Explanation Given: ∫ 1 2 ∫ 1 3 ln ( x y ) y d y d x Calculation: Step 1: ∫ 1 3 ln ( x y ) y d y Compute the indefinite integral ∫ ln ( x y ) y d y Apply u substitution: u = ln(xy), v = 1 y = ln (xy)ln | y | - ∫ 1 y ln | y | d y (a) Apply INTEGRATION by parts: u = 1 y ln | y | , v = 1 = 1 y ln | y | y - ∫ − ln | y | + 1 y 2 y d y = ln | y | - ∫ − ln | y | + 1 y 2 y d y (1) ∫ − ln | y | + 1 y 2 y d y Apply the power rule, ∫ x a d x = x a + 1 a + 1 , a ≠ -1 = − 1 2 ( − ln | y | + 1 ) 2 (2) Subtract (2) in (1) = ln | y | - − 1 2 ( − ln | y | + 1 ) 2 Simplify, = 1 2 ln | y | 2 + 1 2 (b) Subtract (b) in (a) = ln(xy)ln | y | - 1 2 ln | y | 2 − 1 2 + C Step 2: Compute the boundaries ∫ 1 3 ln ( x y ) y d y Formula used: ∫ a b f ( x ) d x = F ( b ) − F ( a ) = lim x → b − ( F ( x ) ) - lim x → a + ( F ( x ) ) lim y → 1 + ( ln ( x y ) ln | y | − 1 2 ln | y | 2 − 1 2 ) = − 1 2 Value y = 1 lim y → 3 − ( ln ( x y ) ln | y | − 1 2 ln | y | 2 − 1 2 ) When value y = 3 = ln ( x ...

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ISBN: 9781319055844

Author: FRANZOSA

Publisher: VST

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Chapter 16.1, Problem 35E

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Evaluate the iterated integral $\int_{1}^{2} \int_{1}^{3} \frac{\ln (x y)}{y} d y d x$

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Chapter 16.1, Problem 34E

Chapter 16.1, Problem 36E

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Chapter 16 Solutions

EBK CALCULUS

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Evaluate the iterated integral ∫ 1 2 ∫ 1 3 ln ( x y ) y d y d x

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