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

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.