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Answers without the blur. Sign up and see all textbooks for free! Q. 45

Expert-verified Found in: Page 1004 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Evaluate each of the double integrals in Exercises 37–54 as iterated integrals. $\int {\int }_{R}\frac{x}{x+y}dA,\phantom{\rule{0ex}{0ex}}whereR=\left\{\left(x,y\right)|1\le x\le 4and0\le y\le 3\right\}$

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

See the step by step solution

## Step 1. Given Information:

Given double integrals :

$\int {\int }_{R}\frac{x}{x+y}dA,\phantom{\rule{0ex}{0ex}}whereR=\left\{\left(x,y\right)|1\le x\le 4and0\le y\le 3\right\}$

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

## Step 2. Solution:

Using Fubini's Theorem

$\int {\int }_{R}\frac{x}{x+y}dA={\int }_{1}^{4}{\int }_{0}^{3}\frac{x}{x+y}dydx\phantom{\rule{0ex}{0ex}}Evaluationprocedurefortheiteratedintegralweget\phantom{\rule{0ex}{0ex}}={\int }_{1}^{4}\left({\int }_{0}^{3}\frac{x}{x+y}dy\right)dx\phantom{\rule{0ex}{0ex}}={\int }_{1}^{4}x{\int }_{0}^{3}\frac{1}{x+y}dydx$

First we solve:

${\int }_{0}^{3}\frac{1}{x+y}dy\phantom{\rule{0ex}{0ex}}={\left[\mathrm{ln}\left(x+t\right)\right]}_{0}^{3}\phantom{\rule{0ex}{0ex}}=\mathrm{ln}\left(x+3\right)-\mathrm{ln}x\phantom{\rule{0ex}{0ex}}Sointegralbecomes:\phantom{\rule{0ex}{0ex}}={\int }_{1}^{4}x\left[\mathrm{ln}\left(x+3\right)-\mathrm{ln}x\right]dx\phantom{\rule{0ex}{0ex}}={\int }_{1}^{4}x\mathrm{ln}\left(x+3\right)dx-{\int }_{1}^{4}x\mathrm{ln}xdx\phantom{\rule{0ex}{0ex}}={I}_{1}-{I}_{2}\phantom{\rule{0ex}{0ex}}Solve{I}_{1}wehave\phantom{\rule{0ex}{0ex}}Putx+3=tsowegetdx=dt\phantom{\rule{0ex}{0ex}}whenx=4thent=7\phantom{\rule{0ex}{0ex}}whenx=1thent=4\phantom{\rule{0ex}{0ex}}{I}_{1}becomes{\int }_{4}^{7}\left(t-3\right)\mathrm{ln}\left(t\right)dt\phantom{\rule{0ex}{0ex}}ApplyintegrationByparts:\phantom{\rule{0ex}{0ex}}={\left[\mathrm{ln}\left(t\right)\left(\frac{{t}^{2}}{2}-3t\right)-\int \frac{1}{2}\left(t-6\right)dt\right]}_{4}^{7}\phantom{\rule{0ex}{0ex}}={\left[\mathrm{ln}\left(t\right)\left(\frac{{t}^{2}}{2}-3t\right)-\frac{1}{2}\left(\frac{{t}^{2}}{2}-6t\right)\right]}_{4}^{7}\phantom{\rule{0ex}{0ex}}=8\mathrm{ln}\left(2\right)+\frac{3}{4}+\frac{7}{2}ln\left(7\right)\phantom{\rule{0ex}{0ex}}Solve{I}_{2}wehave\phantom{\rule{0ex}{0ex}}{\int }_{1}^{4}x\mathrm{ln}xdx\phantom{\rule{0ex}{0ex}}ApplyintegrationByparts:\phantom{\rule{0ex}{0ex}}={\left[\mathrm{ln}\left(x\right)\frac{{x}^{2}}{2}-\int \frac{x}{2}dx\right]}_{1}^{4}\phantom{\rule{0ex}{0ex}}={\left[\mathrm{ln}\left(x\right)\frac{{x}^{2}}{2}-\frac{{x}^{2}}{4}\right]}_{1}^{4}\phantom{\rule{0ex}{0ex}}=16\mathrm{ln}\left(2\right)-\frac{15}{4}\phantom{\rule{0ex}{0ex}}Sowehave\phantom{\rule{0ex}{0ex}}{I}_{1}-{I}_{2}=8\mathrm{ln}\left(2\right)+\frac{3}{4}+\frac{7}{2}ln\left(7\right)-16\mathrm{ln}\left(2\right)+\frac{15}{4}\phantom{\rule{0ex}{0ex}}=-8\mathrm{ln}\left(2\right)+\frac{9}{2}+\frac{7}{2}\mathrm{ln}\left(7\right)$ ### Want to see more solutions like these? 