Explanation Given: The iterated integrals ∫ 0 1 ∫ 0 1 y − x ( x + y ) 3 d y d x And ∫ 0 1 ∫ 0 1 y − x ( x + y ) 3 d x d y have different values. Proof: Consider ∫ 0 1 ∫ 0 1 y − x ( x + y ) 3 d y d x And ∫ 0 1 ∫ 0 1 y − x ( x + y ) 3 d x d y Show that these iterated integrals have different values. First evaluate, ∫ 0 1 ∫ 0 1 y − x ( x + y ) 3 d y d x ∫ 0 1 ∫ 0 1 y − x ( x + y ) 3 d y d x = ∫ 0 1 ( ∫ 0 1 y − x ( x + y ) 3 d y ) d x = ∫ 0 1 ( ∫ 0 1 y + x − 2 x ( x + y ) 3 d y ) d x = ∫ 0 1 ( ( ∫ 0 1 1 ( x + y ) 2 − 2 x ( x + y ) 3 ) d y ) d x = ∫ 0 1 ( ∫ 0 1 d y ( x + y ) 2 − 2 x ∫ 0 1 d y ( x + y ) 3 ) d x Further simplify, ∫ 0 1 ∫ 0 1 y − x ( x + y ) 3 d y d x = ∫ 0 1 ( ∫ 0 1 d y ( x + y ) 2 − 2 x ∫ 0 1 d y ( x + y ) 3 ) d x = ∫ 0 1 ( − ( 1 x + y | 0 1 ) − 2 x − 2 ( 1 ( x + y ) 2 ) 0 1 ) d x = ∫ 0 1 ( − ( 1 x + 1 − 1 x ) + x ( 1 ( x + 1 ) 2 − 1 x 2 ) ) d x = ∫ 0 1 ( − 1 x + 1 + 1 x + x ( x + 1 ) 2 − 1 x ) d x Further simplify, ∫ 0 1 ∫ 0 1 y − x ( x + y ) 3 d y d x = ∫ 0 1 ( − 1 x + 1 + 1 x + x ( x + 1 ) 2 − 1 x ) d x = ∫ 0 1 ( − 1 x + 1 + x + 1 − 1 ( x + 1 ) 2 ) d x = ∫ 0 1 ( − 1 x + 1 + 1 x + 1 − 1 ( x + 1 ) 2 ) d x = − ∫ 0 1 d x ( x + 1 ) 2 Now integrate above integral with respect to x

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ISBN: 9781524916817

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

Publisher: Kendall Hunt Publishing

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Chapter 12.1, Problem 59PS

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To show:the iterated integrals ∫01 ∫ 0 1 y−x ( x+y ) 3 dydx and ∫01 ∫ 0 1 y−x ( x+y ) 3 dxdy have different values and explain why this does not contradict Fubini’s theorem

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Chapter 12.1, Problem 58PS

Chapter 12.1, Problem 60PS

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Chapter 12 Solutions

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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 iterated integrals ∫ 0 1 ∫ 0 1 y − x ( x + y ) 3 d y d x and ∫ 0 1 ∫ 0 1 y − x ( x + y ) 3 d x d y have different values and explain why this does not contradict Fubini’s theorem

Solution Summary: The author explains that the iterated integrals displaystyle 'int' - _01 — y-x(x+y)3

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To show:the iterated integrals ∫01 ∫ 0 1 y−x ( x+y ) 3 dydx and ∫01 ∫ 0 1 y−x ( x+y ) 3 dxdy have different values and explain why this does not contradict Fubini’s theorem

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Chapter 12 Solutions