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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 $x=5$)?

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  • $\begingroup$ In Calc I, the statement was "bounded with finitely many step discontinuities". There are many bounded non-step discontinuities. $\endgroup$ Commented Nov 16, 2019 at 23:39
  • $\begingroup$ @EricTowers Would this be considered a step discontinuity? The limits from the left and right are the same, so there's no "step" $\endgroup$
    – Buddhapus
    Commented Nov 16, 2019 at 23:46
  • $\begingroup$ 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. $\endgroup$ Commented Nov 17, 2019 at 2:29

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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.

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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 $x = 0$.

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

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