As far as I can tell, if you start with the integral form of the remainder of a Taylor polynomial then you can derive the Lagrange form by an application of the mean value theorem for integrals. From Spivak's Calculus:
If $f^{\left({n + 1}\right)}$ is continuous on $[a, x]$, then
$$\displaystyle R_{n,a} \left({x}\right) = \int_a^x \dfrac {f^{\left({n + 1}\right)}\left({t}\right)}{n!} \left({x - t}\right)^n \, \mathrm d t$$
Let m and M be the minimum and maximum of $\dfrac {f^{\left({n + 1}\right)}}{n!}$ on $[a, x]$, then $R_{n,a}(x)$ satisfies
$$m\int_a^x \left({x - t}\right)^n \, \mathrm d t \le R_{n,a} \left({x}\right) \le M\int_a^x \left({x - t}\right)^n \, \mathrm d t$$
so we can write
$$\displaystyle R_{n,a} \left({x}\right) = \alpha \cdot \dfrac {\left({x - a}\right)^{n+1}} {{n+1}!}$$
Now here is my question. Shouldn't this imply that there is a number $x^* \in \left[{a \,.\,.\, x}\right]$ such that
$$\displaystyle R_{n,a} \left({x}\right) = \dfrac {f^{\left({n + 1}\right)} \left({x^*}\right)} {\left({n + 1}\right)!} \left({x - a}\right)^{n + 1}$$
in accordance with the mean value theorem for integrals?
Yet all sources I reviewed, including Spivak, state that $x^* \in \left({a \,.\,.\, x}\right)$.