# Which integral limits to choose when dealing with open intervals?

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.

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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

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