# Ex 8.1, 10 - Chapter 8 Class 12 Application of Integrals (Term 2)

Last updated at Dec. 12, 2019 by Teachoo

Last updated at Dec. 12, 2019 by Teachoo

Transcript

Ex 8.1, 10 Find the area bounded by the curve ๐ฅ2=4๐ฆ and the line ๐ฅ=4๐ฆ โ 2 Here, ๐ฅ2=4๐ฆ 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, ๐ฅ=4๐ฆโ2 Putting in equation of curve , we get ๐ฅ^2=4๐ฆ (4๐ฆโ2)^2=4๐ฆ 16๐ฆ^2+4โ16๐ฆ=4๐ฆ 16๐ฆ^2โ16๐ฆโ4๐ฆ+4=0 16๐ฆ^2โ20๐ฆ+4=0 4[4๐ฆ^2โ5๐ฆ+1]=0 4๐ฆ^2โ5๐ฆ+1=0 4๐ฆ^2โ4๐ฆโ๐ฆ+1=0 4๐ฆ(๐ฆโ1)โ1(๐ฆโ1)=0 (4๐ฆโ1)(๐ฆโ1)=0 So, y = 1/4 , y = 1 For ๐=๐/๐ ๐ฅ=4๐ฆโ2 ๐ฅ=4(1/4)โ2 ๐ฅ =โ1 So, point is (โ1, 1/4) For y = 1 ๐ฅ=4๐ฆโ2 ๐ฅ=4(1)โ2 ๐ฅ =2 So, point is (2, 1) As Point A is in 2nd Quadrant โด A = (โ1 , 1/4) & Point B is in 1st Quadrant โด B = (2 , 1) Finding required area Required Area = Area APBQ โ Area APOQBA Area APBQ Area APBQ = โซ_(โ1)^2โ๐ฆ๐๐ฅ Here, ๐ฆ โ Equation of Line ๐ฅ=4๐ฆโ2 ๐ฅ+2=4๐ฆ ๐ฆ=(๐ฅ + 2)/4 Area APBQ = โซ_(โ1)^2โ(๐ฅ + 2)/4 ๐๐ฅ = 1/4 โซ1_(โ1)^2โ(๐ฅ+2)๐๐ฅ = 1/4 [๐ฅ^2/2+2๐ฅ]_(โ1)^2 = 1/4 [(2^2/2+2(2))โ(ใ(โ1)ใ^2/2+2(โ1))] = 1/4 [(2+4)โ(1/2โ2))] = 1/4 [6โ1/2+2] = 1/4ร15/2 = 15/8 Area APOQBA Area APOQBA = โซ_(โ1)^2โใ๐ฆ ๐๐ฅใ Here, ๐ฆ โ Equation of Parabola ๐ฅ^2=4๐ฆ 4๐ฆ=๐ฅ^2 ๐ฆ=1/4 ๐ฅ^2 Area APOQBA = 1/4 โซ1_(โ1)^2โใ๐ฅ^2 ๐๐ฅใ = 1/4 [๐ฅ^3/3]_(โ1)^2 = 1/4 [((2)^3 โ (โ1)^3)/3] = 1/4 [(8 โ (โ1))/3] = 1/4 [(8 + 1)/3] = 3/4 Now, Required Area = Area APBQ โ Area APOQBA = 15/8 โ 3/4 " = " (15 โ 6)/8 = 9/8 โด Required Area = ๐/๐ Square units

Chapter 8 Class 12 Application of Integrals (Term 2)

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.