Let $ f:[0,1] \rightarrow \mathbb{R}$ of class $C^{3}$. Show that: $$\frac{1}{n}\sum ^{n-1}_{k=0} f\Bigl(\frac{k}{n}\Bigr) =\int ^{1}_{0} f(t)\, dt-\frac{1}{2n}\int ^{1}_{0} f'( t)\, dt+\ \frac{1}{12n^{2}}\int ^{1}_{0} f''( t)\,dt+O\Bigl(\frac{1}{n^{3}}\Bigr)$$
I thought of using Taylor's theorem and introducing an integral and a sum on $$ R_{3}( x)=f( x)-f\Bigl(\frac{k}{n}\Bigr)-\Bigl(x-\frac{k}{n}\Bigr)f' \Bigl(\frac{k}{n}\Bigr)−\frac{\Bigl(x−\frac{k}{n}\Bigr)^{2}}{2} f'' \Bigl(\frac{k}{n}\Bigr)-\frac{\Bigl( x−\frac{k}{n}\Bigr)^{3}}{6} f''' \Bigl(\frac{k}{n}\Bigr)$$ Which yields \begin{multline}\sum ^{n-1}_{k=0}\int ^{\frac{k+1}{n}}_{\frac{k}{n}} R_{3}( x)\,dx=\int ^{1}_{0} f( x)\,dx−\frac{1}{n}\sum ^{n-1}_{k=0} f\Bigl(\frac{k}{n}\Bigr) − \frac{1}{2n^{\!2}}\sum ^{n-1}_{k=0}f'\Bigl(\frac{k}{n}\Bigr)\\ −\frac{1}{6n^{3}}\sum ^{n-1}_{k=0} f''\Bigl(\frac{k}{n}\Bigr) -\frac{1}{24n^{3}}\sum ^{n-1}_{k=0} f''' \Bigl(\frac{k}{n}\Bigr)\end{multline} But I'm stuck here.
Taylor expansion in $\left[\frac{k}{n},\frac{k+1}{n}\right]$: $$ f(t)= f\left(\frac{k}{n}\right)+ f'\left(\frac{k}{n}\right)\left(t-\frac{k}{n}\right)+ \frac{1}{2}f''\left(\frac{k}{n}\right)\left(t-\frac{k}{n}\right)^2+ O\left(\frac{1}{n^2}\right) $$ and integrating over the same interval $$ \int_{k/n}^{(k+1)/n}f(t)dt= \frac{1}{n}f\left(\frac{k}{n}\right)+ \frac{1}{2n^2}f'\left(\frac{k}{n}\right)+ \frac{1}{6n^3}\frac{1}{2}f''\left(\frac{k}{n}\right)+ O\left(\frac{1}{n^3}\right)\tag1 $$ the same holds for $f'$ and $f'',$ so: $$ \int_{k/n}^{(k+1)/n}f'(t)dt= \frac{1}{n}f'\left(\frac{k}{n}\right)+ \frac{1}{2n^2}f''\left(\frac{k}{n}\right)+ O\left(\frac{1}{n^2}\right)\tag2 $$ $$ \int_{k/n}^{(k+1)/n}f''(t)dt= \frac{1}{n}f''\left(\frac{k}{n}\right)+ O\left(\frac{1}{n}\right)\tag3 $$ If we combine the last three formulae in this way $$ (1)-\frac{1}{2n}(2)+\frac{1}{12n^2}(3) $$ and sum for $k=0,\ldots,n-1,$ we should arrive to the result.