Integration property of the Taylor operator In Tom Apostol's Calculus Vol 1, in Theorem $7.2$ it is stated that the Taylor operator $T_n$ has the following integration property:
$$
T_{n+1}g(x) = \int_a^x T_nf(t)dt
$$
where
$$
g(x) = \int_a^x f(t) dt
$$
It is assumed that the function $f$ is differentiable $n$ times at a point $a$.
My question is: is that lower bound of the integral $a$ the same $a$ at which the Taylor operator is taken (i.e. at which the function $f$ is approximated), or it can also be a different constant $b$, and it is just important that the function $f$ is integrable over $[a, b]$?
If that lower integral bound is a different constant from the approximation point, then how to prove that property? If not, can you provide a counter example when it does not hold?
If the lower bound constant is the same constant at which the function $f$ is approximated, we can prove the theorem because $g'(a) = f(a)$. Because of uniqueness of Taylor polynomials and because the two functions on both sides of the equality agree at all derivatives at the point $a$, we have the equality mentioned in the theorem.

If we assume that the two constants are different, we get to a contradiction if the theorem is correct, so the two constants have to be the same.
$$
G(x) = T_{n+1}g(x) = \sum_{k=0}^{n+1} \frac{g^{(k)}(a)}{k!} (x-a)^k \\
F(x) = \int_b^x T_nf(t)dt = \int_b^x \sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(t-a)^k dt = \sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!} \int_b^x (t-a)^k dt = \sum_{k=0}^{n}\frac{f^{(k)}(a)}{(k+1)!} [(x-a)^{k+1} - (b-a)^{k+1}]
$$
Then $\forall f.G(a) = 0$ and $\exists f. F(a) \ne 0$, so there would be a contradiction if the two constants would be different.
So we can prove the theorem to hold for the same constant $a$ in both the integral's lower bound and at the approximation point in the Taylor's polynomial.
