Explanation Given information: The function f is continuous and differentiable on the interval [ 0 , 1 ] . The value of f ( 1 ) = f ( 0 ) = − 1 6 . The given inequality is 0 ≤ ∫ 0 1 ( f ' ( x ) + x − 1 2 ) 2 d x . Calculation: The inequality 0 ≤ ∫ 0 1 ( f ' ( x ) + x − 1 2 ) 2 d x can be expressed as follows: 0 ≤ ∫ 0 1 ( f ' ( x ) + x − 1 2 ) 2 d x = ∫ 0 1 ( f ' ( x ) ) 2 d x + ∫ 0 1 ( 2 x − 1 ) f ' ( x ) d x + ∫ 0 1 ( x − 1 2 ) 2 d x (1) Evaluate the integral ∫ 0 1 ( x − 1 2 ) 2 d x as follows: ∫ 0 1 ( x − 1 2 ) 2 d x = [ 1 3 ( x − 1 2 ) 3 ] 0 1 = 1 3 ( 1 − 1 2 ) 3 − 1 3 ( 0 − 1 2 ) 3 = 1 3 ( 1 8 ) − 1 3 ( − 1 8 ) = 1 24 + 1 24 ∫ 0 1 ( x − 1 2 ) 2 d x = 1 12 Evaluate the integral ∫ 0 1 ( 2 x − 1 ) f ' ( x ) d x as follows: Let consider u = 2 x − 1 . Differentiate u with respect to x . d u = 2 d x Let consider d v = f ' ( x ) . Integrate once

University Calculus: Early Transcendentals, Loose-leaf Edition (4th Edition)

4th Edition

ISBN: 9780135164860

Author: Joel R. Hass, Christopher Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr.

Publisher: PEARSON

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Chapter 8, Problem 31AAE

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To show the inequality ∫01(f'(x))2dx≥2∫01f(x)dx+14.

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Chapter 8, Problem 30AAE

Chapter 8, Problem 32AAE

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University Calculus: Early Transcendentals, Loose-leaf Edition (4th Edition)

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To show the inequality ∫ 0 1 ( f ' ( x ) ) 2 d x ≥ 2 ∫ 0 1 f ( x ) d x + 1 4 .

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To show the inequality ∫01(f'(x))2dx≥2∫01f(x)dx+14.

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Chapter 8 Solutions