How to calculate the area between $y=e^{-x}$, $y=x$ and $x=0$ My problem is that little point that comes from the equation $$e^{-x} = x$$
I can't solve that one. Is there another way without knowing that point or a way to calculate it?
Thanks in advance!

 A: We want to figure out when $e^{-x} = x$. Multiplying by $e^x$ gives
$$
xe^x = 1
$$
The solution to this equation is defined as the $\Omega$-constant, and shares many interesting properties. So we have $\Omega e^\Omega=1$ or $e^{\Omega} = 1/\Omega$ or $\Omega=e^{-\Omega}$. Several fast approximations can be found at the link above. The integral is then
\begin{align*}
A 
= \int_0^\Omega x\,\mathrm{d}x + \int_\Omega^\infty e^{-x}\,\mathrm{d}x
= e^{-\Omega} + \frac{1}{2}\Omega^2 = \Omega\left( 1+\frac{\Omega}{2}\right)\tag{1}
\end{align*}
Where we used that $e^{-\Omega}=\Omega$. For simplicity let us assume that $\Omega \approx \frac{5}{6}\log 2$ then
$$
A \approx \frac{25}{72}(\log 2)^2+\frac{1}{2}2^{1/6}
$$
If we really want to be fancy we can now use that $\log 2 \approx \frac 7{10}$ and $\frac{1}{2}2^{1/6}\approx \frac{1}{2}\left(1+1\cdot\frac{1}{6} \right)$. Where the last approximation comes from $(1+x)^{a} \approx 1 + ax$ where $|x|\leq 1$ and $a<1$. I will leave the rest to you
EDIT: The continued fraction expansion of $\Omega$ is given as
$$[0; 1, 1, 3, 4, 2, 10, 4, 1, 1, 1, 1, 2, 7, 306, 1, 5, 1, 2, 1, 5,\ldots]$$
Hence 
$$
\Omega = \frac{1}{1 + \cfrac{1}{1 + \cfrac{1}{3+\cfrac{1}{4+\cfrac{1}{1+\frac{1}{\cdots}}}}}} \approx 38/67
$$
Inserting this into $(1)$ and using $e^{-\Omega}\approx 1/\Omega$ we have
$$
    A \approx \exp\left(-\frac{38}{67}\right) + \frac{1}{2}\left(\frac{38}{67}\right)^2 
$$
Which is correct to about $10$ digits. Now $e^{-\Omega}=\Omega$ only holds for the exact Omega constant. Otherwise it is an approximation. Hence a tad worse version would have been
$$
    A \approx \frac{38}{67} + \frac{1}{2}\left(\frac{38}{67}\right)^2 = \frac{3268}{4489}
$$
Which is only correct for a handful digits ($3$).
