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Chapter 8 Class 12 Application of Integrals

Serial order wise

Last updated at March 16, 2023 by Teachoo

Example 6 Find the area of the region bounded by the two parabolas π¦=π₯2 and π¦2 = π₯ Drawing figure Here, we have parabolas π¦^2=π₯ π₯^2=π¦ Area required = Area OABC Finding Point of intersection B Solving π¦2 = π₯ π₯2 =π¦ Put (2) in (1) π¦2 = π₯ (π₯^2 )^2=π₯ π₯^4βπ₯=0 π₯(π₯^3β1)=0 Finding y β coordinate For π=π π¦=π₯^2=0^2= 0 So, coordinates are (0 , 0) For π=π π¦=π₯^2=1^2=1 So, coordinates are (1 , 1) Since point B lies in 1st quadrant So, co-ordinate of B is (1 , 1) Finding Area Area OABC = Area OABD β Area OCBD Finding Area OABD Area OABD =β«_0^1βγπ¦ ππ₯γ Here, π¦^2=π₯ π¦=Β±βπ₯ As OABD is in 1st quadrant, value of y is positive β΄ π¦=βπ₯ Area OBQP =β«_0^1βγβπ₯ ππ₯γ =β«_0^1βγβπ₯ ππ₯γ =β«_0^1βγπ₯^(1/2) ππ₯γ = [π₯^(1/2 + 1)/(1/2 + 1)]_0^1 = [π₯^(3/2)/(3/2)]_0^1 = 2/3 [π₯^(3/2) ]_0^1 =2/3 [(1)^(3/2)β(0)^(3/2) ] =2/3 [1β0] =2/3 Area OCBD Area OCBD =β«_0^1βγπ¦ ππ₯γ Here, π₯^2=π¦ π¦=π₯^2 Area OAQP =β«_0^1βγπ₯^2 ππ₯γ =[π₯^(2 + 1)/(2 + 1)]_0^1 =1/3 [π₯^3 ]_0^1 =1/3 [1^3β0^3 ] =π/π Therefore, Area OABC = Area OABD β Area OCBD = 2/3β1/3 = π/π square units