On the Lagrange Interpolation Error for the Trapezoidal Quadrature Rule 
This is part of a document on numerical analysis. I have included its link below. In the derivation for the error in the linear interpolation approximation on page 84, the author "brings" the ξ(x) term of the integral outside using the mean value theorem generalized for integrals. I cannot see how this theorem is used; I am specifically referring to the highlighted conditions from a Wikipedia article in the other image.
Conditions for mean value theorem for integrals.
I haven't found anything on this thus far on the internet. I appreciate any feedback.
Thanks.

This is the weighted or extended version of the mean value theorem, $\int_a^bw(x)g(x)dx=g(c)\int_a^bw(x)dx$ if $w$ has a uniform sign. But you are right, one needs that $f$ is continuous for that.

Here we can show that the function $g$ in
$$
f(x)-P_1(x)=(x-a)(x-b)g(x)
$$
is continuous on $[a,b]$ if $f$ is continuously differentiable there and we know from the interpolation error formula that its values are equal to $g(x)=\frac12f''(\xi_x)$ for some $\xi_x\in(a,b)$. There is no need for the relation $x\mapsto\xi_x$ to be continuous as well.
Proof of the continuity of $g$: Expanding the linear interpolation one finds an alternative way to write $g(x)$ as
\begin{align}
g(x)(x-a)(x-b)&=f(x)-f(a)-(x-a)\frac{f(b)-f(a)}{b-a}
\\
&=\frac{(f(x)-f(a))(b-x)-(x-a)(f(b)-f(x))}{b-a}
\\[1em]
\implies
g(x)&=\frac{\frac{f(b)-f(x)}{b-x}-\frac{f(x)-f(a)}{x-a}}{b-a}=[a,x,b]f
\end{align}
so that for the limits of $g$ to exist in $a$ and $b$ we need that the difference quotients of $f$ converge there, that is, that $f$ is differentiable in these points.

Now inserted into the mean value theorem you get
$$
\int_a^b(x-a)(x-b)g(x)dx=g(c)\int_a^b(x-a)(x-b)dx=\frac12f''(\xi_c)\cdot (-\frac16)(b-a)^3.
$$
