I am looking into the integral: $$I=\int_0^1 x^x(1-x)^{2x}\,dx\neq\frac{3}{8}$$
How might you prove this to be true? What's tough is that the integral $$3/8\lt I<0.37503$$ numerically. I managed to prove this by means of Riemann sums, classically, but ideas like Taylor expansion are extremely difficult in this case...
This is not a solution but approximation. First, note that $\frac{3}{8}=0.3750000000$. The function $f(x)=x^x(1-x)^{2x}$ is convex on $(0,c_1)\cup(c_2,1)$ and concave on $(c_1,c_2)$, where $c_1\approx 0.2718247$ and $c_2\approx 0.5243816$.
So that applying the hermite-hadamard inequality \begin{eqnarray} f\left( {\frac{{a + b}}{2}} \right) \le \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx} \le \frac{{f\left( a \right) + f\left( b \right)}}{2}, \end{eqnarray} which hold for all convex functions $f$ defined on a real interval $[a,b]$. The inequality is reversed if $f$ is concave. The inequality is sharp in both sides.
In our case $[a,b]=[0,1]$.
Therefore, On $(0,c_1)$, we have
\begin{eqnarray} f\left( {\frac{{0 + c_1}}{2}} \right) \le \frac{1}{{c_1 - 0}}\int\limits_0^{c_1} {f\left( x \right)dx} \le \frac{{f\left( 0 \right) + f\left( c_1 \right)}}{2} \end{eqnarray} and we write \begin{eqnarray} 0.1991787227\le \int\limits_0^{c_1} {f\left( x \right)dx} \le 0.2149827495\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1) \end{eqnarray}
On $(c_2,1)$, we have
\begin{eqnarray} 0.04452261402\le \int\limits_{c_2}^{1} {f\left( x \right)dx} \le 0.07962031056\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2) \end{eqnarray} On $(c_1,c_2)$, $f$ is concave and thus the Hermite-Hadamard inequality is reversed, then we have \begin{eqnarray} 0.1155270131\le \int\limits_{c_1}^{c_2} {f\left( x \right)dx} \le 0.1164615612\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(3) \end{eqnarray}
Adding the inequalities (1)-(3), we get
\begin{eqnarray} 0.3581809649\le \int\limits_{0}^{1} {f\left( x \right)dx} \le 0.4110646213 \end{eqnarray} Due to sharpness of H.-H. inequality, this is the best possible analytic approximation even that the approximation of $c_1,c_2$ is almost accurate.
Using Maple I get the numerical solution 0.3750261533 which is not $\frac{3}{8}$.