by Teachoo

Last updated at Dec. 12, 2019 by Teachoo

Transcript

Misc 14 Using the method of integration find the area of the region bounded by lines: 2๐ฅ + ๐ฆ = 4, 3๐ฅโ2๐ฆ=6 and ๐ฅโ3๐ฆ+5=0 Plotting the 3 lines on the graph 2๐ฅ + ๐ฆ = 4 3๐ฅ โ 2๐ฆ = 6 ๐ฅ โ 3๐ฆ + 5 = 0 Find intersecting Points A & B Point A Point A is intersection of lines x โ 3y + 5 = 0 & 2x + y = 4 Now, x โ 3y + 5 = 0 x = 3y โ 5 Putting x = 3y โ 5 in 2x + y = 4 2(3y โ 5) + y = 4 6y โ 10 + y = 4 7y = 14 y = 2 Putting y = 2 in x โ 3y + 5 = 0 x โ 3(2) + 5 = 0 x โ 6 + 5 = 0 x = 1 So, point A (1, 2) Point B Point B is intersection of lines x โ 3y + 5 = 0 & 3x โ 2y = 6 Now, x โ 3y + 5 = 0 x = 3y โ 5 Putting x = 3y โ 5 in 3x โ 2y = 6 3(3y โ 5) โ 2y = 6 9y โ 15 โ 2y = 6 7y = 21 y = 3 Putting y = 3 in x โ 3y + 5 = 0 x โ 3(3) + 5 = 0 x โ 9 + 5 = 0 x = 4 So, point B is (4, 3) Finding area Area Required = Area ABED โ Area ACD โ Area CBE Area ABED Area ABED =โซ_1^4โใ๐ฆ ๐๐ฅใ ๐ฆโ Equation of AB ๐ฅ โ 3๐ฆ+5=0 ๐ฅ+5=3๐ฆ (๐ฅ + 5)/3=๐ฆ ๐ฆ=(๐ฅ + 5)/3 Therefore, Area ABED =โซ_1^4โใ((๐ฅ+5)/3) ๐๐ฅใ =1/3 โซ_1^4โใ(๐ฅ+5) ๐๐ฅใ =1/3 [๐ฅ^2/2+5๐ฅ]_1^4 =1/3 [4^2/2+5.4โ[1^2/2+5.1]] =1/3 [8+20โ1/2โ5] =1/3 [45/2] =15/2 Area ACD Area ACD =โซ_1^2โใ๐ฆ ๐๐ฅใ ๐ฆโ Equation of line AC 2๐ฅ+๐ฆ=4 ๐ฆ=4โ2๐ฅ Area ACD =โซ_1^2โใ(4โ2๐ฅ" " ) ๐๐ฅใ =[4๐ฅโ(2๐ฅ^2)/2]_1^2 =[4๐ฅโ๐ฅ^2 ]_1^2 =[4.2โ2^2โ[4.1โ1^2 ]] =[8โ4โ4+1] = 1 Area CBE Area CBE =โซ_2^4โใ๐ฆ ๐๐ฅใ ๐ฆโ Equation of line BC 3๐ฅ+2๐ฆ=6 3๐ฅโ6=2๐ฆ (3๐ฅ โ 6)/2=๐ฆ ๐ฆ=(3๐ฅ โ 6)/2 Therefore, Area CBE =โซ_2^4โใ((3๐ฅ โ 6)/2) ๐๐ฅใ =1/2 โซ_2^4โใ(3๐ฅโ6) ๐๐ฅใ =1/2 [(3๐ฅ^2)/2โ6๐ฅ]_2^4 =1/2 [ใ3.4ใ^2/2โ6.4โ[ใ3.2ใ^2/2โ6.2]] =1/2 [24โ24โ6+12] =3 Hence Area Required = Area ABED โ Area ACD โ Area CBE =15/2โ1โ3 =15/2โ4 =(15 โ 8)/2 =๐/๐ 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.