Definite Integration - By Substitution

Chapter 7 Class 12 Integrals
Concept wise

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### Transcript

Ex 7.10, 9 The value of the integral โซ_(1/3)^1โใ (๐ฅ โ๐ฅ^3 )^(1/3)/๐ฅ^4 ใ ๐๐ฅ is 6 (B) 0 (C) 3 (D) 4 โซ_(1/3)^1โใ (๐ฅ โ ๐ฅ^3 )^(1/3)/๐ฅ^4 ใ ๐๐ฅ Taking common ๐ฅ^3 from numerator = โซ_(1/3)^1โใ ((๐ฅ^3 )^(1/3) (1/๐ฅ^2 โ1)^(1/3))/๐ฅ^4 ใ ๐๐ฅ = โซ_(1/3)^1โใ (๐ฅ (1/๐ฅ^2 โ1)^(1/3))/๐ฅ^4 ใ ๐๐ฅ = โซ_(1/3)^1โใ ( (1/๐ฅ^2 โ1)^(1/3))/๐ฅ^3 ใ ๐๐ฅ Let t = 1/๐ฅ^2 โ1 ๐๐ก/๐๐ฅ=(โ2)/๐ฅ^3 (โ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, โซ_(1/3)^1โใ ( (1/๐ฅ^2 โ1)^(1/3))/๐ฅ^3 ใ ๐๐ฅ = 1/2 โซ_8^0โใ๐ก^(1/3) ๐๐กใ = (โ1)/2 [๐ก^(1/3 + 1)/(1/3 + 1)]_8^0 = (โ1)/2 [ใ3๐กใ^(4/3 )/4]_8^0 Putting limits = (โ1)/2 (0โ(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.