Explanation Given information: The two curves has a function of y = 1 − 2 x 2 and y = | x | . Find the intersection points of the curves by equating the curves as shown below: f ( x ) = g ( x ) (1) Here, the top curve function is f ( x ) and the bottom curve function is g ( x ) . Substitute 1 − 2 x 2 for f ( x ) and | x | for g ( x ) in Equation (1). 1 − 2 x 2 = | x | x = 1 − 2 x 2 ( x ≥ 0 ) 2 x 2 + x − 1 = 0 Solve the equation. 2 x 2 + x − 1 = 0 2 x 2 + 2 x − 1 x − 1 = 0 2 x ( x + 1 ) − 1 ( x + 1 ) = 0 ( x + 1 ) ( 2 x − 1 ) = 0 Calculate the solutions. x + 1 = 0 x = − 1 2 x − 1 = 0 2 x = 1 x = 1 2 The condition for x is greater than or equal to zero ( x ≥ 0 ) . Hence, the region lies between x = 0 to x = 1 2 . Procedure to sketch the region bounded by the two curves is explained below: Draw the graph for the function y = 1 − 2 x 2 by substituting different values for x . Similarly in the same graph plot for the function y = | x | by substituting different values for x . Shade the region lies between x = 0 and x = 1 2 . The region enclosed by the curves y = 1 − 2 x 2 and y = | x | is shown in Figure (1)

9th Edition

ISBN: 9780357375808

Author: Stewart

Publisher: CENGAGE L

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Chapter 6, Problem 3E

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The area of the region enclosed by the curves.

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Chapter 6, Problem 2E

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Chapter 6 Solutions

CALCULUS: EARLY TRANSCENDENTALS

Ch. 6.1 - (a) Set up an integral for the area of the shaded...Ch. 6.1 - (a) Set up an integral for the area of the shaded...Ch. 6.1 - Find the area of the shaded region.Ch. 6.1 - Find the area of the shaded region.Ch. 6.1 - Find the area of the shaded region. 5.Ch. 6.1 - Find the area of the shaded region. 6.Ch. 6.1 - Set up, but do not evaluate, an integral...Ch. 6.1 - Prob. 8E Ch. 6.1 - Set up, but do not evaluate, an integral...Ch. 6.1 - Set up, but do not evaluate, an integral...

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The area of the region enclosed by the curves.

Solution Summary: The author determines the area of the region enclosed by the curves by equating them.

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The area of the region enclosed by the curves.

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Chapter 6 Solutions