I have the following double integral
$$\int_0^1 \int_{e^x}^{e^{2x}} x \ln(y) dydx$$
I don't know how to plot a graph here so this is my graph attempt based on Wolfram Alpha 
I guess it should be
$$\int_0^{e^2} \int_0^{\ln(\frac{1}{\sqrt{y}})} x \ln(y) dxdy$$
The upper limit on $y$ is cause I assume $\ln(\sqrt{y})-\ln(y)$. Like in the area under the curve I did the same $\ln(\sqrt{y})$ isolated from $y=e^{2x}$ and similarly $\ln(y)$ from $y=e^x$
There's somethig wrong in my approach? The answer given for the first double integral is 1.37898
Please if you know a better way to analize this type of integral without the need to graph, show me.
Hint: You cann't solve these type of integrals without graphs. Think about it's area and find that $$\int_1^e\int_{\frac12\ln y}^{\ln y}x\ln y dx dy+\int_e^{e^2}\int_{\frac12\ln y}^{1}x\ln y dx dy$$
Edit: Look at the graph again
as you see in the strip $0\leqslant x\leqslant1$, the area is between two curves $y=e^x$ and $y=e^{2x}$. This view is seen from $x$-restriction to $y$-restriction. Other view is from $y$-restriction to $x$-restriction, in this case the area isn't uniform so you should look at it in two portion, $1\leqslant y\leqslant e$ and $e\leqslant y\leqslant e^2$. With $1\leqslant y\leqslant e$ (blue color), the area lies between $\dfrac12\ln y\leqslant x\leqslant \ln y$ while in $e\leqslant y\leqslant e^2$ (orange color), the area is between $\dfrac12\ln y\leqslant x\leqslant 1$.