Explanation Given: • y = ln x • y = ( ln x ) 2 Key Concepts Used: If we are given two functions y = f ( x ) a n d y = g ( x ) such that f ( x ) ≥ g ( x ) for all x in an interval [ a , b ] , then the graph of f ( x ) lies above the graph of g ( x ) , and the area between the graphs is equal to the integral of f ( x ) − g ( x ) : Area between the graphs= ∫ f ( x ) ⋅ d x − a b ∫ g ( x ) ⋅ d x a b = ∫ ( f ( x ) − g ( x ) ) ⋅ d x a b . To find the area between the graphs of two given functions, first we must determine which graph lies on top. The first step is to sketch the region, especially, find any points of intersection. Then set up the integrals and evaluate. Here in this problem, we have to find the area between the graphs of two given functions: y = ln x and y = ( ln x ) 2 . First we have to find point of intersections between the given two functions, if any exists. For that we have to equate the given two equations of the functions. Formula Used: Area between the graphs= ∫ f ( x ) ⋅ d x − a b ∫ g ( x ) ⋅ d x a b = ∫ ( f ( x ) − g ( x ) ) ⋅ d x a b . Integration by Parts Formula: ∫ u ( x ) ⋅ v ' ( x ) ⋅ d x = u ( x ) ⋅ v ( x ) − ∫ u ' ( x ) ⋅ v ( x ) ⋅ d x Calculation: Description: Area between the graphs= ∫ ( f ( x ) − g ( x ) ) ⋅ d x a b . Step 1 : Write the equations of the given two functions. y = ln x y = ( ln x ) 2 Step 2 : Find points of intersection between the given two functions, if any exist. For that we have to equate the given two equations of the functions. ( ln x ) 2 = ln x ( ln x ) 2 − ln x = 0 ln x ( ln x − 1 ) = 0 ln x = 0 , ln x = 1 Taking exponential on both sides, e ln x = e 0 , e ln x = e 1 x = 1 , x = e ( B e c a u s e e ln x = x ; e 0 = 1 , e 1 = e ) Thereby two graphs intersect at x = 1 , x = e . Hence graph of y = ln x lies above the graph of y = ( ln x ) 2 . Step 3 : From the above step we identified that graph of y = ln x lies above the graph of y = ( ln x ) 2 . Now we can apply the formula to find the area between the two graphs. Area between the graphs= ∫ ( f ( x ) − g ( x ) ) ⋅ d x a b . Here a = 1 , b = e a n d f ( x ) = ln x , g ( x ) = ( ln x ) 2 . Area= ∫ ( ln x − ( ln x ) 2 1 e ) ⋅ d x = ∫ ln x ⋅ d x − 1 e ∫ ( ln x ) 2 ⋅ d x 1 e Step 4 : By using integration by parts, we have to evaluate the integral ∫ ( ln x ) 2 ⋅ d x 1 e . For that we have to assign u ( x ) a n d v ' ( x ) . From that find the other two unknown values ∫ ( ln x ) 2 ⋅ d x 1 e u ( x ) = ( ln x ) 2 u ' ( x ) = d u d x = d d x ( ln x ) 2 = 2 ⋅ ln x ⋅ d d x ( ln x ) = 2 ⋅ ln x ⋅ 1 x = 2 ⋅ ln x x v ' ( x ) = 1 v ( x ) = ∫ v ' ( x ) ⋅ d x = ∫ 1 ⋅ d x = x Calculation Continued: Step 5 : Substitute the values of u ( x ) , u ' ( x ) , v ( x ) , v ' ( x ) in the formula of integral by parts to find the integral...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 8.1, Problem 80E

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Area enclosed by given two functions

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Chapter 8.1, Problem 79E

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Ch. 8.1 - Prob. 1PQ Ch. 8.1 - Prob. 2PQ Ch. 8.1 - Prob. 3PQ Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

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Area enclosed by given two functions

Solution Summary: The author explains how to determine the area between the graphs of two given functions.

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