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 . $