Let $f(x)$ be a differentiable function on [$a,b$] and $T$ partition of [$a,b$], $T=${$x_0,x_1,\ldots,x_n$}. For arbitrary points $u_i$ we want to calculate the difference:
$$\left\lvert{\int_a^b{f(x)dx}-R(T)}\right\rvert$$
For any point $x \in $ [$x_i,x_{i+1}$] by the mean value theorem there exists $v_i \in (x_i,x_{i+1})$ such that:
$$f(x)-f(u_i)=f'(v_i)(x-u_i)$$
If we let $M=\sup\left\lvert{f'(x)}\right\rvert$. Then: $$\left\lvert{f(x)-f(u_i)}\right\rvert \le M\cdot \left\lvert{x_{i+1}-x_i}\right\rvert$$ And
$$\left\lvert \int_{x_i}^{x_{i+1}}{f(x)dx}-f(u_i)(x_{i+1}-x_i)\right\rvert = \left\lvert{\int_{x_i}^{x_{i+1}}{(f(x)-f(u_i))\ dx}}\right\rvert=\int_{x_i}^{x_{i+1}}{M \Delta x_i} =M(\Delta x_i)^2$$
This is then used as a result for estimating the error but that part is not important.
I don't quite understand how we entered under the integral with $f(u_i)(x_{i+1}-x_i)$ and then how we got $\int_{x_i}^{x_{i+1}}{M\cdot \Delta x_i}$.
For the first part if we want to put a function under the integral then we can do that by writing the derrivative of the function. I just don't see it how $f(u_i)$ is derrivative of $f(u_i)(x_{i+1}-x_i)$.
Also in the second part I assume there should be an inequality sign between the integrals $\left\lvert{\int_{x_i}^{x_{i+1}}{(f(x)-f(u_i))dx}}\right\rvert $ and $\int_{x_i}^{x_{i+1}}{M\cdot \Delta x_i}$.