Value of the integral $\int_{a}^{a}f(x)dx?$ I am confused about the value of the integral $\int_{a}^{a}f(x)dx,$ is it $0$ or not always? In Riemann integration case (i.e. $f$ is bounded ) its zero by definition. What about other cases for example $\int_{0}^{0}\frac{1}{x}? $ Is it zero? Please help. Thanks a lot.
It's always zero...except that there are some conventions in which this notation is used to denote something that's not zero. 

So let me clarify "always": for the Riemann integral, and for the  Lesbegue integral with respect to the standard measure on the real line, and for functions $f: \mathbb R \to \mathbb R$, it's always zero. This property is very useful, for it helps ensure that 
$$
\int_a^b + \int_b^c = \int_a^c
$$
in general. 

So why do I say "except..."?

Well, for instance, if $f$ is a "delta function" (which isn't really a function, but instead a distribution), then this integral can denote something nonzero (or at least
$$
\int_{0^{-}}^{0^{+}} f(x) dx
$$
can be nonzero). 

And if your measure "dx" on the real line happens to assign positive measure to the point $0$, then again the integral can denote something nonzero. But it'd be pretty odd to use "dx" to denote such a measure.
You can say like this $$\int_0^0{1\over x}dx=\lim_{c\to0+}\int_{-c}^{c}{1\over x}dx=\lim_{c\to0+}\left[\ln|x|\right]_{-c}^c=\lim_{c\to 0+}\{\ln|c|-\ln|-c|\}=0$$