Defined integral as the limit of a Riemann sum is not always zero I am first learning calculus and my mathematics notation lingo is not the best. I read that the defined integral could be written as the limit of a Riemann sum. Thus:
$$
\int_{a}^{b} f(x) \,\mathrm dx = \lim_{n \rightarrow \infty}\sum_{i=1}^\infty f(x_{i})\left(\frac{b-a}{n}\right)
$$
My question is, as $n\rightarrow\infty$, why doesn't $\dfrac{b-a}{n}$ equal zero, making the equation always equal to zero?
 A: That's not exactly what the definition is. The limit should be
$$\lim_{n\to\infty} \sum_{i=1}^n f(x_i)\left(\frac{b-a}{n}\right)$$
 i.e., you are summing over $n$ elements. Sure, each of the summands becomes smaller and smaller, but the number of the summands increases at the same rate, so the sum will not always be zero.
In fact, take a look at what happens when $f(x)=1$ for all $x$. In that case, 
$$\begin{align}\lim_{n\to\infty} \sum_{i=1}^n f(x_i)\left(\frac{b-a}{n}\right)&=\lim_{n\to\infty}\sum_{i=1}^n 1\cdot\left(\frac{b-a}{n}\right)\\&=\lim_{n\to\infty}\left((b-a)\cdot\sum_{i=1}^n\frac{1}{n}\right)\\&=\lim_{n\to\infty}((b-a)\cdot 1) = \lim_{n\to\infty}(b-a)=b-a\neq 0\end{align}$$

Also, a further word of warning not entirely on topic, but related to your question:
If the Riemann integral of a function over an interval $[a,b]$ exists, then the integral is equal to $$\lim_{n\to\infty} \sum_{i=1}^n f(x_i)\left(\frac{b-a}{n}\right).$$
However, the existance of the limit above does not, in itself, guarantee that a function is Riemann integrable. There exist functions for which
$$\lim_{n\to\infty} \sum_{i=1}^n f(x_i)\left(\frac{b-a}{n}\right)$$
exists, but $\int_a^b f(x)dx$ does not exist.
