Chapter 7 Class 12 Integrals
Concept wise

ย  Ex 7.9, 9 (MCQ) - Value of integral (x - x^3)^1/3 / x^4 - Ex 7.9 - Ex 7.9

part 2 - Ex 7.9, 9 (MCQ) - Ex 7.9 - Serial order wise - Chapter 7 Class 12 Integrals
part 3 - Ex 7.9, 9 (MCQ) - Ex 7.9 - Serial order wise - Chapter 7 Class 12 Integrals

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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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