Explanation Given information: The provided curve is: y = x 4 − x 2 and y = x 2 − x 4 . Formula used: Area between two graphs: If f ( x ) ≥ g ( x ) for all x in [ a , b ] (so that the graph of f doesn’t move below that of g ), then the area of the region between the graphs of f and g are given and between x = a and x = b is given by: A = ∫ a b [ f ( x ) − g ( x ) d x ] Or A = Integral of ( Top − Bottom ) Calculation: Consider the provided equation: y = x 4 − x 2 and y = x 2 − x 4 Draw the graph of the above function to check whether one is above the other or they cross: Calculate the values at different values of x : Substitute x = 0 in the function y = x 4 − x 2 . y = ( 0 ) 4 − ( 0 ) 2 = 0 Substitute x = 1 in the function y = x 4 − x 2 . y = ( 1 ) 4 − ( 1 ) 2 = 0 Substitute x = 2 in the function y = x 4 − x 2 . y = ( 2 ) 4 − ( 2 ) 2 = 16 − 4 = 12 Substitute x = − 1 in the function y = x 4 − x 2 . y = ( − 1 ) 4 − ( − 1 ) 2 = 0 The table provided below shows the values of the function at different values of x . x y = x 4 − x 2 Coordinates ( x , y ) 0 0 ( 0 , 0 ) 1 0 ( 1 , 0 ) 2 12 ( 2 , 12 ) − 1 0 ( − 1 , 0 ) Now, consider y = x 2 − x 4 Substitute x = 0 in the function y = x 2 − x 4 . y = ( 0 ) 2 − ( 0 ) 4 = 0 Substitute x = 1 in the function y = x 2 − x 4 . y = ( 1 ) 2 − ( 1 ) 4 = 0 Substitute x = 2 in the function y = x 2 − x 4 . y = ( 2 ) 2 − ( 2 ) 4 = 4 − 16 = − 12 Substitute x = − 1 in the function y = x 2 − x 4 . y = ( − 1 ) 2 − ( − 1 ) 4 = 0 The table provided below shows the values of the function at different values of x . x y = x 2 − x 4 Coordinates ( x , y ) 0 0 ( 0 , 0 ) 1 0 ( 1 , 0 ) 2 − 12 ( 2 , − 12 ) − 1 0 ( − 1 , 0 ) Use the above tables and information to draw the graph of y = x 4 − x 2 and y = x 2 − x 4 in the same graph so as to check whether they cut or not: It can be seen the graph cuts each other. It can be deduced from the above graph that the lower region g ( x ) = x 2 − x 4 and upper region is f ( x ) = x 4 − x 2 . Since, interval is not provided, which means crossing point is to be consider as interval Calculation for cross point of the graph to get: x 4 − x 2 = x 2 − x 4 2 x 4 − 2 x 2 = 0 2 x 2 ( x 2 − 1 ) = 0 Thus 2 x 2 = 0 x = 0 And x 2 = 1 x = ± 1 So, consider the interval as [ − 1 , 0 ] and [ 0 , 1 ]

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ISBN: 9781337291408

Author: Stefan Waner, Steven Costenoble

Publisher: Brooks Cole

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Chapter 7.2, Problem 30E

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To calculate: The area of the region enclosed y=x4−x2 and y=x2−x4.

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Chapter 7.2, Problem 29E

Chapter 7.2, Problem 31E

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The area of the region enclosed y = x 4 − x 2 and y = x 2 − x 4 .

Solution Summary: The author calculates the area of the region enclosed by the equation y=x4-X2.

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To calculate: The area of the region enclosed y=x4−x2 and y=x2−x4.

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Chapter 7 Solutions