Regarding the first question, consider discretization points $x_0=1,x_1,\dots,x_N=10$. By computing the first and the second derivative
\begin{align}
(x^x)'&=x^x\left(\log x+1\right)\\
(x^x)''&=x^x\left(\log x+1\right)\left(\log x+1+\frac{1}{x}\right)
\end{align}
you immediately get that the function is increasing and convex on $[1,9]$. From this the trapezoidal method is upper bound to the unknown integral:
\begin{align}
I_{\text{UB}}(N)=\sum_{i=1}^N\frac{x_{i-1}^{x_{i-1}}+x_i^{x_i}}{2}(x_i-x_{i-1})=\underbrace{\frac{x_N-x_0}{N}}_{\frac{9}{N}}\sum_{i=1}^N\frac{x_{i-1}^{x_{i-1}}+x_i^{x_i}}{2}
\end{align}
By considering the Taylor approximation of the first order on each subinterval made in the left point $x_{i-1}$, we get the lower bound:
\begin{align}
I_{\text{LB}}(N)=&\sum_{i=1}^Nx_{i-1}^{x_{i-1}}\left[1+\frac{x_i-x_{i-1}}{2}\left(\log x_{i-1}+1\right)\right](x_i-x_{i-1})=\nonumber\\
&\underbrace{\frac{x_N-x_0}{N}}_{\frac{9}{N}}\sum_{i=1}^Nx_{i-1}^{x_{i-1}}\left[1+\frac{x_i-x_{i-1}}{2}(\log x_{i-1}+1)\right]
\end{align}
The error is given by
\begin{align}
\text{e}(N)=I_{\text{UB}}(N)-I_{\text{LB}}(N)~\to_{N\to \infty}~0
\end{align}
where $I_{LB}<\text{err}<I_{UB}$. Numerically:
\begin{align}
N= 10^5, &~I_{\text{LB}} = 3057488867.500303, ~I_{\text{UB}} = 3057488934.374307, ~\text{err} = 66.87400\\
N= 10^6, &~I_{\text{LB}} = 3057488911.636012, ~I_{\text{UB}} = 3057488912.304783, ~\text{err} = 0.66877\\
N=10^7, &~I_{\text{LB}} = 3057488912.077400, ~I_{\text{UB}} = 3057488912.084087, ~\text{err} = 0.00668
\end{align}