Explanation Given Information: The provided functions are y = x 2 and y = 4 x + 5 . Formula Used: The 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 and between x = a and x = b is given by: A = ∫ a b [ f ( x ) − g ( x ) d x ] Calculation: The provided functions are y = x 2 and y = 4 x + 5 . The points can be obtained by keeping them equal: x 2 = 4 x + 5 x 2 − 4 x − 5 = 0 x 2 − 5 x + x − 5 = 0 x ( x − 5 ) + 1 ( x − 5 ) = 0 Further simplify, x ( x − 5 ) + 1 ( x − 5 ) = 0 ( x + 1 ) ( x − 5 ) = 0 Therefore, x = − 1 or x = 5 . Substitute x = − 1 in the equation y = x 2 . y = x 2 = ( − 1 ) 2 = 1 Now, substitute x = 5 in the equation y = x 2 . y = x 2 = ( 5 ) 2 = 25 Therefore, both graphs intersect at points ( − 1 , 1 ) and ( 5 , 25 ) . Consider the equation y = x 2 . The graph is obtained as follows: Substitute x = 0 in y = x 2 . y = ( 0 ) 2 = 0 Substitute x = 1 in y = x 2 . y = ( 1 ) 2 = 1 Substitute x = 3 in y = x 2 . y = ( 3 ) 2 = 9 Summarized all points in a table, x y (x,y) 0 0 ( 0 , 0 ) 1 1 ( 1 , 1 ) 3 9 ( 3 , 9 ) Now, consider the equation y = 4 x + 5 , Substitute x = 0 in y = 4 x + 5 . y = 4 ( 0 ) + 5 = 5 Substitute x = 1 in y = 4 x + 5 . y = 4 ( 1 ) + 5 = 9 Substitute x = 3 in y = 4 x + 5

12th Edition

ISBN: 9780357865095

Author: HARSHBARGER

Publisher: CENGAGE L

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Chapter 13, Problem 20RE

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To calculate: The area under the graph of y=x2 and y=4x+5 using the definite integral method.

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Chapter 13, Problem 19RE

Chapter 13, Problem 21RE

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The area under the graph of y = x 2 and y = 4 x + 5 using the definite integral method.

Solution Summary: The author calculates the area under the graph of y=x2 and = 4x+5 using the definite integral method.

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To calculate: The area under the graph of y=x2 and y=4x+5 using the definite integral method.

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