Let $f:[a,b] \rightarrow \mathbb{R}$ be a continuous function. For each $n \in \mathbb{N}$, let $x_1^n$, ...$x_n^n$ be points in $\mathbb{R}$ such that for each $j \leq n$, we have
$$a + {(j-1)(b-a)\over n} \leq x_j^n \lt a + {j(b-a)\over n}$$.
Prove that $$\int_a^b f(x)dx = \lim_{n\rightarrow \infty} \sum_{j=1}^{n}f(x_j^n)\frac {b-a}n$$
The idea.
Since $f$ is continuous on $[a, b]$, we know already that $f$ must be integrable.
Thus, we may find a sequence of tagged partitions whose length (norm/mesh) goes to $0$ and find the value the Riemann sums tend to. (See here for reference.)
The partitions.
Define the $n^{\text{th}}$ partition to be the finite sequence $$a < a + \dfrac{b-a}{n} < a + \dfrac{2(b-a)}{n} < \cdots < a + \dfrac{(n-1)(b-a)}{n} < b.$$
Note that the mesh of this partition is exactly $\dfrac{1}{n}$, which tends to $0$.
Thus, we may choose any tags we like and calculate the Riemann sums. The limit of the Riemann sums must then converge to the integral. (Note that we had this liberty of choosing the tags since we knew, a priori, that the integral exists.)
The tags.
We will choose the tags to be $x_1^n, \ldots, x_n^n$. It is then clear that each tag does lie in the appropriate sub-interval. (By the inequality given to you.)
The Riemann sums.
Thus, we may calculate the $n^{\text{th}}$ Riemann sum as $$\sum_{j=1}^nf(x_j^n)\left(\dfrac{b-a}{n}\right).$$ (Note that $\left(\dfrac{b-a}{n}\right)$ is the size of each subinterval.)
Thus, taking the limit of the above sequence of sums gives us the integral, which is precisely what you required.