Calculate the integral, draw the region A and express it appropriately to interchange the order of integration $$\int_Af=\int_{-1}^1\int _{-2|x|}^{|x|}e^{x+y}\,dy\,dx$$ Let's first see what $A$ is like
Note that in our drawing of A $ -1 \leq x \leq 1 $ y $ -2 \leq y \leq 1 $
We have already found the new outer limits, now we will find the inner ones. On the other hand, let us note that we can express our integral as the sum of two integrals, taking the left part as $ B $ and the right part as $ C $, so $$\int_Af=\int_Bf+\int_Cf= \int_{-1}^1\int _{2x}^{-x} e^{x+y} \,dy\,dx+\int_{-1}^1\int _{-2x}^x e^{x+y} \, dy \, dx$$ Let's start by changing the order of integration in $B$, we know that $ -2 \leq y \leq 1 $ if $ y = -x $ then $ -y = x $, on the other hand if $ y = 2x $ then $\frac{y}{2} = x$. Then $$\int_Bf=\int_{-1}^1\int _{2x}^{-x} e^{x+y} \, dy \, dx = \int_{-2}^1\int_{\frac{y}{2}}^{-y}e^{x+y} \, dx \, dy$$ In the same way for the integral in $C,$ $ y = x $ and if $ y = -2x $ then $ - \frac {y} {2} = x $ then
$$\int_Cf=\int_{-1}^1\int_{-2x}^x e^{x+y} \, dy \, dx = \int_{-2}^1 \int_{-\frac{y}{2}}^y e^{x+y} \, dx \, dy$$ then $$\int_Af = \int_{-2}^1\int_{\frac{y}{2}}^{-y} e^{x+y} \, dx \, dy + \int_{-2}^1\int_{-\frac{y}{2}}^y e^{x+y} \, dx \, dy$$ However, at the time of integral I do not get the desired result, I suppose that the way in which I change the integration limits is not adequate, can someone help me to change the integration limits appropriately?
According to your drawing, we have that $$\begin{align}\int_A f&=\int_{-1}^1\int _{-2|x|}^{|x|}e^{x+y}\,dy\,dx\\ &=\overbrace{\int_{-2}^0\int_{-1}^{y/2}e^{x+y}\,dx\,dy}+\overbrace{\int_{-2}^0\int_{-y/2}^{1}e^{x+y}\,dx\,dy}\\\qquad&+ \underbrace{\int_{0}^1\int_{-1}^{-y}e^{x+y}\,dx\,dy}_B+\underbrace{\int_{0}^1\int_{y}^{1}e^{x+y}\,dx\,dy}_A.\end{align}$$ In both cases the result is $$\frac{e^2}{2}+e^{-1}-\frac{5}{6}+\frac{e^{-3}}{3}.$$ A third way to express the same integral is $$\int_{-2}^1\int_{-1}^{1}e^{x+y}\,dx\,dy-\int_{-2}^0\int_{y/2}^{-y/2}e^{x+y}\,dx\,dy-\int_{0}^1\int_{-y}^{y}e^{x+y}\,dx\,dy.$$