Let f be a non-negative function in [ 0 , 1 ] and twice differentiable in ( 0 , 1 ) . If ∫ 0 x 1 - f ' ( t ) 2 dt = ∫ 0 x f ( t ) dt , 0 ≤ x ≤ 1 and f ( 0 ) = 0 , then lim x → 0 1 x 2 ∫ 0 x f ( t ) dt

Question

Let f be a non-negative function in [ 0 , 1 ] and twice differentiable in ( 0 , 1 ) . If &#8747; 0 x 1 - f ' ( t ) 2 dt = &#8747; 0 x f ( t ) dt , 0 &#8804; x &#8804; 1 and f ( 0 ) = 0 , then lim x &#8594; 0 1 x 2 &#8747; 0 x f ( t ) dt :

Quantrex Academy · Accepted Answer

Correct answer: D. equals 1 2

Newton - Leibnitz rule d d x &#8747; f x g x h x d x &#160; = &#160; g ' x h g ( x ) &#160; - &#160; &#160; f ' x h f ( x ) Given that &#8747; 0 x 1 - f ' ( t ) 2 d t = &#8747; 0 x f ( t ) d t Differentiate w . r . t . &#160; x 1 - f ' ( x ) 2 = f ( x ) &#160; Squaring on both sides &#8658; 1 - f ' ( x ) 2 = ( f ( x ) ) 2 &#160;&#160; &#8658; f ' ( x ) 2 = 1 - ( f ( x ) ) 2 f x &#160; = &#160; y &#160; &#8658; f ' x &#160; = &#160; d y d x &#8658; d y d x 2 = 1 - y 2 &#160;&#160; &#8658;

Let f be a non-negative function in [ 0 , 1 ] and twice differentiable in ( 0 , 1 ) . If ∫ 0 x 1 - f ' ( t ) 2 dt = ∫ 0 x f ( t ) dt , 0 ≤ x ≤ 1 and f ( 0 ) = 0 , then lim x → 0 1 x 2 ∫ 0 x f ( t ) dt :

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Step-by-step solution

More from Definite Integration

Let f be a non-negative function in [ 0 , 1 ] and twice differentiable in ( 0 , 1 ) . If &#8747; 0 x 1 - f &#39; ( t ) 2 dt = &#8747; 0 x f ( t ) dt , 0 &#8804; x &#8804; 1 and f ( 0 ) = 0 , then lim x &#8594; 0 1 x 2 &#8747; 0 x f ( t ) dt :

Options

Correct answer

Step-by-step solution

More from Definite Integration

Let f be a non-negative function in [ 0 , 1 ] and twice differentiable in ( 0 , 1 ) . If ∫ 0 x 1 - f ' ( t ) 2 dt = ∫ 0 x f ( t ) dt , 0 ≤ x ≤ 1 and f ( 0 ) = 0 , then lim x → 0 1 x 2 ∫ 0 x f ( t ) dt :