Is it possible for an integral between $a$ and $a$ to have a value other than 0 Does there exist a function $f: \mathbb{R} \to \mathbb{R}$ with a constant $a \in \mathbb{R}$ such that
$$\int_{a}^{a} f(x) \, dx  \neq 0 \quad$$
holds?
 A: As others have explained, in general the answer is 'no'.
However in physics it is sometimes nice to have something 'function-like' that is zero everywhere but in one point (a null set) and still has a integral of 1. This is not a function but a more generalized object - a distribution, the Dirac Distribution.
A: Assuming that the integral is defined, the answer in NO.
$$\int_{a}^{a} f(x) dx =\int_{\{a\}} f(x) dx$$ and since the integral on the right is taken over the set of measure $0$, the integral is $0$.
However, notice that if instead of Lebesgue measure we put a counting measure on $\mathbb{R}$ (call it $\mu$), then
$$\int_{a}^{a} f(x) d \mu =\int_{\{a\}} f(x) d \mu=f(a)$$ Thus, any measurable function that is non-zero on $a$ will give you a non-zero integral.
A: No, because $\{a\}$ is a null set.
To be more precise, for any  function $f \colon \mathbb{R} \to \mathbb{R}$ you have that $$ f \cdot 1_{\{a\}}  = f(a) \cdot 1_{\{a\}}  $$ is Lebesgue measurable. Thus the integral is always defined and zero because of
$$ \int_\mathbb{R}  f(x) \cdot 1_{\{a\}}(x) \, dx = \int_{\{a\}} f(a) \, dx = f(a) \cdot \lambda(\{a\}) = 0.$$
Remark 1: 
For any null set $A$, and any Lebesgue measurable function $f$ holds
$$ \int_A f(x) \, dx = 0 $$
and every Riemann integrable function is Lebesgue measurable.
Remark 2: 
For any null set $A$ and any function $f \colon \mathbb{R} \to \mathbb{R}$ you have that $1_A \cdot f$ is Lebesgue measurable and thus $$\int_A f(x) \, dx = 0.$$ 
A: No... you are considering the area of a rectangle of width 0, so it has to be 0
A: No.
Think about the geometric definition of the integral and see what happens if the interval of integration is just one point.
A: Yes, it can be undefined:
$$\int_0^0 \frac 1x dx = undefined \ne 0$$
I'm sure it's not thing you were thinking of, but it goes to show that it's not always $0$. Sometimes, it just doesn't exist.
A: If the left hand side expression has sense at all, then its value must be zero because of the definition of integration.
If you think about area under the graph, then the area is zero because there is no area inside a line (the line going from the point $a$ in the $x$ axis to the corresponding point in the graph, which might be only a point if $f(a)=0$). In this case there is no point in thinking of $f(a)=\infty$ because this wouldn't make sense with Riemann integration. It can't be considered and improper integral either, because there is no subinterval where the integral can be thought of as a proper integral.
If we go further and consider abstract integration, then we have to use Lebesgue integration and according to the Lebesgue measure of the set over which the integral is taken, $\mu (\{a\})=0$, the integral is zero.
