Explanation Given information: The provided curve is: g ( x ) = 1 − x 2 and h ( x ) = x 2 + x . Formula used: Area between two curve: If f and g are continuous functions on [ a , b ] and if f ( x ) ≥ g ( x ) on [ a , b ] , then the area of the region bounded by y = f ( x ) , y = g ( x ) , x = a , and x = b is: A = ∫ a b [ f ( x ) − g ( x ) d x ] Or A = Integral of ( Top − Bottom ) Calculation: Consider the provided equation: g ( x ) = 1 − x 2 and h ( x ) = x 2 + x Draw the graph of the above function to check whether g ( x ) ≥ h ( x ) : Calculate the values at different values of x : Substitute x = 0 in the function g ( x ) = 1 − x 2 . y = 1 − ( 0 ) 2 = 1 Substitute x = 1 in the function g ( x ) = 1 − x 2 . y = 1 − ( 1 ) 2 = 0 Substitute x = 2 in the function g ( x ) = 1 − x 2 . y = 1 − ( 2 ) 2 = − 3 Substitute x = − 1 in the function g ( x ) = 1 − x 2 . y = 1 − ( − 1 ) 2 = 0 The table provided below shows the values of the function at different values of x . x g ( x ) = 1 − x 2 Coordinates ( x , y ) 0 1 ( 0 , 1 ) 1 0 ( 1 , 0 ) 2 − 3 ( 2 , − 3 ) − 1 0 ( − 1 , 0 ) Now, consider h ( x ) = x 2 + x Substitute x = 0 in the function h ( x ) = x 2 + x . y = ( 0 ) 2 + ( 0 ) = 0 Substitute x = 1 in the function h ( x ) = x 2 + x . y = ( 1 ) 2 + ( 1 ) = 2 Substitute x = 2 in the function h ( x ) = x 2 + x . y = ( 2 ) 2 + ( 2 ) = 6 Substitute x = − 1 in the function h ( x ) = x 2 + x . y = ( − 1 ) 2 + ( − 1 ) = 0 The table provided below shows the values of the function at different values of x

WebAssign Printed Access Card for Harshbarger/Reynolds' Mathematical Applications for the Management, Life, and Social Sciences, 12th Edition, Multi-Term

12th Edition

ISBN: 9781337630542

Author: Ronald J. Harshbarger, James J. Reynolds

Publisher: Cengage Learning

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Chapter 13.3, Problem 20E

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To calculate: The area of the region between g(x)=1−x2 and h(x)=x2+x.

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Chapter 13.3, Problem 19E

Chapter 13.3, Problem 21E

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Chapter 13 Solutions

WebAssign Printed Access Card for Harshbarger/Reynolds' Mathematical Applications for the Management, Life, and Social Sciences, 12th Edition, Multi-Term

Ch. 13.1 - CHECKPOINT 1. For the interval [0, 2]. determine...Ch. 13.1 - Prob. 2CP Ch. 13.1 - Prob. 1E Ch. 13.1 - Prob. 2E Ch. 13.1 - Prob. 3E Ch. 13.1 - Prob. 4E Ch. 13.1 - Prob. 5E Ch. 13.1 - Prob. 6E Ch. 13.1 - Prob. 7E Ch. 13.1 - Prob. 8E

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The area of the region between g ( x ) = 1 − x 2 and h ( x ) = x 2 + x .

Solution Summary: The author calculates the area of the region between g(x)=1-x

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To calculate: The area of the region between g(x)=1−x2 and h(x)=x2+x.

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Chapter 13 Solutions