# 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 $$x=5$$)?

• In Calc I, the statement was "bounded with finitely many step discontinuities". There are many bounded non-step discontinuities. – Eric Towers Nov 16 at 23:39
• @EricTowers Would this be considered a step discontinuity? The limits from the left and right are the same, so there's no "step" – Buddhapus Nov 16 at 23:46
• A a function with a step discontinuity is defined at the step. You have a removable discontinuity: a point where the function is not defined but the limits from the left and right agree. – Eric Towers Nov 17 at 2:29

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 $$x = 0$$.
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.
$$\int_{-5}^{5} f(x)dx$$ 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.