Q. 43 - Calculus

Question

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

$$\begin{gathered} \int {\int }_{R}x{e}^{xy} dA, \\ where \{R=(x,y)| 0\le x\le 1 and 0\le y\le ln5\} \end{gathered}$$

Step-by-Step Solution

Answer

The value is $\frac{4}{\ln 5}-1$

Step 1. Given Information:

Given double integrals :

$$\begin{gathered} \int {\int }_{R}x{e}^{xy} dA, \\ where \{R=(x,y)| 0\le x\le 1 and 0\le y\le ln5\} \end{gathered}$$

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

Step 2. Solution:

$$\begin{gathered} U\sin g Fubini\text{'}s Theorem \\ \int {\int }_{R}x{e}^{xy} dA={\int }_0^1{\int }_0^{ln5}x{e}^{xy} dy dx \\ Evaluation procedure for the iterated integral we get \\ ={\int }_0^1\left({\int }_0^{\ln 5}x{e}^{xy} dy\right) dx \\ ={\int }_0^{1}x\left({\int }_0^{\ln 5}{e}^{xy} dy\right) dx \end{gathered}$$

$$\begin{gathered} First we solve \\ {\int }_0^{\ln 5}{e}^{xy} dy \\ Put xy=t so we have dy=\frac{dt}{x} \\ When y=\ln \left(5\right) then t=x\ln \left(5\right) \\ When y=0 then t=0 \\ So integral becomes: \\ ={\int }_0^{x\ln 5}\frac{e^t}{x} dt \\ =\frac{1}{x}{\int }_0^{x\ln 5}{e}^t dt \\ By Fundamental Theorem of Calculus we have \\ =\frac{1}{x}{\left[e^t\right]}_0^{x\ln 5} \\ Evaluation of the inner antiderivative we get \\ =\frac{1}{x}\left[e^{\ln 5^x}-1\right] \\ So we have \\ ={\int }_0^{1}x\frac{1}{x}\left[5^x-1\right] dx \\ ={\int }_0^1\left[5^x-1\right] dx \end{gathered}$$

$$\begin{gathered} By Fundamental Theorem of Calculus we have \\ ={\int }_0^1{5}^x dx-{\int }_0^{1}dx \\ ={\int }_0^1{e}^{\ln 5^x} dx-{\int }_0^{1}dx \\ ={\int }_0^1{e}^{x\ln 5} dx-{\int }_0^{1}dx \\ ={\left[\frac{e^{x\ln 5}}{\ln 5}\right]}_0^1-{\left[x\right]}_0^1 \\ =\frac{e^{\ln 5}-1}{\ln 5}-1 \\ =\frac{4}{\ln 5}-1 \end{gathered}$$