Explanation Given: The functions are y = x 3 − 2 x 2 + x and y = x 2 − x . Formula used: Area of the region A = ∫ a b [ ( Upper function ) − ( Lower function ) ] d x Where a ≤ x ≤ b Calculation: The functions are y = x 3 − 2 x 2 + x and y = x 2 − x . The graphs of y = x 3 − 2 x 2 + x and y = x 2 − x intersect where x 3 − 2 x 2 + x = x 2 − x x 3 − 2 x 2 + x − x 2 + x = 0 x 3 − 3 x 2 + 2 x = 0 x ( x 2 − 3 x + 2 ) = 0 Find the factors of the above equation. x ( x 2 − 2 x − x + 2 ) = 0 x ( x ( x − 2 ) − 1 ( x − 2 ) ) = 0 x ( x − 2 ) ( x − 1 ) = 0 That is. x = 0 or ( x − 2 ) = 0 or ( x − 1 ) = 0 x = 0 or x = 2 or x = 1 The graph of the function is: Thus, the enclosed area lies over the interval [ 0 , 1 ] and [ 1 , 2 ] . The region enclosed by the graphs of y = x 3 − 2 x 2 + x and y = x 2 − x over the interval [ 0 , 2 ] . The graph of y = x 3 − 2 x 2 + x lies above the graph of y = x 2 − x for the interval [ 0 , 1 ] and the graph of y = x 2 − x lies above the graph of y = x 3 − 2 x 2 + x for the interval [ 1 , 2 ] Now, find the area...

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 6, Problem 3CRE

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Compute the area of the region enclosed by the graphs of the function.

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Chapter 6, Problem 2CRE

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Chapter 6 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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Compute the area of the region enclosed by the graphs of the function.

Solution Summary: The graph of the function is: The enclosed area lies over the interval [ 0, 1 ] and [ 1, 2 ].

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Compute the area of the region enclosed by the graphs of the function.

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Chapter 6 Solutions