How to show that $\frac{\pi}{5}\leq\int_0^1 x^x\,dx\leq\frac{\pi}{4}$ Show that:
$$\frac{\pi}{5}\leq\int_0^1 x^x\,dx\leq\frac{\pi}{4}$$
All I've got so far is that the minimum of $x^x$ is $e^{-1/e}$. At this point I could 
compare $\pi/5$ to  $e^{-1/e}$ but I'm required to prove both sides without using the calculator. This is all I've got at the moment.
 A: Changing variables $x\mapsto e^{-x}$ yields
$$
\begin{align}
\int_0^1(x\log(x))^n\,\mathrm{d}x
&=\int_\infty^0(-xe^{-x})^n\,\mathrm{d}e^{-x}\\
&=(-1)^n\int_0^\infty x^ne^{-(n+1)x}\,\mathrm{d}x\\
&=\frac{(-1)^n}{(n+1)^{n+1}}\int_0^\infty x^ne^{-x}\,\mathrm{d}x\\
&=\frac{(-1)^nn!}{(n+1)^{n+1}}\tag{1}
\end{align}
$$
Pluging $(1)$ into $\displaystyle e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$ gives us
$$
\int_0^1x^x\,\mathrm{d}x=\sum_{n=0}^\infty\frac{(-1)^n}{(n+1)^{n+1}}\tag{2}
$$
As an alternating series with decreasing absolute values, we know that by using $(2)$,
$$
\begin{align}
\int_0^1x^x\,\mathrm{d}x
&>1-\frac14\\
&=\frac34\\
&>\pi/5\tag{3}
\end{align}
$$
and
$$
\begin{align}
\int_0^1x^x\,\mathrm{d}x
&<1-\frac14+\frac{1}{27}-\frac{1}{256}+\frac{1}{3125}\\
&=\frac{16922537}{21600000}\\
&<\pi/4\tag{4}
\end{align}   
$$
A: It's already been mentioned in the comments that the minimum of the integrand (which is $(1/\mathrm e)^{1/\mathrm e}$, not $\mathrm e^{1/\mathrm e}$) is greater than $\pi/5$. However, proving that $(1/\mathrm e)^{1/\mathrm e}\gt\pi/5$ without a calculator would probably be rather tedious. A bound for which this would be slightly easier can be obtained by using the convexity of the exponential function:
$$
\begin{align}
\int_0^1x^x\mathrm dx=\int_0^1\exp(x\log x)\,\mathrm dx\ge\exp\left(\int_0^1x\log x\,\mathrm dx\right)=\exp\left(-\frac14\right)\gt\frac\pi5\;.\end{align}
$$
You still need to evaluate a couple of terms of some series whose error bounds you know in order to prove the last inequality, but it should be a bit easier than for $(1/\mathrm e)^{1/\mathrm e}$.
