Another error of Taylor polynomial

Let $f$ be a function, $b<d$, $n\in\mathbb{N}$, $f^{(n)}$ is continuous on $[b,d]$, $f^{(n+1)}$ exists on $(b,d)$. The remainder of its $n$-th degree Taylor polynomial at $a\in [b,d]$ is $$R_n(x):=f(x)-f(a)-(x-a)f'(a)-\frac{(x-a)^2}{2}f''(a)-\dots-\frac{(x-a)^n}{n!}f^{(n)}(a).$$ We prove that for all $a,x\in(b,d)$ $$R_n(x)=\int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^ndt.$$

This formula is useful in proving that a function has Taylor representation at $x=a$.

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Proof:

For $n=0$, this reduces to the Fundamental Theorem of Calculus. For $n\in\mathbb{N}$, notice $$R'_n(x)=f'(x)-f'(a)-(x-a)f''(a)-\dots-\frac{(x-a)^{n-1}}{(n-1)!}f^{(n)}(a).$$ Applying inductive hypothesis on $f'$, we get $$R'_n(x)=\int_{a}^{x}\frac{f^{(n+1)}(t)}{(n-1)!}(x-t)^{n-1}dt.$$ With $R_n(a)=0$ $$R_n(x)=\int_{a}^xR'_n(y)dy=\int_{a}^x\int_{a}^y\frac{f^{(n+1)}(t)}{(n-1)!}(y-t)^{n-1}dtdy=\int_{a}^x\int_{t}^x\frac{f^{(n+1)}(t)}{(n-1)!}(y-t)^{n-1}dydt.$$ This evaluates to $$R_n(x)=\int_{a}^x\frac{f^{(n+1)}(t)}{n!}(y-t)^{n}\bigg|_{y=t}^{y=x}dt=\int_{a}^x\frac{f^{(n+1)}(t)}{n!}(x-t)^{n}dt.$$