Explanation Given: Vertices ( x 1 , y 1 ) , ( x 1 , y 2 ) , ( x 2 , y 2 ) a n d ( x 2 , y 1 ) , where x 1 < x 2 and y 1 < y 2 Calculation: Consider the vertices of the rectangle, ( x 1 , y 1 ) , ( x 1 , y 2 ) , ( x 2 , y 2 ) , ( x 2 , y 1 ) where x 1 < x 2 , y 1 < y 2 Consider the fundamental theorem of calculus states that, if f and F are real valued continuous functions on the closed interval [ a , b ] such that F is continuous on all point of the interval and the derivative of F is f for almost all points of the interval, then, F ' ( x ) = f ( x ) . Also, ∫ a b f ( x ) d x = F ( b ) − F ( a ) Consider the integral, ∫ ∫ R ∂ 2 f ∂ y ∂ x = ∫ ∫ ∂ 2 f ∂ y ∂ x d x d y Integrate with respect to x keeping y as constant and get, ∫ ∫ ∂ 2 f ∂ y ∂ x d x d y = ∫ f ( x , y ) x 1 x 2 d y = ∫

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Question

Chapter 12.1, Problem 57PS

To determine

To show:the given condition ∫ ∫ R ∂ 2 f ∂y∂x dA=f(x1,y1)−f(x2,y1)+f(x2,y2)−f(x1,y2)

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Chapter 12.1, Problem 56PS

Chapter 12.1, Problem 58PS

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Chapter 12 Solutions

Calculus

Ch. 12.1 - Prob. 1PS Ch. 12.1 - Prob. 2PS Ch. 12.1 - Prob. 3PS Ch. 12.1 - Prob. 4PS Ch. 12.1 - Prob. 5PS Ch. 12.1 - Prob. 6PS Ch. 12.1 - Prob. 7PS Ch. 12.1 - Prob. 8PS Ch. 12.1 - Prob. 9PS Ch. 12.1 - Prob. 10PS

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the given condition ∫ ∫ R ∂ 2 f ∂ y ∂ x d A = f ( x 1 , y 1 ) − f ( x 2 , y 1 ) + f ( x 2 , y 2 ) − f ( x 1 , y 2 )

Solution Summary: The author explains how the fundamental theorem of calculus states that if f and F are real valued continuous functions on the closed interval, and the derivative of F is

To show:the given condition ∫ ∫ R ∂ 2 f ∂y∂x dA=f(x1,y1)−f(x2,y1)+f(x2,y2)−f(x1,y2)

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Chapter 12 Solutions