Ex 7.10, 9 - Value of integral (x - x3)1/3 / x4 - Ex 7.10

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

Transcript

Ex7.10, 9 Choose the correct answer The value of the integral โˆซ_(1/3)^1โ–’ใ€– (๐‘ฅ โˆ’๐‘ฅ^3 )^(1/3)/๐‘ฅ^4 ใ€— ๐‘‘๐‘ฅ is 6 (B) 0 (C) 3 (D) 4 โˆซ_(1/3)^1โ–’ใ€– (๐‘ฅ โˆ’๐‘ฅ^3 )^(1/3)/๐‘ฅ^4 ใ€— ๐‘‘๐‘ฅ Taking common ๐‘ฅ^3 from numerator = โˆซ_(1/3)^1โ–’ใ€– ((๐‘ฅ^3 )^(1/3) (1/๐‘ฅ^2 โˆ’1)^(1/3))/๐‘ฅ^4 ใ€— ๐‘‘๐‘ฅ = โˆซ_(1/3)^1โ–’ใ€– (๐‘ฅ (1/๐‘ฅ^2 โˆ’1)^(1/3))/๐‘ฅ^4 ใ€— ๐‘‘๐‘ฅ = โˆซ_(1/3)^1โ–’ใ€– ( (1/๐‘ฅ^2 โˆ’1)^(1/3))/๐‘ฅ^3 ใ€— ๐‘‘๐‘ฅ Let t = 1/๐‘ฅ^2 โˆ’1 ๐‘‘๐‘ก/๐‘‘๐‘ฅ=(โˆ’2)/๐‘ฅ^3 (โˆ’๐‘‘๐‘ก)/2=๐‘‘๐‘ฅ/๐‘ฅ^3 Thus, when x varies from 1/3 to 1, t varies form 0 to 8 Substituting values, 1/2 โˆซ_8^0โ–’ใ€–๐‘ก^(1/3) ๐‘‘๐‘กใ€— = (โˆ’1)/2 [๐‘ก^(1/3 + 1)/(1/3+1)]_8^0 = (โˆ’1)/2 [ใ€–3๐‘กใ€—^(4/3 )/4]_8^0 Putting limits, = (โˆ’1)/2 (0โˆ’(3(8)^(4/3))/4) = 1/2 (3/4) (8)^(4/3) = 1/2 (3/4) (2^3 )^(4/3) = 1/2 (3/4) (2^4 ) = 6 So, (A) is the correct answer.

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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.