2
$\begingroup$

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 $x=5$)?

$\endgroup$
  • $\begingroup$ In Calc I, the statement was "bounded with finitely many step discontinuities". There are many bounded non-step discontinuities. $\endgroup$ – Eric Towers Nov 16 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 Nov 16 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. $\endgroup$ – Eric Towers Nov 17 at 2:29
1
$\begingroup$

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 $x = 0$.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.