The Taylor remainder $R_n(x)$ equals an $n$th-derivative term at an intermediate point in Lagrange and Cauchy form, and an integral of $f^{(n)}$ when $f^{(n)}$ is continuous.Suggest a slogan
Let \(n\geq1\), \(p\in\RR\), \(\delta>0\), and let \(f\colon(p-\delta,p+\delta)\to\RR\) be \(n\) times differentiable. For \(x\in(p-\delta,p+\delta)\) put \[ R_n(x)\coloneqq f(x)-\sum_{k=0}^{n-1} \frac{f^{(k)}(p)}{k!}(x-p)^k . \] Let \(x\in(p-\delta,p+\delta)\) with \(x\neq p\).
(Lagrange form) There exists \(\xi\) strictly between \(p\) and \(x\) such that \[ R_n(x)=\frac{f^{(n)}(\xi)}{n!}(x-p)^n . \]
(Cauchy form) There exists \(\eta\) strictly between \(p\) and \(x\) such that \[ R_n(x)=\frac{f^{(n)}(\eta)}{(n-1)!}(x-\eta)^{n-1}(x-p) . \]
(Integral form) If moreover \(f^{(n)}\) is continuous on \((p-\delta,p+\delta)\), then \[ R_n(x)=\int_p^x \frac{f^{(n)}(t)}{(n-1)!}(x-t)^{n-1}\,dt . \]
Part 1 is in [@Rud76].