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Ex 7.9, 9 The value of the integral ∫_(1/3)^1▒〖 (𝑥 −𝑥^3 )^(1/3)/𝑥^4 〗 𝑑𝑥 is 6 (B) 0 (C) 3 (D) 4 ∫_(1/3)^1▒〖 (𝑥 − 𝑥^3 )^(1/3)/𝑥^4 〗 𝑑𝑥 Taking common 𝑥^3 from numerator = ∫_(1/3)^1▒〖 ((𝑥^3 )^(1/3) (1/𝑥^2 −1)^(1/3))/𝑥^4 〗 𝑑𝑥 = ∫_(1/3)^1▒〖 (𝑥 (1/𝑥^2 −1)^(1/3))/𝑥^4 〗 𝑑𝑥 = ∫_(1/3)^1▒〖 ( (1/𝑥^2 −1)^(1/3))/𝑥^3 〗 𝑑𝑥 Let t = 1/𝑥^2 −1 𝑑𝑡/𝑑𝑥=(−2)/𝑥^3 (−𝑑𝑡)/2=𝑑𝑥/𝑥^3 Thus, when x varies from 1/3 to 1, t varies form 0 to 8 Substituting values, ∫_(1/3)^1▒〖 ( (1/𝑥^2 −1)^(1/3))/𝑥^3 〗 𝑑𝑥 = 1/2 ∫_8^0▒〖𝑡^(1/3) 𝑑𝑡〗 = (−1)/2 [𝑡^(1/3 + 1)/(1/3 + 1)]_8^0 = (−1)/2 [〖3𝑡〗^(4/3 )/4]_8^0 Putting limits = (−1)/2 (0−(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 and Computer Science at Teachoo