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

Transcript

Ex 8.1, 10 Find the area bounded by the curve ๐‘ฅ2=4๐‘ฆ and the line ๐‘ฅ=4๐‘ฆ โ€“ 2 Here, ๐‘ฅ2=4๐‘ฆ is a parabola And, x = 4y โ€“ 2 is a line which intersects the parabola at points A and B We need to find Area of shaded region First we find Points A and B Finding points A and B Points A & B are the intersection of curve and line We know that, ๐‘ฅ=4๐‘ฆโˆ’2 Putting in equation of curve , we get ๐‘ฅ^2=4๐‘ฆ (4๐‘ฆโˆ’2)^2=4๐‘ฆ 16๐‘ฆ^2+4โˆ’16๐‘ฆ=4๐‘ฆ 16๐‘ฆ^2โˆ’16๐‘ฆโˆ’4๐‘ฆ+4=0 16๐‘ฆ^2โˆ’20๐‘ฆ+4=0 4[4๐‘ฆ^2โˆ’5๐‘ฆ+1]=0 4๐‘ฆ^2โˆ’5๐‘ฆ+1=0 4๐‘ฆ^2โˆ’4๐‘ฆโˆ’๐‘ฆ+1=0 4๐‘ฆ(๐‘ฆโˆ’1)โˆ’1(๐‘ฆโˆ’1)=0 (4๐‘ฆโˆ’1)(๐‘ฆโˆ’1)=0 So, y = 1/4 , y = 1 For ๐’š=๐Ÿ/๐Ÿ’ ๐‘ฅ=4๐‘ฆโˆ’2 ๐‘ฅ=4(1/4)โˆ’2 ๐‘ฅ =โˆ’1 So, point is (โ€“1, 1/4) For y = 1 ๐‘ฅ=4๐‘ฆโˆ’2 ๐‘ฅ=4(1)โˆ’2 ๐‘ฅ =2 So, point is (2, 1) As Point A is in 2nd Quadrant โˆด A = (โˆ’1 , 1/4) & Point B is in 1st Quadrant โˆด B = (2 , 1) Finding required area Required Area = Area APBQ โ€“ Area APOQBA Area APBQ Area APBQ = โˆซ_(โˆ’1)^2โ–’๐‘ฆ๐‘‘๐‘ฅ Here, ๐‘ฆ โ†’ Equation of Line ๐‘ฅ=4๐‘ฆโˆ’2 ๐‘ฅ+2=4๐‘ฆ ๐‘ฆ=(๐‘ฅ + 2)/4 Area APBQ = โˆซ_(โˆ’1)^2โ–’(๐‘ฅ + 2)/4 ๐‘‘๐‘ฅ = 1/4 โˆซ1_(โˆ’1)^2โ–’(๐‘ฅ+2)๐‘‘๐‘ฅ = 1/4 [๐‘ฅ^2/2+2๐‘ฅ]_(โˆ’1)^2 = 1/4 [(2^2/2+2(2))โˆ’(ใ€–(โˆ’1)ใ€—^2/2+2(โˆ’1))] = 1/4 [(2+4)โˆ’(1/2โˆ’2))] = 1/4 [6โˆ’1/2+2] = 1/4ร—15/2 = 15/8 Area APOQBA Area APOQBA = โˆซ_(โˆ’1)^2โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— Here, ๐‘ฆ โ†’ Equation of Parabola ๐‘ฅ^2=4๐‘ฆ 4๐‘ฆ=๐‘ฅ^2 ๐‘ฆ=1/4 ๐‘ฅ^2 Area APOQBA = 1/4 โˆซ1_(โˆ’1)^2โ–’ใ€–๐‘ฅ^2 ๐‘‘๐‘ฅใ€— = 1/4 [๐‘ฅ^3/3]_(โˆ’1)^2 = 1/4 [((2)^3 โˆ’ (โˆ’1)^3)/3] = 1/4 [(8 โˆ’ (โˆ’1))/3] = 1/4 [(8 + 1)/3] = 3/4 Now, Required Area = Area APBQ โ€“ Area APOQBA = 15/8 โ€“ 3/4 " = " (15 โˆ’ 6)/8 = 9/8 โˆด Required Area = 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.