Short Answer

Evaluate each of the double integrals in Exercises 37–54 as iterated integrals.
$\int \int_{R} \frac{x}{x + y} d A, w h e r e R = {(x, y) | 1 \leq x \leq 4 a n d 0 \leq y \leq 3}$

The value is $- 8 \ln (2) + \frac{9}{2} + \frac{7}{2} \ln (7)$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given Information:

Given double integrals :

$\int \int_{R} \frac{x}{x + y} d A, w h e r e R = {(x, y) | 1 \leq x \leq 4 a n d 0 \leq y \leq 3}$

We want to evaluate each of the double integrals as iterated integrals.

Step 2. Solution:

Using Fubini's Theorem

\int \int_{R} \frac{x}{x + y} d A = \int_{1}^{4} \int_{0}^{3} \frac{x}{x + y} d y d x E v a l u a t i o n p r o c e d u r e f o r t h e i t e r a t e d i n t e g r a l w e g e t = \int_{1}^{4} (\int_{0}^{3} \frac{x}{x + y} d y) d x = \int_{1}^{4} x \int_{0}^{3} \frac{1}{x + y} d y d x

First we solve:

$\int_{0}^{3} \frac{1}{x + y} d y = {[\ln (x + t)]}_{0}^{3} = \ln (x + 3) - \ln x S o i n t e g r a l b e c o m e s : = \int_{1}^{4} x [\ln (x + 3) - \ln x] d x = \int_{1}^{4} x \ln (x + 3) d x - \int_{1}^{4} x \ln x d x = I_{1} - I_{2} S o l v e I_{1} w e h a v e P u t x + 3 = t s o w e g e t d x = d t w h e n x = 4 t h e n t = 7 w h e n x = 1 t h e n t = 4 I_{1} b e c o m e s \int_{4}^{7} (t - 3) \ln (t) d t A p p l y i n t e g r a t i o n B y p a r t s : = {[\ln (t) (\frac{t^{2}}{2} - 3 t) - \int \frac{1}{2} (t - 6) d t]}_{4}^{7} = {[\ln (t) (\frac{t^{2}}{2} - 3 t) - \frac{1}{2} (\frac{t^{2}}{2} - 6 t)]}_{4}^{7} = 8 \ln (2) + \frac{3}{4} + \frac{7}{2} l n (7) S o l v e I_{2} w e h a v e \int_{1}^{4} x \ln x d x A p p l y i n t e g r a t i o n B y p a r t s : = {[\ln (x) \frac{x^{2}}{2} - \int \frac{x}{2} d x]}_{1}^{4} = {[\ln (x) \frac{x^{2}}{2} - \frac{x^{2}}{4}]}_{1}^{4} = 16 \ln (2) - \frac{15}{4} S o w e h a v e I_{1} - I_{2} = 8 \ln (2) + \frac{3}{4} + \frac{7}{2} l n (7) - 16 \ln (2) + \frac{15}{4} = - 8 \ln (2) + \frac{9}{2} + \frac{7}{2} \ln (7)$

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Calculus

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Short Answer

Evaluate each of the double integrals in Exercises 37–54 as iterated integrals.
$\int \int_{R} \frac{x}{x + y} d A, w h e r e R = {(x, y) | 1 \leq x \leq 4 a n d 0 \leq y \leq 3}$

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Calculus

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Evaluate each of the double integrals in Exercises 37–54 as iterated integrals. ∫∫Rxx+ydA,whereR={(x,y) | 1≤x≤4 and 0≤y≤3}

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Evaluate each of the double integrals in Exercises 37–54 as iterated integrals.
$\int \int_{R} \frac{x}{x + y} d A, w h e r e R = {(x, y) | 1 \leq x \leq 4 a n d 0 \leq y \leq 3}$