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 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</

6th Edition

ISBN: 9781119399322

Author: Hughes-Hallett

Publisher: WILEY

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Chapter 6.1, Problem 17P

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 16P

Chapter 6.1, Problem 18P

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Chapter 6 Solutions

APPLIED CALCULUS (WILEY PLUS)

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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 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.

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 .

Solution Summary: The author explains the theorem of integral, where the value of the antiderivative 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.