a) Explanation Given: A function f is given such that f ′ ( x ) = f ( x ) . Concept used: Integration of a function can be defined by anti derivative of that function. For any function f ( x ) = a x + C , derivative of the function can be evaluated as f ′ ( x ) = a d d x ( x ) + d d x ( C ) = a ( 1 ) + 0 = a Therefore, antiderivative of a is given by ∫ a d x = a x + C Here, a is any constant and C is integrating constant. On other hand, ∫ a d x = a ∫ d x = a x + C Formula used: Fundamental theorem of Calculus I: ∫ a b f ( x ) d x = F ( b ) − F ( a ) Where F ′ ( x ) = f ( x ) Calculation: Now, consider the integral ∫ a b f ( x ) d x b) To determine To show: That ∫ a b [ f ( x ) ] 2 d x = 1 2 { [ f ( b ) ] 2 − [ f ( a ) ] 2 } .

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Question

Chapter 5.4, Problem 56PS

To determine

To show:That ∫abf(x)dx=f(b)−f(a) .

To determine

To show: That ∫ab[f( x)]2dx=12{[f(b)]2−[f(a)]2} .

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Chapter 5.4, Problem 55PS

Chapter 5.4, Problem 57PS

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Chapter 5 Solutions

Calculus

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That ∫ a b f ( x ) d x = f ( b ) − f ( a ) .

Solution Summary: The author explains that integration of a function can be defined by antiderivative of that function.