Question

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

$\int {\int }_R{e}^{x+y}dA,$

where$R=\left\{\left(x,y\right)|0\le x\le 1and0\le y\le 1\right\}$.

Short answer

The value of double integral is :-

$\int {\int }_R{e}^{x+y}dA={\left(e-1\right)}^2$

where,$R=\left\{\left(x,y\right)|0\le x\le 1and0\le y\le 1\right\}$

Step-by-step solution

Step 1. Given Information 

We have given the following double integral :-

$\int {\int }_R{e}^{x+y}dA,$

where $R=\left\{\left(x,y\right)|0\le x\le 1and0\le y\le 1\right\}$.

We have to evaluate this double integral.

Step 2. Use iterated integrals

The given double integral is :-

$\int {\int }_R{e}^{x+y}dA,$

where $R=\left\{\left(x,y\right)|0\le x\le 1and0\le y\le 1\right\}$.

Then by using Fubini's Theorem, we can writ this double integral as following :-

role="math" localid="1650413616232" $\int {\int }_R{e}^{x+y}dA={\int }_0^1{\int }_0^1{e}^{x+y}dxdy$

We can write this as following :-

${\int }_0^1{\int }_0^1{e}^{x+y}dxdy={\int }_0^1{\int }_0^1{e}^x{e}^{y}dxdy$

Then by using iterated integrals, we have :-

role="math" localid="1650413690814" ${\int }_0^1{\int }_0^1{e}^x{e}^{y}dxdy={\int }_0^1\left({\int }_0^1{e}^x{e}^{y}dx\right)dy$

Now we can solve this integral as following :-

$$\begin{gathered} {\int }_0^1\left({\int }_0^1{e}^x{e}^{y}dx\right)dy \\ ={\int }_0^1{e}^y{{\left[e^x\right]}^1}_{0}dy \\ ={\int }_0^1{e}^y\left[e^1-e^0\right]dy \\ ={\int }_0^1{e}^y\left[e-1\right]dy \\ =\left[e-1\right]{{\left[e^y\right]}^1}_0 \\ =\left[e-1\right]\left[e^1-e^0\right] \\ =\left[e-1\right]\left[e-1\right] \\ ={\left(e-1\right)}^2 \end{gathered}$$