Method for evaluating Darboux integrals by a sequence of partitions? Suppose we have been asked to evaluate the integral of a bounded function, $f$ over the closed interval $[a,b]$ 
Then, further suppose that $f$ is integrable i.e $sup \ L(f,P) = inf \ U(f,P)$ where the $sup$ and $inf$ are taken over all possible partitions $P$. 
Then, how do we actually evaluate the integral by using a sequence of partitions, and why does that method work? (Suppose it's too difficult to consider the sup or inf over all partitions)
 A: The Riemann integral defined as $I = \sup_P L(f,P) = \inf_P U(f,P)$ is equivalent to the integral defined as $I = \lim_{\|P\| \to 0} S(f,P)$ where $S(f,P)$ is a sum of the form $\sum_{j=1}^n f(c_j)(x_j - x_{j-1})$.  When $f$ is Riemann integrable, that limit is unique regardless of how the tags $c_j \in [x_{j-1},x_j]$ are chosen or how the partition points are distributed as $\|P\| = \max_{1 \leqslant j \leqslant n} (x_j - x_{j-1}) \to 0$. This is proved here.
To evaluate the limit, pick partitions and tags as convenient.  For example, we can chose $c_j = x_j = a + j(b-a)/n$, in which case the conditions $(b-a)/n = \|P\| \to 0$ and $n \to \infty$ coincide.  Whence,
$$I = \int_a^b f(x) \, dx = \lim_{n \to \infty} \frac{b-a}{n} \sum_{j=1}^n f(a + j(b-a)/n).$$
Generally, for all but the simplest functions, it may be difficult if not impossible to compute the definite integral (analytically) as a limit of Riemann sums. The fundamental theorem of calculus is often the more expedient approach if, of course, we can find a closed form for the anti-derivative. 
Some non-trivial computations with Riemann sums are are given for $xe^x$ and for a more difficult example.
