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The problem stands as it follows: If $D = [-1,1] \times [-1,2]$ show
$$1\leq \int\int_D \frac{dxdy}{x^2 + y^2 +1} \leq 6$$

I've tried to compute de Integral so far and this is what I've got: $$I = \int_{-1}^2\frac{1}{\sqrt{y^2 +1}}\Biggl(\arctan\biggr(\frac{1}{\sqrt{y^2 +1}}\biggl) - \arctan\biggr(\frac{-1}{\sqrt{y^2 +1}}\biggl)\Biggr)dy$$

Now, looking to compute this integral seems pretty annoying, but I don't think it will give me the result I'm looking for.

Am I going through the right way or am I doing wrong?

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  • $\begingroup$ I dont know if it works (I dont have paper and pen) but the function is continuous and $D$ is compact. Maybe this help. $\endgroup$ – Lucas Corrêa Feb 24 at 22:35
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For all $x,y\in [-1,1]\times [-1,2]$, $1\le x^2+y^2+1\le 1^2+2^2+1=6$. Therefore, $\frac16\le \frac1{x^2+y^2+1}\le 1$. Integrate those inequalities, and $$1=\int_{-1}^1\int_{-1}^2\frac16\,dy\,dx \le \int_{-1}^1\int_{-1}^2\frac1{x^2+y^2+1}\,dy\,dx\le \int_{-1}^1\int_{-1}^2 1\,dy\,dx=6$$ There it is. If we're looking to show an inequality, start with simple estimates rather than trying to evaluate things exactly.

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