Chapter 7 Class 12 Integrals
Concept wise

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Ex 7.9, 9 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, โˆซ_(1/3)^1โ–’ใ€– ( (1/๐‘ฅ^2 โˆ’1)^(1/3))/๐‘ฅ^3 ใ€— ๐‘‘๐‘ฅ = 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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.