Q 53 - Calculus

Question

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

$\int {\int }_R{x}^2{e}^{xy}dA,$

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

Step-by-Step Solution

Answer

The value of double integral is :-

$\int {\int }_R{x}^2{e}^{xy}dA=\frac{1}{2}$

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

Step 1. Given Information

We have given the following double integral :-

$\int {\int }_R{x}^2{e}^{xy}dA,$

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

We have to evaluate this double integral.

Step 2. Use iterated integrals

The given double integral is :-

$\int {\int }_R{x}^2{e}^{xy}dA,$

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

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 :-

$\int {\int }_R{x}^2{e}^{xy}dA={\int }_0^1{{\int }_0^1}_R{x}^2{e}^{xy}dydx$

Then by using iterated integrals, we have :-

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

Now we can solve this integral as following :-

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

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

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