Question

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

$$\begin{gathered} \int {\int }_{R}x{e}^{xy}dA, \\ where\{R=(x,y)|0\le x\le 1and0\le y\le ln5\} \end{gathered}$$

Short answer

The value is$\frac{4}{\ln 5}-1$

Step-by-step solution

Step 1. Given Information:

Given double integrals :

$$\begin{gathered} \int {\int }_{R}x{e}^{xy}dA, \\ where\{R=(x,y)|0\le x\le 1and0\le y\le ln5\} \end{gathered}$$

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

Step 2. Solution:

$$\begin{gathered} U\sin gFubini\text{'}sTheorem \\ \int {\int }_{R}x{e}^{xy}dA={\int }_0^1{\int }_0^{ln5}x{e}^{xy}dydx \\ Evaluationprocedurefortheiteratedintegralweget \\ ={\int }_0^1\left({\int }_0^{\ln 5}x{e}^{xy}dy\right)dx \\ ={\int }_0^{1}x\left({\int }_0^{\ln 5}{e}^{xy}dy\right)dx \end{gathered}$$

$$\begin{gathered} Firstwesolve \\ {\int }_0^{\ln 5}{e}^{xy}dy \\ Putxy=tsowehavedy=\frac{dt}{x} \\ Wheny=\ln \left(5\right)thent=x\ln \left(5\right) \\ Wheny=0thent=0 \\ Sointegralbecomes: \\ ={\int }_0^{x\ln 5}\frac{e^t}{x}dt \\ =\frac{1}{x}{\int }_0^{x\ln 5}{e}^{t}dt \\ ByFundamentalTheoremofCalculuswehave \\ =\frac{1}{x}{\left[e^t\right]}_0^{x\ln 5} \\ Evaluationoftheinnerantiderivativeweget \\ =\frac{1}{x}\left[e^{\ln 5^x}-1\right] \\ Sowehave \\ ={\int }_0^{1}x\frac{1}{x}\left[5^x-1\right]dx \\ ={\int }_0^1\left[5^x-1\right]dx \end{gathered}$$

$$\begin{gathered} ByFundamentalTheoremofCalculuswehave \\ ={\int }_0^1{5}^{x}dx-{\int }_0^{1}dx \\ ={\int }_0^1{e}^{\ln 5^x}dx-{\int }_0^{1}dx \\ ={\int }_0^1{e}^{x\ln 5}dx-{\int }_0^{1}dx \\ ={\left[\frac{e^{x\ln 5}}{\ln 5}\right]}_0^1-{\left[x\right]}_0^1 \\ =\frac{e^{\ln 5}-1}{\ln 5}-1 \\ =\frac{4}{\ln 5}-1 \end{gathered}$$