Example 14 - Prove that y2 = 4x, x2 = 4y divide area of square

Example 14 - Chapter 8 Class 12 Application of Integrals - Part 2
Example 14 - Chapter 8 Class 12 Application of Integrals - Part 3
Example 14 - Chapter 8 Class 12 Application of Integrals - Part 4
Example 14 - Chapter 8 Class 12 Application of Integrals - Part 5
Example 14 - Chapter 8 Class 12 Application of Integrals - Part 6
Example 14 - Chapter 8 Class 12 Application of Integrals - Part 7
Example 14 - Chapter 8 Class 12 Application of Integrals - Part 8
Example 14 - Chapter 8 Class 12 Application of Integrals - Part 9

  1. Chapter 8 Class 12 Application of Integrals (Term 2)
  2. Serial order wise

Transcript

Example 14 Prove that curves 𝑦2=4π‘₯ and π‘₯2=4𝑦 divide the area of the square bounded by π‘₯=0, π‘₯=4, 𝑦=4 and 𝑦=0 into three equal parts Drawing figure Here, we have parabolas 𝑦^2=4π‘₯ π‘₯^2=4𝑦 And, Square made by the lines x = 4, y = 4, x = 0, y = 0 We need to prove that area of square is divided into 3 parts by the curve So, we need to prove Area OPQA = Area OAQB = Area OBQR Area OPQA Area OPQA = ∫_0^4▒〖𝑦 𝑑π‘₯γ€— Here, 4𝑦=π‘₯^2 𝑦=π‘₯^2/4 So, Area OPQA =∫_0^4β–’γ€–π‘₯^2/4 𝑑π‘₯γ€— = 1/4 [π‘₯^3/3]_0^4 =1/12Γ—[4^3βˆ’0^3 ] =1/12 Γ— [64βˆ’0] =64/12=16/3 Area OPQA is the area by curve x2 = 4y in the x-axis from x = 0 to x = 4 Area OBQR Since Area is on π‘¦βˆ’π‘Žπ‘₯𝑖𝑠 , we use formula ∫1β–’γ€–π‘₯ 𝑑𝑦〗 Area OBQR = ∫_0^4β–’γ€–π‘₯ 𝑑𝑦〗 Here, 𝑦^2=4π‘₯ π‘₯=𝑦^2/4 So, Area OBQR =∫_0^4▒〖𝑦^2/4 𝑑𝑦〗 = 1/4 [𝑦^3/3]_0^4=1/12Γ—[4^3βˆ’0^3 ]=1/12 Γ—[64βˆ’0]=16/3 Area OBQR is the area by curve y2 = 4x in the y-axis from y = 0 to y = 4 Area OAQB Area OAQB = Area OBQP – Area OAQP Finding Area OBQP Area OBQP =∫_0^4▒〖𝑦 𝑑π‘₯γ€— Here, 𝑦^2=4π‘₯ 𝑦=±√4π‘₯ As OBQP is in 1st quadrant, value of y is positive ∴ 𝑦=√4π‘₯ Area OBQP =∫_0^4β–’γ€–βˆš4π‘₯ 𝑑π‘₯γ€— =2∫_0^4β–’γ€–βˆšπ‘₯ 𝑑π‘₯γ€— =2∫_0^4β–’γ€–π‘₯^(1/2) 𝑑π‘₯γ€— =2 Γ— [π‘₯^(1/2+1)/(1/2+1)]_0^4 =2 Γ— [π‘₯^(3/2)/(3/2)]_0^4 =2 Γ— 2/3 [π‘₯^(3/2) ]_0^4 =4/3 [(4)^(3/2)βˆ’(0)^(3/2) ] =4/3 [8βˆ’0] =32/3 Area OAQP Area OAQP =∫_0^4▒〖𝑦 𝑑π‘₯γ€— Here, π‘₯^2=4𝑦 𝑦=π‘₯^2/4 Area OAQP =∫_0^4β–’γ€–π‘₯^2/4 𝑑π‘₯γ€— =1/4 [π‘₯^(2+1)/(2+1)]_0^4 =1/(4 Γ—3) [π‘₯^3 ]_0^4 =1/12 [4^3βˆ’0^3 ] Area OAQP Area OAQP =∫_0^4▒〖𝑦 𝑑π‘₯γ€— Here, π‘₯^2=4𝑦 𝑦=π‘₯^2/4 Area OAQP =∫_0^4β–’γ€–π‘₯^2/4 𝑑π‘₯γ€— =1/4 [π‘₯^(2+1)/(2+1)]_0^4 =1/(4 Γ—3) [π‘₯^3 ]_0^4 =1/12 [4^3βˆ’0^3 ] =1/12 Γ— [64βˆ’0] =16/3 ∴ Area OAQB = Area OBQD – Area OAQP = 32/3βˆ’16/3 = 16/3 So, Area OAQB = Area OAQP = Area OBRQ = πŸπŸ”/πŸ‘ square units Hence Proved.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.