Explanation Second Fundamental Theorem of Calculus: Let f ( x ) be continuous on an interval I , let a ∈ I and for x ∈ I , define F ( x ) = ∫ a x f ( t ) d t for all x ∈ I Then F ( x ) is differentiable on I and F ' ( x ) = f ( x ) for all x ∈ I In other words, if f is continuous on an interval I , let a ∈ I and for x ∈ I , then d d x ( ∫ a x f ( t ) d t ) = f ( x ) Chain Rule: If f is continuous on an open interval I containing the values of differentiable functions y = a ( x ) & y = b ( x ) then d d x [ ∫ a ( x ) b ( x ) f ( t ) d t ] = f ( b ( x ) ) b ' ( x ) − f ( a ( x ) ) a ' ( x ) Given: H ( x ) = ∫ 4 x 2 9 1 t d t Formula used: Second Fundamental Theorem of Calculus: Let f ( x ) be continuous on an interval I , let a ∈ I and for x ∈ I , define F ( x ) = ∫ a x f ( t ) d t for all x ∈ I Then F ( x ) is differentiable on I and F ' ( x ) = f ( x ) for all x ∈ I In other words, if f is continuous on an interval I , let a ∈ I and for x ∈ I , then d d x ( ∫ a x f ( t ) d t ) = f ( x ) Chain Rule: If f is continuous on an open interval I containing the values of differentiable functions y = a ( x ) & y = b ( x ) , then d d x [ ∫

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 5, Problem 91CRE

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To Determine:

The value of $H' (1)$ , where $H (x) = \int_{4 x^{2}}^{9} \frac{1}{t} d t$ .

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Chapter 5, Problem 90CRE

Chapter 5, Problem 92CRE

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Chapter 5 Solutions

Calculus - Standalone book

Ch. 5.1 - Prob. 1PQ Ch. 5.1 - Prob. 2PQ Ch. 5.1 - Prob. 3PQ Ch. 5.1 - Prob. 4PQ Ch. 5.1 - Prob. 5PQ Ch. 5.1 - Prob. 1E Ch. 5.1 - Prob. 2E Ch. 5.1 - Prob. 3E Ch. 5.1 - Prob. 4E Ch. 5.1 - Prob. 5E

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The value of H ' ( 1 ) , where H ( x ) = ∫ 4 x 2 9 1 t d t .

Solution Summary: The author explains the Second Fundamental Theorem of Calculus: Let f ( x ) be continuous on an interval I, let a I and for y.

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To Determine:

The value of $H' (1)$ , where $H (x) = \int_{4 x^{2}}^{9} \frac{1}{t} d t$ .

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Chapter 5 Solutions

The value of H ' ( 1 ) , where H ( x ) = ∫ 4 x 2 9 1 t d t .

Solution Summary: The author explains the Second Fundamental Theorem of Calculus: Let f ( x ) be continuous on an interval I, let a I and for y.

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To Determine: The value of H'(1), where H(x)=∫4x291tdt.

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Chapter 5 Solutions

To Determine:

The value of $H' (1)$ , where $H (x) = \int_{4 x^{2}}^{9} \frac{1}{t} d t$ .