Alternative representation of Taylor's Theorem of one variable In the text Vector Calculus (Jerrold E. Marsden), the author states that for a smooth function of one variable $f: \Bbb R \rightarrow \Bbb R$ that Taylor's theorem asserts that:
$f(x_0 + h) = f(x_0) + f'(x_0)\cdot h + \frac {f''(x_0)}{2} h^2 + ... + \frac {f^{(k)}(x_0)}{k!}h^k + R_k(x_0,h)$ where $R_k(x_0,h) = \int_{x_0}^{x_0+h} \frac {x_0 +h - \tau} {k!} f^{k+1}(\tau) dt$
Normally I know taylor's theorem stated as follows: Let $k ≥ 1$ be an integer and let the function f : R → R be $k$ times differentiable at the point $a \in \Bbb R$. Then there exists a function $h_k : \Bbb R \rightarrow \Bbb R$ such that
${\displaystyle f(x)=f(a)+f'(a)(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+\cdots +{\frac {f^{(k)}(a)}{k!}}(x-a)^{k}+h_{k}(x)(x-a)^{k}}$. Has the author substituted $x_0 = a$ ? I am unsure how this first formula is obtained. Any insights appreciated.
 A: If $f$ is $k+1$ times continuously differentiable on $[x_0,x_0+h]$ then
$$\frac{d}{dt}\sum_{j=1}^{k}\frac{h^j}{j!}f^{(j)}[x_0 + th](1-t)^j \\=\sum_{j=1}^{k}\frac{h^{j+1}}{j!}f^{(j+1)}[x_0 + th](1-t)^j - \sum_{j=1}^{k}\frac{h^j}{(j-1)!}f^{(j)}[x_0 + th](1-t)^{j-1}\\= \frac{h^{k+1}}{k!}f^{(k+1)}[x_0 + th](1-t)^{k}- hf'[x_0+th]. $$
Integrating over $[0,1]$,
$$-\sum_{j=1}^{k}\frac{h^j}{j!}f^{(j)}(x_0) \\= \frac{h^{k+1}}{k!}\int_0^1 f^{(k+1)}[x_0 + th](1-t)^{k} \, dt-h\int_0^1f'[x_0+th] \, dt\\=\frac{h^{k+1}}{k!}\int_0^1 f^{(k+1)}[x_0 + th](1-t)^{k} \, dt-f(x_0+h) + f(x_0).$$
Rearranging,
$$f(x_0+h) = f(x_0) + \sum_{j=1}^{k}\frac{h^j}{j!}f^{(j)}(x_0) + \frac{h^{k+1}}{k!}\int_0^1 f^{(k+1)}[x_0 + th](1-t)^{k} \, dt\\ = f(x_0) + \sum_{j=1}^{k}\frac{h^j}{j!}f^{(j)}(x_0) + \frac{1}{k!}\int_{x_0}^{x_0+h} f^{(k+1)}(\tau)(x_0 + h - \tau)^{k} \, d\tau$$
That is the derivation of the Taylor series with the integral form of the remainder. 
Another approach is to use repeated integration by parts on
$$f(x_0+h) = f(x_0) + \int_{x_0}^{x_0 + h} f'(t) \, dt.$$
Carrying this out for one step should provide some insight on how the integral remainder develops.  With $u = f'(t)$ and $dv = dt$ we have $du = f''(t) dt$ and $v = t +c$.  Notice that I am going to use the trick of retaining the integration constant $c$, to be determined later.
Thus,
$$f(x_0+h) = f(x_0) + \left. (t+c) f'(t)\right|_{x_0}^{x_0 + h} - \int_{x_0}^{x_0 + h} (t+c)f''(t) \, dt \\ = f(x_0) + (x_0+h +c) f'(x_0 + h ) - (x_0 + c)f'(x_0) - \int_{x_0}^{x_0 + h} (t+c)f''(t) \, dt. $$
Now choose $c = - x_0 - h$ to obtain
$$f(x_0+h) = f(x_0) + f'(x_0)h + \int_{x_0}^{x_0 + h} (x_0 + h - t) f''(t) \, dt$$
