Q. 45 - Calculus

Question

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

$$\begin{gathered} \int {\int }_R\frac{x}{x+y}dA, \\ whereR=\left\{\right(x,y) | 1\le x\le 4 and 0\le y\le 3\} \end{gathered}$$

Step-by-Step Solution

Answer

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

Step 1. Given Information:

Given double integrals : 

$$\begin{gathered} \int {\int }_R\frac{x}{x+y}dA, \\ whereR=\left\{\right(x,y) | 1\le x\le 4 and 0\le y\le 3\} \end{gathered}$$

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

Step 2. Solution:

Using Fubini's Theorem

$$\begin{gathered} \int {\int }_R\frac{x}{x+y}dA={\int }_1^4{\int }_0^3\frac{x}{x+y}dy dx \\ Evaluation procedure for the iterated integral we get \\ ={\int }_1^4\left({\int }_0^3\frac{x}{x+y}dy\right) dx \\ ={\int }_1^{4}x{\int }_0^3\frac{1}{x+y}dy dx \end{gathered}$$

First we solve:

$$\begin{gathered} {\int }_0^3\frac{1}{x+y}dy \\ ={\left[\ln (x+t)\right]}_0^3 \\ =\ln (x+3)-\ln x \\ So integral becomes: \\ ={\int }_1^{4}x\left[\ln (x+3)-\ln x\right]dx \\ ={\int }_1^{4}x\ln (x+3)dx-{\int }_1^{4}x\ln xdx \\ =I_1-I_2 \\ Solve I_1 we have \\ Put x+3=t so we get dx=dt \\ when x=4 then t=7 \\ when x=1 then t=4 \\ {I}_1 becomes {\int }_4^7(t-3) \ln \left(t\right) dt \\ Apply integration By parts: \\ ={\left[\ln \left(t\right)\left(\frac{t^2}{2}-3t\right)-\int \frac{1}{2}\left(t-6\right)dt\right]}_4^7 \\ ={\left[\ln \left(t\right)\left(\frac{t^2}{2}-3t\right)-\frac{1}{2}\left(\frac{t^2}{2}-6t\right)\right]}_4^7 \\ =8\ln \left(2\right)+\frac{3}{4}+\frac{7}{2}ln\left(7\right) \\ Solve I_2 we have \\ {\int }_1^{4}x\ln xdx \\ Apply integration By parts: \\ ={\left[\ln \left(x\right)\frac{x^2}{2}-\int \frac{x}{2}dx\right]}_1^4 \\ ={\left[\ln \left(x\right)\frac{x^2}{2}-\frac{x^2}{4}\right]}_1^4 \\ =16 \ln \left(2\right)-\frac{15}{4} \\ So we have \\ {I}_1-I_2=8\ln \left(2\right)+\frac{3}{4}+\frac{7}{2}ln\left(7\right)-16 \ln \left(2\right)+\frac{15}{4} \\ =-8\ln \left(2\right)+\frac{9}{2}+\frac{7}{2}\ln \left(7\right) \end{gathered}$$