Chapter 7 Class 12 Integrals

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Important Questions for exams Class 12

Ex 7.9, 9 (MCQ) - Chapter 7 Class 12 Integrals (Important Question)

Last updated at July 14, 2026 by

  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 āˆ«_(1/3)^1ā–’怖 (š‘„ āˆ’š‘„^3 )^(1/3)/š‘„^4 怗 š‘‘š‘„ is 6 (B) 0 (C) 3 (D) 4 āˆ«_(1/3)^1ā–’怖 (š‘„ āˆ’ š‘„^3 )^(1/3)/š‘„^4 怗 š‘‘š‘„ Taking common š‘„^3 from numerator = āˆ«_(1/3)^1ā–’怖 ((š‘„^3 )^(1/3) (1/š‘„^2 āˆ’1)^(1/3))/š‘„^4 怗 š‘‘š‘„ = āˆ«_(1/3)^1ā–’怖 (š‘„ (1/š‘„^2 āˆ’1)^(1/3))/š‘„^4 怗 š‘‘š‘„ = āˆ«_(1/3)^1ā–’怖 ( (1/š‘„^2 āˆ’1)^(1/3))/š‘„^3 怗 š‘‘š‘„ Let t = 1/š‘„^2 āˆ’1 š‘‘š‘”/š‘‘š‘„=(āˆ’2)/š‘„^3 (āˆ’š‘‘š‘”)/2=š‘‘š‘„/š‘„^3 Thus, when x varies from 1/3 to 1, t varies form 0 to 8 Substituting values, āˆ«_(1/3)^1ā–’怖 ( (1/š‘„^2 āˆ’1)^(1/3))/š‘„^3 怗 š‘‘š‘„ = 1/2 āˆ«_8^0ā–’怖š‘”^(1/3) š‘‘š‘”怗 = (āˆ’1)/2 [š‘”^(1/3 + 1)/(1/3 + 1)]_8^0 = (āˆ’1)/2 [怖3š‘”怗^(4/3 )/4]_8^0 Putting limits = (āˆ’1)/2 (0āˆ’(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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