## University Calculus: Early Transcendentals (3rd Edition)

$4$
Here, we can see that $f(x)=\dfrac{1}{|\sqrt {x-1}|}$ is not defined at $1$. However, $f(x)=\dfrac{1}{|\sqrt x-1|}=\dfrac{1}{-\sqrt {x-1}}$ for all $x \in [0,1)$ and $f(x)=\dfrac{1}{|\sqrt {x-1}|}=\dfrac{1}{\sqrt {x-1}}$ for all $x \in (1,2]$ Thus, we have $\int_{0}^{1} \dfrac{1}{|\sqrt {x-1}|}+\int_{1}^{2} \dfrac{1}{|\sqrt {x-1}|}=\lim\limits_{p \to 1^{-}} \int_{0}^{p} \dfrac{1}{|\sqrt {1-x}|}+\lim\limits_{q \to 1^{+}} \int_{q}^{2} \dfrac{1}{|\sqrt {x-1}|}$ This implies that $\lim\limits_{p \to 1^{-}} (-2 \sqrt{1-p}+2)+\lim\limits_{q \to 1^{+}} (2-2\sqrt {q-1})=2+2=4$