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 $x=5$)?
Best Answer
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.