Explanation Given: The functions are y = x 2 + 2 x , y = x 2 − 1 and h ( x ) = x 2 + x − 2 . 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 2 + 2 x , y = x 2 − 1 and h ( x ) = x 2 + x − 2 . We can rewrite the function as: y = x 2 + x − 2 The graphs of y = x 2 + 2 x and y = x 2 + x − 2 intersect where x 2 + 2 x = x 2 + x − 2 x 2 + 2 x − x 2 − x + 2 = 0 x + 2 = 0 That is. ( x + 2 ) = 0 x = − 2 The graphs of y = x 2 − 1 and y = x 2 + x − 2 intersect where x 2 − 1 = x 2 + x − 2 x 2 − 1 − x 2 − x + 2 = 0 − x + 1 = 0 That is. − x + 1 = 0 x = 1 The graphs of y = x 2 − 1 and y = x 2 + 2 x intersect where x 2 − 1 = x 2 + 2 x x 2 − 1 − x 2 − 2 x = 0 − 2 x − 1 = 0 That is. − 2 x − 1 = 0 − 2 x = 1 x = − 1 2 The graph of the function is: Thus, the enclosed area lies over the interval [ − 2 , − 1 2 ] and [ − 1 2 , 1 ] . The region enclosed by the graphs of y = x 2 + 2 x , y = x 2 − 1 and h ( x ) = x 2 + x − 2 over the interval [ − 2 , 1 ] . The graph of y = x 2 + 2 x lies above the graph of y = x 2 + x − 2 for the interval [ − 2 , − 1 2 ] and the graph of y = x 2 − 1 lies above the graph of y = x 2 + x − 2 for the interval [ − 1 2 , 1 ] ...

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 6, Problem 4CRE

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