Taylor formula with integral remainder

Let $m\ge 1$, 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, andRepeated 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 $a\le x\le b$, the remainder satisfiesIf the highest derivative is bounded, direct integration improves this to $|R(x)|\le \Vert f^{(m)}{\Vert }_{\infty }(x-a{)}^m/m!$. For $f(x)=x^{3/2}$ on $[0,1]$, $f^{\prime }(0)=0$ and $f^{\prime \prime }(t)=3/(4\sqrt{t})$ 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.