Integrate a piece-wise continuous function not defined at endpoint What if I had a piece-wise continuous function that was defined in the following way:
$$f(x) =
\begin{cases} 
      2x-3 & x < 0 \\
      2x-3 & x > 0.
\end{cases}$$
In my Calc I class we say that a function $f$ is integrable if it is continuous, or bounded with finitely many discontinuities (a piece-wise function). Does this apply to the function above (is the function above integrable, given that there's a hole at $x=0$, if I need to integrate from like $x=-5$ to $x=5$)?
 A: No matter what value you prescribe to $f(0)$, the function you have defined is still bounded on each closed interval and discontinuous only at the point 0, hence integrable by your definition. In fact more is true: The integral of your function is equal to the integral of $2x-3$ on any closed interval.
A: $\int_{-5}^{5} f(x)dx$ is to be taken as $\int_{-5}^{5} g(x)dx$ where $g(x)=f(x)$ for $x \neq 0$ and $g(0)$ is any arbitrary value, say $g(0)=0$. The value of $g(0)$ has no effect on the integral. 
A: In Calc I, you defined integrable to mean that the limit as the norm of a partition goes to zero of a Riemann sum of your function exists.  For this function, any choice of sample points that samples $x = 0$ is undefined, making any Riemann sum that samples $x = 0$ be undefined.  This behaviour persists as the partition is refined, so the resulting limit does not avoid the undefined-ness.  The limit does not exist.
You may have also studied improper integrals (typically a Calc II topic).  If so, you know to write your example integral as
$$  \lim_{\ell \rightarrow  0^-} \int_{-5}^{\ell} f(x) \,\mathrm{d}x + \lim_{r \rightarrow 0^+} \int_{r}^{5} f(x) \,\mathrm{d}x  \text{.}  $$
Both of these integrals exist for each choice of $\ell$ and $r$ and both of these limits exist, so you end up with a value for your integral -- it just skips over the one point at $x = 0$.
