Consider the following exercise:
If $f$ is uniformly continuous in $[a,b]$ then show that $$\int_{a}^{b}f(x)dx=\lim_{n\to\infty}\frac{b-a}{n}\sum_{i=0}^{n-1}f\left(a+\frac{i(b-a)}{n}\right).$$
The definition of Riemann integrable I have is:
A function $f$ is Riemann integrable in $[a,b]$ if and only if there is a number $L\in\mathbb{R}$ such that for every $\varepsilon>0$ there is a partition $P_{\varepsilon}$ such that the inequality $$\left|S(,f,P;\xi_0,...,\xi_{n-1})-L\right|<\varepsilon$$ holds for every partition $P=\{a_0,...,a_n\}$ finer than $P_{\varepsilon}$ and every election of points $\xi_i\in\{a_i,...,a_{i+1}\}.$ The number $L$ is defined as $$L=\int_{a}^{b}f(x)dx.$$ And $S(f,P;\xi_0,...,\xi_{n-1})$ denotes the Riemann sum of $f$ with respect to a partition $P$.
Is it necessary to use that definition of integral to prove the exercise? If so, how can I do it? Any help is welcome.
Further hint:
With $x_i = a+\frac{i(b-a)}{n}$, we have $x_{i+1}- x_i = \frac{b-a}{n}$ and
$$\left|\int_a^b f(x) \, dx - \frac{b-a}{n}\sum_{i=0}^{n-1}f\left(a+\frac{i(b-a)}{n}\right)\right|= \left|\sum_{i=0}^{n-1}\int_{x_i}^{x_{i+1}} f(x) \, dx - \frac{b-a}{n}\sum_{i=0}^{n-1}f\left(x_i\right)\right|\\= \left|\sum_{i=0}^{n-1}\int_{x_i}^{x_{i+1}} [f(x)-f(x_i)] \, dx \right| \leqslant \ldots $$