We consider the truncation error in the left Riemann sum (for the right Riemann sum the derivation is analogous and the result is the same) and the midpoint Riemann sum of a 'neat' function \(f(x)\) on the interval \([a,b]\). We assume \(n\) tags that are \(\displaystyle h=\frac{b-a}{n}\) apart. We write the Riemann sum with the letter \(R\). The analysis will show that the midpoint Riemann sum is a better numerical integration method than the left point or right point Riemann sum.

Suppose \(M\) is the maximum of \(|f'|\) on \([a,b]\). Then we have for the left Riemann sum \(R\) : \[\left|\int_a^bf(x)\,\dd x - R\right|\le \tfrac{1}{2}M(b-a)h\] In other words: the truncation error is linear in the mesh size \(h\)

The tags are \(s_k=a+(k-1)h\) for \(k=1,\ldots n\) and the left Riemann sum is given by \[R=\sum_{k=1}^{n}f(a+(k-1)h)\cdot h= \bigl( f(a)+f(a+h)+\cdots f(b-h)\bigr)\cdot h\] We \(s_{n+1}=b\). We first consider the sub-interval \([s_{k},s_{k+1}]\) and note that it follows from Taylor's theorem that \[f(x)=f(s_{k})+f'(\xi_x)(x-s_{k})\] \(x\) and that \(\xi_x\) in this interval ( \(\xi\) depends on \(x\), but this does not really matter). Therefore: \[\left|\int_{s_{k}}^{s_{k+1}}\bigl(f(x)-f(s_{k})\bigr)\,\dd x\right|\le M\cdot \int_{s_{k}}^{s_{k+1}} (x-s_{k})\,\dd x=M\cdot \tfrac{1}{2}h^2\] so that \[\left|\int_{s_{k}}^{s_{k+1}}f(x)\,\dd x - f(s_{k})h\right|\le \tfrac{1}{2}M h^2\] But then: \[\begin{aligned}\left|\int_a^b f(x)\,\dd x - R\right|&=\left|\sum_{k=1}^{n}\int_{s_{k}}^{s_{k+1}}f(x)\,\dd x-\sum_{k=1}^{n}f(s_{k})h\right|\\ \\ &=\left|\sum_{k=1}^{n}\left(\int_{s_{k}}^{s_{k+1}}f(x)\,\dd x-f(s_{k})h\right)\right|\\ \\ &\le \sum_{k=1}^{n}\left|\int_{s_{k}}^{s_{k+1}}f(x)\,\dd x - f(s_{k})h\right|\\ \\ &\le \sum_{k=1}^{n}\tfrac{1}{2}M h^2 = \tfrac{1}{2}M h^2 n=\tfrac{1}{2}M(b-a)h\end{aligned}\]

Suppose \(M\) is the maximum of \(|f'|\) on \([a,b]\). Then we have for the right Riemann sum \(R\): \[\left|\int_a^bf(x)\,\dd x - R\right|\le \tfrac{1}{2}M(b-a)h\] In other words: the truncation error is linear in the mesh size \(h\)

Suppose \(M\) is the maximum of \(|f''|\) on \([a,b]\). Then we have for the midpoint Riemann sum \(R\): \[\left|\int_a^bf(x)\,\dd x - R\right|\le \tfrac{1}{24}M(b-a)h^2\] In other words: the truncation error is quadratic in the mesh size \(h\).

The scattering points are \(s_k=a+(k-\tfrac{1}{2})h\) for \(k=1,\ldots n\) and the midpoint Riemann sum is given by \[R=\sum_{k=1}^{n}f(a+(k-\tfrac{1}{2}h)\cdot h= \bigl( f(a+\tfrac{1}{2}h)+f(a+\tfrac{3}{2}h)+\cdots f(b-\tfrac{1}{2}\,h)\bigr)\cdot h\] The points of the distribution are \(x_k=a+k\cdot h\) for \(k=0,1,\ldots n\). We first consider the subinterval \([x_{k-1},x_{k}]\) for \(k=1,\ldots n\) and note that it follows from Taylor's theorem that \[f(x)=f(s_{k})+f'(s_{k})(x-s_{k})+\tfrac{1}{2}f''(\xi_x)(x-s_{k-1})^2\] for \(x\) and \(\xi_x\) in this subinterval ( \(\xi_x\) depends on \(x\), but this does not really matter). Then: \[\begin{aligned}\left|\int_{x_{k-1}}^{x_{k}}\bigl(f(x)-f(s_{k})\bigr)\,\dd x\right|&= \left|\int_{x_{k-1}}^{x_{k}} f'(s_k)(x-s_{k})\,\dd x+\int_{x_{k-1}}^{x_{k}}\tfrac{1}{2}f''(\xi_x) (x-s_{k})^2\,\dd x\right|\\ \\ &\le\tfrac{1}{2}M\cdot \tfrac{1}{3}\cdot2\cdot (\tfrac{1}{2}h)^3=\tfrac{1}{24}Mh^3\end{aligned}\] This result is obtained if you realise that the first integral is equal to \(0\) because \(s_k\) is exactly in the middle between \(x_{k-1}\) and \(x_k\) and that the second integral can be estimated. So: \[\left|\int_{x_{k-1}}^{x_{k}}f(x)\,\dd x - f(s_{k})h\right|\le \tfrac{1}{24}Mh^3\] But then: \[\begin{aligned}\left|\int_a^b f(x)\,\dd x - R\right|&=\left|\sum_{k=1}^{n}\int_{x_{k-1}}^{x_{k}}f(x)\,\dd x-\sum_{k=1}^{n}f(s_{k})h\right|\\ \\ &=\left|\sum_{k=1}^{n}\left(\int_{x_{k-1}}^{x_{k}}f(x)\,\dd x-f(s_{k})h\right)\right|\\ \\ &\le \sum_{k=1}^{n}\left|\int_{x_{k-1}}^{x_{k}}f(x)\,\dd x - f(s_{k})h\right|\\ \\ &\le \sum_{k=1}^{n}\tfrac{1}{24}\,M\cdot h^3 = \tfrac{1}{24}M\cdot h^3 n=\tfrac{1}{24}M(b-a)h^2\end{aligned}\]