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Ex 7.10, 9 - Value of integral (x - x^3)^1/3 / x^4 - Ex 7.10

Ex 7.10, 9 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.10, 9 - Chapter 7 Class 12 Integrals - Part 3

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Transcript

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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.