I am reading "Calculus 4th Edition" by Michael Spivak.
He derived the following proposition on p.423:
If $f^{n+1}$ is continuous on $[a, x]$, then the following equality holds: $$f(x) = f(a)+f'(a)(x-a)+\frac{f^{''}(a)}{2}(x-a)^2+\cdots+\frac{f^{n}(a)}{n!}(x-a)^n+\int_{a}^{x}\frac{f^{n+1}(t)}{n!}(x-t)^ndt.$$
Next he wrote like the following:
If $m$ and $M$ are the minimum and maximum of $\frac{f^{n+1}(x)}{n!}$ on $[a, x]$, then $$m \frac{(x-a)^{n+1}}{n+1} = m\int_{a}^{x}(x-t)^ndt \leq \int_{a}^{x}\frac{f^{n+1}(t)}{n!}(x-t)^ndt \leq M\int_{a}^{x}(x-t)^ndt = M \frac{(x-a)^{n+1}}{n+1}.$$
So we can write $$\int_{a}^{x}\frac{f^{n+1}(t)}{n!}(x-t)^ndt = \alpha\frac{(x-a)^{n+1}}{n+1}$$ for some number $\alpha$ between $m$ and $M$.
Next he wrote like the following:
Since we've assumed that $f^{(n+1)}$ is continuous, this means that for some $t$ in $(a,x)$ we can also write $$\int_{a}^{x}\frac{f^{n+1}(t)}{n!}(x-t)^ndt = \frac{f^{(n+1)}(t)}{n!} \frac{(x-a)^{n+1}}{n+1}=\frac{f^{(n+1)}(t)}{(n+1)!}(x-a)^{n+1}.$$
I think it is obvious that for some $t \in [a, x]$, $$\int_{a}^{x}\frac{f^{n+1}(t)}{n!}(x-t)^ndt = \frac{f^{(n+1)}(t)}{(n+1)!}(x-a)^{n+1}.$$
But I don't think it is obvious that for some $t \in (a, x)$, $$\int_{a}^{x}\frac{f^{n+1}(t)}{n!}(x-t)^ndt = \frac{f^{(n+1)}(t)}{(n+1)!}(x-a)^{n+1}.$$
Why $t \in (a, x)$?