Ex 8.1

Chapter 8 Class 12 Application of Integrals
Serial order wise

### Transcript

Question 6 The area between ๐ฅ=๐ฆ2 and ๐ฅ = 4 is divided into two equal parts by the line ๐ฅ=๐, find the value of a. Given curve ๐ฆ^2=๐ฅ Let AB represent the line ๐ฅ=๐ CD represent the line ๐ฅ=4 Since the line ๐ฅ=๐ divides the region into two equal parts โด Area of OBA = Area of ABCD 2 ร โซ_0^๐โใ๐ฆ ๐๐ฅใ="2 ร" โซ_๐^4โใ๐ฆ ๐๐ฅใ โซ_๐^๐โใ๐ ๐๐ใ=โซ_๐^๐โใ๐ ๐๐ใ Now, y2 = x y = ยฑ โ๐ฅ Since, the curve is symmetric about x-axis we can take either positive or negative value of ๐ฆ So, lets take ๐ฆ=โ๐ฅ Now, From (1) โซ_0^๐โใ๐ฆ ๐๐ฅใ=โซ_๐^4โใ๐ฆ ๐๐ฅใ โซ_0^๐โโ๐ฅ ๐๐ฅ=โซ_๐^4โโ๐ฅ ๐๐ฅ [๐ฅ^(1/2 + 1)/(1/2 + 1)]_0^๐=[๐ฅ^(1/2+1)/(1/2+1)]_๐^4 [๐ฅ^((1+2)/2) ]_0^๐=[๐ฅ^((1+2)/2) ]_๐^4 [๐ฅ^(3/2) ]_0^๐=[๐ฅ^(3/2) ]_๐^4 (๐)^(3/2)โ0=(4)^(3/2)โ(๐)^(3/2) 2(๐)^(3/2)=(4)^(3/2) Taking ใ2/3ใ^๐กโ root on both sides (2)^(2/3) ๐^(3/2 ร 2/3)=4^(3/2 ร 2/3) (2)^(2/3) ๐=4 ๐=(2)^2/(2)^(2/3) ๐=(2)^(2 โ 2/3) ๐=(2)^((6 โ 2)/3) ๐=(2)^(4/3) ๐=(2)^(2 ร 2/3) ๐=[2^2 ]^(2/3) ๐=(๐)^(๐/๐) So, value of a is (4)^(2/3)