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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Misc 29 Evaluate the definite integral โˆซ_0^1โ–’ใ€–๐‘‘๐‘ฅ/(โˆš(1 + ๐‘ฅ) โˆ’ โˆš๐‘ฅ) ใ€— โˆซ_0^1โ–’ใ€–๐‘‘๐‘ฅ/(โˆš(1 + ๐‘ฅ) โˆ’ โˆš๐‘ฅ) ใ€— Rationalizing i.e., multiplying and dividing by (โˆš(1 + ๐‘ฅ)+โˆš๐‘ฅ) = โˆซ_0^1โ–’ใ€–๐‘‘๐‘ฅ/(โˆš(1 + ๐‘ฅ) โˆ’ โˆš๐‘ฅ) ใ€—ร—(โˆš(1 + ๐‘ฅ) + โˆš๐‘ฅ)/(โˆš(1 + ๐‘ฅ) + โˆš๐‘ฅ) . ๐‘‘๐‘ฅ = โˆซ_0^1โ–’(โˆš(1 + ๐‘ฅ) + โˆš๐‘ฅ)/((โˆš(1 + ๐‘ฅ) )^2 โˆ’ (โˆš๐‘ฅ )^2 ) . ๐‘‘๐‘ฅ = โˆซ_0^1โ–’(โˆš(1 + ๐‘ฅ) + โˆš๐‘ฅ)/(1 + ๐‘ฅ โˆ’ ๐‘ฅ) . ๐‘‘๐‘ฅ = โˆซ_0^1โ–’(โˆš(1 + ๐‘ฅ) + โˆš๐‘ฅ)/1 . ๐‘‘๐‘ฅ = โˆซ_0^1โ–’โˆš(1+๐‘ฅ) . ๐‘‘๐‘ฅ+โˆซ_0^1โ–’โˆš๐‘ฅ . ๐‘‘๐‘ฅ" " = โˆซ_0^1โ–’(1+๐‘ฅ)^(1/2) ๐‘‘๐‘ฅ+โˆซ_0^1โ–’(๐‘ฅ)^(1/2) ๐‘‘๐‘ฅ" " = [(1 + ๐‘ฅ)^(1/2 + 1)/(1/2 + 1)]_0^1 + [ใ€–๐‘ฅ ใ€—^(1/2 + 1)/(1/2 + 1)]_0^1 = [(1 + ๐‘ฅ)^(3/2)/(3/2)]_0^1 + [ใ€–๐‘ฅ ใ€—^(3/2)/(3/2)]_0^1 = ใ€–2/3 [(1+๐‘ฅ)^(3/2) ]ใ€—_0^1 + 2/3 [ใ€–๐‘ฅ ใ€—^(3/2) ]_0^1 = 2/3 [(1+1)^(3/2)โˆ’(1+0)^(3/2) ] + 2/3 [(1)^(3/2)โˆ’(0)^(3/2) ] = 2/3 [(2)^(3/2)โˆ’(1)^(3/2) ] + 2/3 [1โˆ’0] = 2/3 . (2)^(3/2)โˆ’2/3 [1]+2/3 [1] = 2/3 (2)^(3/2) = 2/3 [(2)^(1/2) ]^3 = 2/3 (โˆš2 )^3 = 2/3 . 2 โˆš2 = (๐Ÿ’ โˆš๐Ÿ)/๐Ÿ‘

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.