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  1. Chapter 8 Class 12 Application of Integrals
  2. Serial order wise

Transcript

Misc 14 Using the method of integration find the area of the region bounded by lines: 2๐‘ฅ + ๐‘ฆ = 4, 3๐‘ฅโ€“2๐‘ฆ=6 and ๐‘ฅโ€“3๐‘ฆ+5=0 Plotting the 3 lines on the graph 2๐‘ฅ + ๐‘ฆ = 4 3๐‘ฅ โ€“ 2๐‘ฆ = 6 ๐‘ฅ โ€“ 3๐‘ฆ + 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 =โˆซ_1^4โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— ๐‘ฆโ†’ Equation of AB ๐‘ฅ โ€“ 3๐‘ฆ+5=0 ๐‘ฅ+5=3๐‘ฆ (๐‘ฅ + 5)/3=๐‘ฆ ๐‘ฆ=(๐‘ฅ + 5)/3 Therefore, Area ABED =โˆซ_1^4โ–’ใ€–((๐‘ฅ+5)/3) ๐‘‘๐‘ฅใ€— =1/3 โˆซ_1^4โ–’ใ€–(๐‘ฅ+5) ๐‘‘๐‘ฅใ€— =1/3 [๐‘ฅ^2/2+5๐‘ฅ]_1^4 =1/3 [4^2/2+5.4โˆ’[1^2/2+5.1]] =1/3 [8+20โˆ’1/2โˆ’5] =1/3 [45/2] =15/2 Area ACD Area ACD =โˆซ_1^2โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— ๐‘ฆโ†’ Equation of line AC 2๐‘ฅ+๐‘ฆ=4 ๐‘ฆ=4โˆ’2๐‘ฅ Area ACD =โˆซ_1^2โ–’ใ€–(4โˆ’2๐‘ฅ" " ) ๐‘‘๐‘ฅใ€— =[4๐‘ฅโˆ’(2๐‘ฅ^2)/2]_1^2 =[4๐‘ฅโˆ’๐‘ฅ^2 ]_1^2 =[4.2โˆ’2^2โˆ’[4.1โˆ’1^2 ]] =[8โˆ’4โˆ’4+1] = 1 Area CBE Area CBE =โˆซ_2^4โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— ๐‘ฆโ†’ Equation of line BC 3๐‘ฅ+2๐‘ฆ=6 3๐‘ฅโˆ’6=2๐‘ฆ (3๐‘ฅ โˆ’ 6)/2=๐‘ฆ ๐‘ฆ=(3๐‘ฅ โˆ’ 6)/2 Therefore, Area CBE =โˆซ_2^4โ–’ใ€–((3๐‘ฅ โˆ’ 6)/2) ๐‘‘๐‘ฅใ€— =1/2 โˆซ_2^4โ–’ใ€–(3๐‘ฅโˆ’6) ๐‘‘๐‘ฅใ€— =1/2 [(3๐‘ฅ^2)/2โˆ’6๐‘ฅ]_2^4 =1/2 [ใ€–3.4ใ€—^2/2โˆ’6.4โˆ’[ใ€–3.2ใ€—^2/2โˆ’6.2]] =1/2 [24โˆ’24โˆ’6+12] =3 Hence Area Required = Area ABED โ€“ Area ACD โ€“ Area CBE =15/2โˆ’1โˆ’3 =15/2โˆ’4 =(15 โˆ’ 8)/2 =๐Ÿ•/๐Ÿ 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.