Check sibling questions

Example 6 - Find area bounded by two parabolas y = x2, y2 = x

Example 6 - Chapter 8 Class 12 Application of Integrals - Part 2
Example 6 - Chapter 8 Class 12 Application of Integrals - Part 3
Example 6 - Chapter 8 Class 12 Application of Integrals - Part 4
Example 6 - Chapter 8 Class 12 Application of Integrals - Part 5
Example 6 - Chapter 8 Class 12 Application of Integrals - Part 6
Example 6 - Chapter 8 Class 12 Application of Integrals - Part 7


Transcript

Example 6 Find the area of the region bounded by the two parabolas 𝑦=π‘₯2 and 𝑦2 = π‘₯ Drawing figure Here, we have parabolas 𝑦^2=π‘₯ π‘₯^2=𝑦 Area required = Area OABC Finding Point of intersection B Solving 𝑦2 = π‘₯ π‘₯2 =𝑦 Put (2) in (1) 𝑦2 = π‘₯ (π‘₯^2 )^2=π‘₯ π‘₯^4βˆ’π‘₯=0 π‘₯(π‘₯^3βˆ’1)=0 Finding y – coordinate For 𝒙=𝟎 𝑦=π‘₯^2=0^2= 0 So, coordinates are (0 , 0) For 𝒙=𝟏 𝑦=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 – Area OCBD Finding Area OABD Area OABD =∫_0^1▒〖𝑦 𝑑π‘₯γ€— Here, 𝑦^2=π‘₯ 𝑦=±√π‘₯ As OABD is in 1st quadrant, value of y is positive ∴ 𝑦=√π‘₯ Area OBQP =∫_0^1β–’γ€–βˆšπ‘₯ 𝑑π‘₯γ€— =∫_0^1β–’γ€–βˆšπ‘₯ 𝑑π‘₯γ€— =∫_0^1β–’γ€–π‘₯^(1/2) 𝑑π‘₯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)βˆ’(0)^(3/2) ] =2/3 [1βˆ’0] =2/3 Area OCBD Area OCBD =∫_0^1▒〖𝑦 𝑑π‘₯γ€— Here, π‘₯^2=𝑦 𝑦=π‘₯^2 Area OAQP =∫_0^1β–’γ€–π‘₯^2 𝑑π‘₯γ€— =[π‘₯^(2 + 1)/(2 + 1)]_0^1 =1/3 [π‘₯^3 ]_0^1 =1/3 [1^3βˆ’0^3 ] =𝟏/πŸ‘ Therefore, Area OABC = Area OABD – Area OCBD = 2/3βˆ’1/3 = 𝟏/πŸ‘ square units

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.