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: $$\begin{aligned}\left|\int_a^b f(x)\,\dd x - R\right|&=\left|\sum_{k=1}^{n}\int_{s_{k}}^{s_{k+1}}f(x)\,\dd x-\sum_{k=1}^{n}f(s_{k})h\right|\\ \\ &=\left|\sum_{k=1}^{n}\left(\int_{s_{k}}^{s_{k+1}}f(x)\,\dd x-f(s_{k})h\right)\right|\\ \\ &\le \sum_{k=1}^{n}\left|\int_{s_{k}}^{s_{k+1}}f(x)\,\dd x - f(s_{k})h\right|\\ \\ &\le \sum_{k=1}^{n}\tfrac{1}{2}M h^2 = \tfrac{1}{2}M h^2 n=\tfrac{1}{2}M(b-a)h\end{aligned}$$

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: $$\begin{aligned}\left|\int_a^b f(x)\,\dd x - R\right|&=\left|\sum_{k=1}^{n}\int_{x_{k-1}}^{x_{k}}f(x)\,\dd x-\sum_{k=1}^{n}f(s_{k})h\right|\\ \\ &=\left|\sum_{k=1}^{n}\left(\int_{x_{k-1}}^{x_{k}}f(x)\,\dd x-f(s_{k})h\right)\right|\\ \\ &\le \sum_{k=1}^{n}\left|\int_{x_{k-1}}^{x_{k}}f(x)\,\dd x - f(s_{k})h\right|\\ \\ &\le \sum_{k=1}^{n}\tfrac{1}{24}\,M\cdot h^3 = \tfrac{1}{24}M\cdot h^3 n=\tfrac{1}{24}M(b-a)h^2\end{aligned}$$