Question

Evaluate each of the double integrals in Exercises 37–54 as iterated integrals.

$$\begin{gathered} \int {\int }_R\frac{x}{x+y}dA, \\ whereR=\left\{\right(x,y)|1\le x\le 4and0\le y\le 3\} \end{gathered}$$

Short answer

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

Step-by-step solution

Step 1. Given Information:

Given double integrals :

$$\begin{gathered} \int {\int }_R\frac{x}{x+y}dA, \\ whereR=\left\{\right(x,y)|1\le x\le 4and0\le y\le 3\} \end{gathered}$$

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

Step 2. Solution:

Using Fubini's Theorem

$$\begin{gathered} \int {\int }_R\frac{x}{x+y}dA={\int }_1^4{\int }_0^3\frac{x}{x+y}dydx \\ Evaluationprocedurefortheiteratedintegralweget \\ ={\int }_1^4\left({\int }_0^3\frac{x}{x+y}dy\right)dx \\ ={\int }_1^{4}x{\int }_0^3\frac{1}{x+y}dydx \end{gathered}$$

First we solve:

$$\begin{gathered} {\int }_0^3\frac{1}{x+y}dy \\ ={\left[\ln (x+t)\right]}_0^3 \\ =\ln (x+3)-\ln x \\ Sointegralbecomes: \\ ={\int }_1^{4}x\left[\ln (x+3)-\ln x\right]dx \\ ={\int }_1^{4}x\ln (x+3)dx-{\int }_1^{4}x\ln xdx \\ =I_1-I_2 \\ Solve{I}_{1}wehave \\ Putx+3=tsowegetdx=dt \\ whenx=4thent=7 \\ whenx=1thent=4 \\ {I}_{1}becomes{\int }_4^7(t-3)\ln \left(t\right)dt \\ ApplyintegrationByparts: \\ ={\left[\ln \left(t\right)\left(\frac{t^2}{2}-3t\right)-\int \frac{1}{2}\left(t-6\right)dt\right]}_4^7 \\ ={\left[\ln \left(t\right)\left(\frac{t^2}{2}-3t\right)-\frac{1}{2}\left(\frac{t^2}{2}-6t\right)\right]}_4^7 \\ =8\ln \left(2\right)+\frac{3}{4}+\frac{7}{2}ln\left(7\right) \\ Solve{I}_{2}wehave \\ {\int }_1^{4}x\ln xdx \\ ApplyintegrationByparts: \\ ={\left[\ln \left(x\right)\frac{x^2}{2}-\int \frac{x}{2}dx\right]}_1^4 \\ ={\left[\ln \left(x\right)\frac{x^2}{2}-\frac{x^2}{4}\right]}_1^4 \\ =16\ln \left(2\right)-\frac{15}{4} \\ Sowehave \\ {I}_1-I_2=8\ln \left(2\right)+\frac{3}{4}+\frac{7}{2}ln\left(7\right)-16\ln \left(2\right)+\frac{15}{4} \\ =-8\ln \left(2\right)+\frac{9}{2}+\frac{7}{2}\ln \left(7\right) \end{gathered}$$