Determine the area of the region bounded by $x=e^{1+2y}, x=e^{1-y}, y=-2, y=1$. I have no idea how to approach this question. I have already tried to graph it to get a better understanding of the question. However, I do not know what 'area' I am trying to find. Is it the part I have shaded?

 A: The graph is a good starting point, the intersection is at

*

*$e^{1+2y} =e^{1-y} \implies y=0, \: x=e$
therefore we have the following set up
$$A=\int_{-2}^0\int_{e^{1+2y}}^{e^{1-y}} dx dy+\int_{0}^1\int_{e^{1-y}}^{e^{1+2y}} dx dy$$
A: My approach:
$$x=e^{1+2y} \iff y= f(x) = \frac{\ln x - 1}{2}$$
$$x= e^{1-y} \iff y= g(x) = 1 - \ln x$$
Next step is to find all of intersection points $A, B, C, D \space \& \space E$.
The area, which I marked as $I_A$, is the sum of two integrals:
$$I_A=\int_{x(A)}^{x(B)}(1 - (1- \ln x))dx + \int_{x(B)}^{x(D)}(1 - \frac{\ln x - 1}{2})dx.$$
$I_B$ you will find similarly...
Figure:

A: $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}$
\begin{align} 
f_1(y)&=\exp(1+2y)
,\\
f_2(y)&=\exp(1-y)
\end{align}
\begin{align} 
S_{\mathsf{bot}}
&=
\int_{-2}^0 f_2(y)-f_1(y)\, dy
\\
&=-\exp(1-y)-\tfrac12\,\exp(1+2y)\Big|_{-2}^0
=\e^3+\frac1{2\e^3}-\tfrac32\,\e
,\\
S_{\mathsf{top}}
&=\int_0^1 f_1(y)-f_2(y)\, dy
\\
&=\tfrac12\,\exp(1+2y)+\exp(1-y)\Big|_0^1
=\tfrac12\,\e^3-\tfrac32\,\e+1
,\\
S_{\mathsf{tot}}
&=S_{\mathsf{bot}}
+S_{\mathsf{top}}
=
\frac1{2\,\e^3}+\tfrac32\,\e^3-3\,\e+1
\approx 22.998
.
\end{align}
$\endgroup$
