Explanation Given information: The graph of the derivative f ( x ) : Concept used: According to the theorem of integral With variable upper bound, the value of the antiderivative of a function at point t coincides with the area under the graph of that function increased by the additive constant. Graph: The equation of the given graph is f ( x ) = 1 which is a constant function, therefore its antiderivative F ( x ) must be a linear function. Integrating with respect to ‘ x’, we get F ( x ) = ∫ f ( x ) d x = ∫ 1 d x = x + c Using the condition F ( 0 ) = 0 , we get 0 = 0 + c ⇒ c = 0 Therefore, F ( x ) = x is the required function to sketch

6th Edition

ISBN: 9781119399353

Author: DA

Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT

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Chapter 6.1, Problem 16P

To determine

To graph:

Two functions F such that F'=f, for one case let F(0)=0 and in the other case let F(0)=1.

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Chapter 6.1, Problem 15P

Chapter 6.1, Problem 17P

Chapter 6 Solutions

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To graph: Two functions F such that F ' = f , for one case let F ( 0 ) = 0 and in the other case let F ( 0 ) = 1 .

Solution Summary: The author explains the theorem of integral, where the value of the antiderivative of a function coincides with the area under the graph of that function increased by the additive constant.

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To graph:

Two functions F such that F'=f, for one case let F(0)=0 and in the other case let F(0)=1.

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Chapter 6 Solutions

To graph: Two functions F such that F ' = f , for one case let F ( 0 ) = 0 and in the other case let F ( 0 ) = 1 .

Solution Summary: The author explains the theorem of integral, where the value of the antiderivative of a function coincides with the area under the graph of that function increased by the additive constant.

Videos

To graph: Two functions F such that F'=f, for one case let F(0)=0 and in the other case let F(0)=1.

Want to see the full answer?

Chapter 6 Solutions

To graph:

Two functions F such that F'=f, for one case let F(0)=0 and in the other case let F(0)=1.