Taylor's theorem with an integral remainder

For fixed $f,c,x$, define $I_m=\frac{1}{m!}{\int }_c^x{f}^{(m+1)}(t)(x-t{)}^{m}dt$ for $0\le m\le n$. For $m=0$, [L1] gives $I_0=f(x)-f(c)$. For $m\ge 1$, apply integration by parts [L1] to $u(t)=f^{(m)}(t)$ and $v(t)=(x-t{)}^m\mathrm{/}m!$. Since $v(x)=0$, $v(c)=(x-c{)}^m\mathrm{/}m!$, and $v^{\mathrm{\prime }}(t)=-(x-t{)}^{m-1}\mathrm{/}(m-1)!$, it gives $I_m=I_{m-1}-\frac{f^{(m)}(c)(x-c{)}^m}{m!}\mathrm{.}$ Induction from $I_0$ yields $I_n=f(x)-{\sum }_{j=0}^n{f}^{(j)}(c)(x-c{)}^j\mathrm{/}j!=f(x)-P_n(x)$. The oriented integration-by-parts identity makes the same recurrence valid when $x<c$; if $x=c$, both sides are zero.

Let $d=\mathrm{\mid }x-c\mathrm{\mid }$ and $\psi (t)=f^{(n+1)}(t)(x-t{)}^n\mathrm{/}n!$. This is continuous. The triangle inequality in [L3], applied in both directions, gives $\mid \mathrm{\mid }\psi (s)\mathrm{\mid }-\mathrm{\mid }\psi (t)\mathrm{\mid }\mid \le \mathrm{\mid }\psi (s)-\psi (t)\mathrm{\mid }$, so $\mathrm{\mid }\psi \mathrm{\mid }$ is continuous too and both functions are integrable. By [L2] and [L4], the integral is monotone in the integrand, so $\mid {\int }_c^x\psi (t)dt\mid \le {\int }_{\min (c,x)}^{\max (c,x)}\mathrm{\mid }\psi (t)\mathrm{\mid }dt\le \frac{M}{n!}{\int }_{\min (c,x)}^{\max (c,x)}\mathrm{\mid }x-t{\mathrm{\mid }}^{n}dt\mathrm{.}$ The first inequality follows by integrating $-\mathrm{\mid }\psi \mathrm{\mid }\le \psi \le \mathrm{\mid }\psi \mathrm{\mid }$ on the forward interval and using $\int (-\mathrm{\mid }\psi \mathrm{\mid })=-\int \mathrm{\mid }\psi \mathrm{\mid }$; orientation does not change the absolute value. The last power integral is $d^{n+1}\mathrm{/}(n+1)$ by [L1], using its polynomial antiderivative on the segment to either side of $x$. Thus the bound is $M{d}^{n+1}\mathrm{/}(n+1)!$. Combining this with step 1.1 proves the result.