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

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Chapter 8, Problem 32AAE

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To show the inequality ∫0a(f'(x))2dx≥2∫0af(x)dx−(2ab+a312).

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Chapter 8, Problem 31AAE

Chapter 8, Problem 33AAE

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To show the inequality ∫ 0 a ( f ' ( x ) ) 2 d x ≥ 2 ∫ 0 a f ( x ) d x − ( 2 a b + a 3 12 ) .

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To show the inequality ∫0a(f'(x))2dx≥2∫0af(x)dx−(2ab+a312).

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