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I have this double integral over a domain which I want to solve:

$$ \int\int_B 1 \ dxdy \\B=\{(x,y) \in \mathbb{R}^2 | \frac{1}{4} x^2 -1 \leq y \leq 2-x \} $$

So I tried to solve the inequation to get my limits for the integral which is where I failed:

I have cheated by looking at what wolfram-alpha calculates but I have no idea how I come to that result by myself.

$$ \frac{1}{4} x ^2 -1 \leq y \leq 2-x $$ $$ x^2-4 \leq 4y \quad \land \quad y \leq 2-x $$ $$ x^2 \leq 4y+4 \quad \land \quad x+y \leq 2 $$

I have no idea how I can get further than that.

Also when I try to plug the results which I got from WA into a integration calculator I get not the results which I already had to check my answer for correctness.

I tried to calculate (the values that WA gave me) $$\int_{-6}^2 \int_{\frac{1}{4}x^2-4}^{2-x}1 \ dydx = \frac{136}{3}\neq \frac{-33}{3}$$

How can I solve this inequality and where I my error of thought ?

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It should be $$\iint_B 1 \ dxdy=\int_{-6}^2\left( \int_{\frac{x^2}{4}-1}^{2-x}1 \ dy\right)dx =\int_{-6}^2 \left(3-x-\frac{x^2}{4} \right)dx\\=\left[3x-\frac{x^2}{2}-\frac{x^3}{12}\right]_{-6}^2=\frac{64}{3}.$$ P.S. Note that \begin{align*} &2-x\geq \frac{x^2}{4}-1\Leftrightarrow 2-x-\frac{x^2}{4}+1\geq 0\Leftrightarrow(−1/4)(x^2+4x−12)\geq 0\\ &\Leftrightarrow(x^2+4x−12)\leq 0\Leftrightarrow(x+6)(x-2)\leq 0 \Leftrightarrow x\in[-6,2]. \end{align*}

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  • $\begingroup$ im trying to undestand your answer and I dont understand where the $(x+6)$ comesfrom in the second term and why the $\frac{1}{4}$ disappears in the third term $\endgroup$ – zython Aug 19 '17 at 11:29
  • $\begingroup$ @zython At the second step, putting all the terms on the left side you obtain $(-1/4)(x^2+4x-12)$. Then factorize $x^2+4x-12$ as $(x+6)(x-2)$. Finally note that $(-1/4)A\geq 0$ iff $A\leq 0$ because $-1/4$ is negative. $\endgroup$ – Robert Z Aug 19 '17 at 11:43
  • $\begingroup$ sorry if this question sounds dumb but what do you mean " putting all the terms on the left side" ? $\endgroup$ – zython Aug 19 '17 at 11:49
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    $\begingroup$ @zython More details are added to my P.S. in the answer. $\endgroup$ – Robert Z Aug 19 '17 at 11:55
  • $\begingroup$ thank you for your answer and your time, this answer makes sense to me know $\endgroup$ – zython Aug 19 '17 at 12:02

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