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24 dA - I 3x – Y when D is the parallelogram bounded by x – 2y = 0, x – 2y = 2, - and Зх — у 3D 1, За — у — 2, |

24 dA - I 3x – Y when D is the parallelogram bounded by x – 2y = 0, x – 2y = 2, - and Зх — у 3D 1, За — у — 2, |

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Evaluate the integral \[ I = \int \int_D \frac{x - 2y}{3x - y} \, dA \] when \(D\) is the parallelogram bounded by \[ x - 2y = 0, \quad x - 2y = 2, \] and \[ 3x - y = 1, \quad 3x - y = 2, \] by making an appropriate change of variables. 1. \(I = \frac{2}{5}\) 2. \(I = \frac{2}{5} \ln 2\) 3. \(I = \frac{4}{5} \ln 2\) 4. \(I = 0\) 5. \(I = \frac{4}{5}\)
Transcribed Image Text:Evaluate the integral \[ I = \int \int_D \frac{x - 2y}{3x - y} \, dA \] when \(D\) is the parallelogram bounded by \[ x - 2y = 0, \quad x - 2y = 2, \] and \[ 3x - y = 1, \quad 3x - y = 2, \] by making an appropriate change of variables. 1. \(I = \frac{2}{5}\) 2. \(I = \frac{2}{5} \ln 2\) 3. \(I = \frac{4}{5} \ln 2\) 4. \(I = 0\) 5. \(I = \frac{4}{5}\)
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