I have been give a function, its average value, and the lower bound of an interval, I am trying to find the upper bound but I am a bit lost on how to do that. Here is what I have so far:
$$ f(x)= \frac{(e^x)}{1+e^{2x}},\quad f_{ave}=\frac{\pi}{12}(\frac{1}{\ln(\sqrt(3))}), \quad x:[0,b] $$ $$\int_1^{u(b)}\frac{du}{1+u^2}$$ $$ =\arctan(u) $$ $$= \arctan(e^x)\vert_0^b $$ $$= \arctan(e^b)-\frac{\pi}{4} $$ $$\frac{\pi}{12}(\frac{1}{\ln(\sqrt(3))}) = (\arctan(e^b)-\frac{\pi}{4}) (\frac{1}{b-0}) $$ This is about as far as I can get. Help would be monstrously appreciated.
So we want that: $$\int_{0}^{b}\frac{e^x}{1+e^{2x}}\,dx = \int_{1}^{e^b}\frac{du}{1+u^2}=\arctan(e^b)-\frac{\pi}{4}=\frac{\pi b}{6\log 3} $$ and there is exactly one positive real solution, since $f(b)=\frac{\pi b}{6\log 3}$ is a convex function that equals $0$ at $b=0$ while $g(b)=\arctan(e^b)-\frac{\pi}{4}$ is a concave function that equals $0$ at $b=0$.
An "educated guess" is to try some value of $b$ that is a rational multiple of $\log 3$, hoping to have a simple closed form for both $f(b)$ and $g(b)$.
$$ b = \frac{1}{2}\log 3 =\color{red}{\log\sqrt{3}}$$ works pretty good, since $\arctan\sqrt{3}=\frac{\pi}{3}$.