Proof that for monotone $f$, $\int_{a}^b f(x) dx-\frac{b-a}{n}\sum_{i=1}^n {f\left(a+i \frac{b-a}{n}\right)=O\left(\frac{1}{n}\right)}.$

Question

Let $f:[a,b] \rightarrow \mathbb{R}$ be a monotone function in $[a,b]$. Prove that $$\int_{a}^b f(x) dx-\frac{b-a}{n}\sum_{i=1}^n {f\left(a+i \frac{b-a}{n}\right)=O\left(\frac{1}{n}\right)}.$$ I don't have any idea about how I should do it. I tried to understand the geometric mean in this expression, and I think that for the function to be monotone in $[a,b]$ the function has to be Riemann-integrable. Then I'd use this to apply the limit around $0$ or apply the limit to infinity to find $k,c \in \mathbb{R}$ to prove that these functions are $O\left(\frac{1}{n}\right)$.

Answer 1

Assume that $f$ is nondecreasing. Then

$$L = \frac{b-a}{n}\sum_{i=1}^n f\left(a+(i-1) \frac{b-a}{n}\right) \leqslant\int_{a}^b f(x) \,dx \leqslant \frac{b-a}{n}\sum_{i=1}^n f\left(a+i \frac{b-a}{n}\right) =U,$$

and since $U-L$ forms a telescoping sum,

$$0 \leqslant U- \int_{a}^b f(x) \,dx \leqslant U-L = (f(b)-f(a)) \frac{b-a}{n}$$

Thus, the upper sum approximates the integral with an error of order $1/n$. Use a similar approach if the function is nonincreasing.