An answerer gave a derivation (Where do the factorials come from in the taylor series?) for the standard form of the taylor polynomial series, copied and pasted below for ease of viewing.
I wanted to learn why the use of integrals is justified and, more importantly, what it accomplishes in the derivation. On what grounds could the answerer use integrals?
Start with the fundamental theorem of calculus: $$ f(x) = f(x_0) + \int_{x_0}^x f^\prime(y) \mathrm{d} y $$ and reapply it to $f(y)$: $$ f(x) = f(x_0) + \int_{x_0}^x \left( f^\prime(x_0) + \int_{x_0}^y f^{\prime\prime}(z) \mathrm{d} z \right) \mathrm{d} y = f(x_0) +f^\prime(x_0) \int_{x_0}^x \mathrm{d} y + \underbrace{\int_{x_0}^x \left( \int_{x_0}^y f^{\prime\prime}(z) \mathrm{d} z\right)\mathrm{d} y}_{\mathcal{R}_2(x)} $$ Repeated this with $f^{\prime\prime}(z)$: $$ f(x) = f(x_0) + f^\prime(x_0) \underbrace{\int_{x_0}^x \mathrm{d} y}_{I_1(x)} + f^{\prime\prime}(x_0) \underbrace{\int_{x_0}^x \int_{x_0}^y \mathrm{d}z \mathrm{d} y}_{I_2(x)} + \underbrace{\int_{x_0}^x \int_{x_0}^y \int_{x_0}^z f(w) \mathrm{d} w \mathrm{d} z \mathrm{d} y}_{\mathcal{R}_3(x)} $$ and keeping going we get: $$ f(x) = f(x_0) + f^\prime(x_0) \int_{x_0}^x \mathrm{d} y + \cdots + f^{(k)}(x_0) \underbrace{\int_{x_0}^{x} \int_{x_0}^{y_1} \int_{x_0}^{y_2} \cdots \int_{x_0}^{y_{k-2}} \mathrm{d} y_{k-1} \cdots\mathrm{d} y_3 \mathrm{d} y_2 \mathrm{d} y_1}_{I_k(x)} + \mathcal{R}_{k+1}(x) $$ The iterated integrals $I_k(x)$ are easy to evaluate. They can be defined recursively $$ I_0(x) = 1, \quad I_k(x) = \int_{x_0}^x I_{k-1}(y) \mathrm{d} y $$ Giving $I_k(x) = \frac{1}{k!} (x-x_0)^k$.
