Assume $f$ has continuous derivative $f'$ on [a,b]. Prove the following summation formula, without using partial integration: \begin{equation} \sum_{a< x \le b}f(n)=\int_{a}^{b}f(x)dx+\int_{a}^{b}f'(x)\{x\}dx+f(a)\{a\}-f(b)\{b\} \end{equation} where $\{x\}:=x-[x]$ represents the fraction function.
I have already managed to prove this result using integration by parts, but I am told this is unnecessary.
You can observe that, if $n\in\mathbb{Z}$ and $t\in\left(n-1,n\right)$ you have $$\left\{ t\right\} =t-\left[t\right]=t-n+1$$ where $\left\{ t\right\}$ is the fractional part of $t$ and $\left[t\right]$ is the entire part of $t$. So you have $$\frac{d}{dt}\left(\left\{ t\right\} f\left(t\right)\right)=\left\{ t\right\} f'\left(t\right)+f\left(t\right)$$ so $$\int_{n-1}^{n}\left(\left\{ t\right\} f'\left(t\right)+f\left(t\right)\right)dt=f\left(n\right)$$ hence $$\underset{a<n\leq b}{\sum}f\left(n\right)=\int_{a}^{b}f\left(t\right)dt+\int_{a}^{b}\left\{ t\right\} f'\left(t\right)dt-\left\{ a\right\} f\left(a\right)+\left\{ b\right\} f\left(b\right)$$ because you can use the same argument on the interval $\left(a,\left[a\right]+1\right)$ and $\left(b,\left[b\right]\right)$.