Explanation Given: The provided equations are: y = 6 x − x 2 and y = x 2 − 2 x Formula used: The area of a region is given by A = ∫ x l x h ∫ y l y h d y d x Vertex form of parabola’s equation is given by ( y − k ) = a ( x − h ) 2 Here ( h , k ) is the vertex of the parabola Calculations: The equation y = x 2 − 2 x is the equation of a parabola. Write it in its vertex form y = x 2 − 2 x y + 1 = x 2 − 2 x + 1 ( y − ( − 1 ) ) = ( x − 1 ) 2 Vertex of this parabola is given by V = ( 1 , − 1 ) The equation y = 6 x − x 2 is also the equation of parabola. Write it in its vertex form y = − ( x 2 − 6 x ) ( y − 9 ) = − ( x 2 − 6 x + 9 ) ( y − 9 ) = − ( x − 3 ) 2 So, vertex of this parabola is given by V = ( 3 , 9 ) Find intersection of both parabolas x 2 − 2 x = 6 x − x 2 2 x 2 − 8 x = 0 2 x ( x − 4 ) = 0 This gives x = 0 and x = 4

7th Edition

ISBN: 9780357762554

Author: Larson

Publisher: Cengage

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Chapter 14, Problem 8RE

To determine

To calculate: The area of the region bounded by the graphs of given equation y=6x−x2 and y=x2−2x.

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Chapter 14 Solutions

CALCULUS EARLY TRANSCENDENTAL FUNCTIONS

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The area of the region bounded by the graphs of given equation y = 6 x − x 2 and y = x 2 − 2 x .

Solution Summary: The author explains that the area of a region is given by 643.

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To calculate: The area of the region bounded by the graphs of given equation y=6x−x2 and y=x2−2x.

Chapter 14 Solutions