= Taylor formula with integral remainder
Let $m\geq1$, let $f\in C^{m-1}[a,b]$, and suppose $f^{(m-1)}$ is an <absolutely continuous function>. Its almost-everywhere <weak derivative> $f^{(m)}$ belongs to <L1 space>, and
$$
f(x)=\sum_{j=0}^{m-1}\frac{f^{(j)}(a)}{j!}(x-a)^j+\frac1{(m-1)!}\int_a^x(x-t)^{m-1}f^{(m)}(t)\,dt.
$$
Repeated use of the <fundamental theorem of calculus for Lebesgue integration> and interchange of the finite triangular integration region proves the formula. The highest <derivative> need not be continuous or bounded.
For $a\leq x\leq b$, the remainder satisfies
$$
|R(x)|\leq\frac{(x-a)^{m-1}}{(m-1)!}\|f^{(m)}\|_{L^1(a,x)}.
$$
If the highest <derivative> is bounded, direct integration improves this to $|R(x)|\leq\|f^{(m)}\|_\infty(x-a)^m/m!$. For $f(x)=x^{3/2}$ on $[0,1]$, $f'(0)=0$ and $f''(t)=3/(4\sqrt t)$ is integrable but unbounded. The formula with $m=2$ still gives $f(x)=\int_0^x(x-t)3/(4\sqrt t)\,dt=x^{3/2}$. This is why the integral remainder is useful beyond a bounded-derivative error estimate.
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