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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!

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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
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2 Answers

up vote 2 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.

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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 $$ \int_0^1 {f(x)\,dx} = \mathop {\lim }\limits_{\scriptstyle c \to 0^ + \atop \scriptstyle d \to 1^ -} \int_c^d {f(x)\,dx} = 0 - ( - \pi ) = \pi . $$

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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
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