Respuesta :
I'm guessing the purpose of this exercise is to find the average value of [tex]f(x,y)=e^{\sin(x+y)}[/tex] over the region [tex]D[/tex],
[tex]D=\left\{(x,y)\mid (x,y)\in[-\pi,\pi]^2\right\}[/tex]
which is
[tex]\displaystyle\frac{\displaystyle\iint_Df(x,y)\,\mathrm dx\,\mathrm dy}{\displaystyle\iint_D\mathrm dx\,\mathrm dy}[/tex]
The denominator is the measure (area) of [tex]D[/tex], which is easy to compute:
[tex]\displaystyle\iint_D\mathrm dx\,\mathrm dy=\int_{-\pi}^\pi\int_{-\pi}^\pi\mathrm dx\,\mathrm dy=(2\pi)^2=4\pi^2[/tex]
Not to be confused with the integral in the numerator:
[tex]\displaystyle\iint_Df(x,y)\,\mathrm dx\,\mathrm dy=\int_{-\pi}^\pi\int_{-\pi}^\pi e^{\sin(x+y)}\,\mathrm dx\,\mathrm dy[/tex]
but this integral is difficult to compute, hence the inequalities. We have
[tex]-1\le\sin(x+y)\le1[/tex]
[tex]\implies\dfrac1e\le e^{\sin(x+y)}\le e[/tex]
[tex]\implies\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}e\le\iint_De^{\sin(x+y)}\,\mathrm dx\,\mathrm dy\le\iint_De\,\mathrm dx\,\mathrm dy[/tex]
[tex]\implies\displaystyle\frac{4\pi^2}e\le\iint_De^{\sin(x+y)}\,\mathrm dx\,\mathrm dy\le4\pi^2e[/tex]
We have shown that if we integrate the given function over region D after dividing by [tex]\rm 4\pi^2[/tex] we will get [tex]\rm \frac{1}{e}\leqslant \iint_{D}^{}\frac{1}{4 \pi^2}e^{sin(x+y)}dxdy\leq e[/tex]
It is given that the function:
[tex]\rm f(x,y)=e^{sin(x+y)}[/tex]
over the region [tex]\rmD\left\{ {{(x,y)}|[-\pi ,\pi]\times[-\pi ,\pi]}\right\}[/tex]
It is required to show that:
[tex]\rm \frac{1}{e}\leqslant \iint_{D}\frac{1}{4\pi^2}e^{sin(x+y)}dxdy\leq e[/tex]
What is integration?
It is defined as the mathematical calculation by which we can sum up all the smaller parts into a unit.
We have inequalities in the question:
[tex]\rm -1\leqslant sin(x+y)\leq 1[/tex]
[tex]\rm e^{-1}\leqslant e^{sin(x+y)}\leq e[/tex] [tex]\rm (taking \ power \ of \ e \ both\ side)[/tex]
[tex]\rm \iint_{D}^{}\frac{dxdy}{e}\leqslant \iint_{D}^{}e^{sin(x+y)}dxdy\leq \iint_{D}^{}edxdy[/tex] [tex]\rm \\\\(By\ double \ integrating \ both \ side \ over \ the \ D \ region) \\[/tex]
After integration on both sides and putting limits { (x, y) | [-π, π] × [-π, π] } we will get,
[tex]\rm \frac{ 4\pi^2}{e}\leqslant \iint_{D}^{}e^{sin(x+y)}dxdy\leq 4\pi^2e[/tex] [tex]\rm (Dividing \ by\ 4\pi^2 \ on \ both \ side)[/tex]
[tex]\rm \frac{1}{e}\leqslant \iint_{D}^{}\frac{1}{4 \pi^2}e^{sin(x+y)}dxdy\leq e[/tex]
Thus, we have shown that if we integrate the given function over region D after dividing by [tex]\rm 4\pi^2[/tex] we will get [tex]\rm \frac{1}{e}\leqslant \iint_{D}^{}\frac{1}{4 \pi^2}e^{sin(x+y)}dxdy\leq e[/tex]
Learn more about integration here:
https://brainly.in/question/4630073
