Area bounded by curve and horizontal or vertical line

Chapter 8 Class 12 Application of Integrals
Concept wise

This video is only available for Teachoo black users

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

### Transcript

Question 6 The area between π₯=π¦2 and π₯ = 4 is divided into two equal parts by the line π₯=π, find the value of a. Given curve π¦^2=π₯ Let AB represent the line π₯=π CD represent the line π₯=4 Since the line π₯=π divides the region into two equal parts β΄ Area of OBA = Area of ABCD 2 Γ β«_0^πβγπ¦ ππ₯γ="2 Γ" β«_π^4βγπ¦ ππ₯γ β«_π^πβγπ ππγ=β«_π^πβγπ ππγ Now, y2 = x y = Β± βπ₯ Since, the curve is symmetric about x-axis we can take either positive or negative value of π¦ So, lets take π¦=βπ₯ Now, From (1) β«_0^πβγπ¦ ππ₯γ=β«_π^4βγπ¦ ππ₯γ β«_0^πββπ₯ ππ₯=β«_π^4ββπ₯ ππ₯ [π₯^(1/2 + 1)/(1/2 + 1)]_0^π=[π₯^(1/2+1)/(1/2+1)]_π^4 [π₯^((1+2)/2) ]_0^π=[π₯^((1+2)/2) ]_π^4 [π₯^(3/2) ]_0^π=[π₯^(3/2) ]_π^4 (π)^(3/2)β0=(4)^(3/2)β(π)^(3/2) 2(π)^(3/2)=(4)^(3/2) Taking γ2/3γ^π‘β root on both sides (2)^(2/3) π^(3/2 Γ 2/3)=4^(3/2 Γ 2/3) (2)^(2/3) π=4 π=(2)^2/(2)^(2/3) π=(2)^(2 β 2/3) π=(2)^((6 β 2)/3) π=(2)^(4/3) π=(2)^(2 Γ 2/3) π=[2^2 ]^(2/3) π=(π)^(π/π) So, value of a is (4)^(2/3)