Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let's say I need to find out the length of an arc between the open interval $(a, b)$ with $a,b\in\mathbb{R}$.

  1. How would I set the limits for the integral?
  2. Am I still allowed to use $a$ and $b$?
  3. Adding (subtracting) some small $\epsilon$ to $a$ (from $b$) seems like a dirty way to deal with this situation.

Thanks in advance for your help!

share|cite|improve this question
A single point has $0$ length. Similarly, when you are calculating areas, a single line segment has $0$ area. – André Nicolas Aug 8 '11 at 10:44
@AndréNicolas: Thanks, that was exactly what I thought, but I needed to be sure. :) – HomerS1995 Aug 8 '11 at 11:14
"Between" is the wrong word. Saying "between $a$ and $b$" makes sense; saying "on the interval $(a,b)$" makes sense. But if you say "between the open interval $(a,b)$...." one expects "and" to follow that. Between that interval and something else. – Michael Hardy Aug 8 '11 at 18:08
up vote 3 down vote accepted

Yes, you can include the endpoints $a,b$ in the integral. It doesn't affect the value of the arc length to add or drop the endpoints, so the definite integral will give the right answer.

share|cite|improve this answer

It is perhaps relevant to consider here the notion of Improper integral. For the case of an open interval $(a,b)$, with $a,b \in \mathbb{R}$, consider $$ \int_a^b {f(x)\,dx} = \mathop {\lim }\limits_{\scriptstyle c \to a^ +\atop \scriptstyle d \to b^ -} \int_c^d {f(x)\,dx} . $$ Example: Suppose that $f$ is defined on $(0,1)$ by $$ f(x) = \frac{1}{{\sqrt x \sqrt {1 - x} }}. $$ Note that $f$ is unbounded near $0^+$ and near $1^-$. Nevertheless, the integral $\int_0^1 {f(x)\,dx}$ exists as an improper integral. To evaluate it, first note that the antiderivative of $f$ is given by $$ \int {f(x)\,dx} = - 2\arctan \bigg(\sqrt {\frac{1-x}{x}} \bigg) + C. $$ Hence, by the fundamental theorem of calculus, $$ \int_c^d {f(x)\,dx} = - 2\arctan \bigg(\sqrt {\frac{1-x}{x}} \bigg) \bigg|_c^d , $$ for any $c$ and $d$ such that $0 < c < d < 1$. Now, $$ \mathop {\lim }\limits_{d \to 1^ - } \bigg[ - 2\arctan \bigg(\sqrt {\frac{1-x}{x}} \bigg)\bigg] = -2 \arctan (\sqrt{0}) = 0 $$ (using that $\arctan$ is continuous at $0$) and $$ \mathop {\lim }\limits_{c \to 0^ + } \bigg[ - 2\arctan \bigg(\sqrt {\frac{1-x}{x}} \bigg)\bigg] = - 2\mathop {\lim }\limits_{t \to \infty } \arctan (t) = - 2\frac{\pi }{2} = - \pi $$ (using that $t: = \sqrt {\frac{{1 - x}}{x}} \to \infty $ as $x \to 0^+$). Thus $$ \int_0^1 {f(x)\,dx} = \mathop {\lim }\limits_{\scriptstyle c \to 0^ + \atop \scriptstyle d \to 1^ -} \int_c^d {f(x)\,dx} = 0 - ( - \pi ) = \pi . $$

share|cite|improve this answer
The "\hfill"'s above should be ignored. – Shai Covo Aug 8 '11 at 11:45
Why didn't you just take them out? :-) – joriki Aug 8 '11 at 15:00
@joriki: Thanks for the edit. – Shai Covo Aug 10 '11 at 5:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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