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Last updated at Dec. 12, 2019 by Teachoo

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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.