We consider a function $f : [a,b] \rightarrow \mathbb{R}$ of class $\mathcal{C}^\infty$ and the trapezoidal method $$T_n(f) = h \sum_{i=0}^{n-1} \frac{f(a_i) + f(a_{i+1})}{2},$$ where $h = \frac{b-a}{n}$ and $\forall i \in \llbracket 0, n-1 \rrbracket, a_i = a + ih$.
Using the result of question 38, show that, for all integer $m \geqslant 1$, $$\int_a^b f(x)\,\mathrm{d}x = T_n(f) - \sum_{p=1}^m \frac{\gamma_{2p}}{n^{2p}} + \rho_{2m}(n)$$ where the coefficients $\gamma_{2p}$ are given by $$\gamma_{2p} = \frac{(b-a)^{2p} b_{2p}}{(2p)!}\left(f^{(2p-1)}(b) - f^{(2p-1)}(a)\right)$$ and $\rho_{2m}(n)$ is a remainder integral satisfying the bound $$|\rho_{2m}(n)| \leqslant \frac{C_{2m}}{n^{2m}}$$ where $C_{2m}$ is a constant to be determined depending only on $m$, $a$ and $b$.