91Ó°ÊÓ

We have given the following double integral :-

$∫{∫}_R{x}^2{e}^{xy}dA,$

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

We have to evaluate this double integral.

The given double integral is :-

$∫{∫}_R{x}^2{e}^{xy}dA,$

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

In order to solve this double integral we will firstly integrated with$y$.

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

role="math" localid="1650586863367" $∫{∫}_R{x}^2{e}^{xy}dA={∫}_0^1{{∫}_0^1}_R{x}^2{e}^{xy}dydx$

Then by using iterated integrals, we have :-

${∫}_0^1{{∫}_0^1}_R{x}^2{e}^{xy}dydx={∫}_0^1\left({{∫}_0^1}_R{x}^2{e}^{xy}dy\right)dx$

Now we can solve this integral as following :-

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

Now use product rule of integration $∫f\left(x\right)g\left(x\right)dx=f\left(x\right)∫g\left(x\right)dx-∫\left[\frac{d}{dx}f\left(x\right)\left(∫g\left(x\right)\right)dx\right]dx$

$$\begin{gathered} ={{\left[\left(x{e}^x-e^x\right)-\frac{x^2}{2}\right]}^1}_0 \\ =\left[\left(e-e-\frac{1}{2}\right)-\left(0-e^0-0\right)\right] \\ =-\frac{1}{2}+1 \\ =\frac{1}{2} \end{gathered}$$