Answer
By the comparison test, the integral $\int_{0}^{1} \dfrac{|\sin x|}{\sqrt x} \ dx$ converges.
Work Step by Step
We are given the function
$f(x)=\int_{0}^{1} \dfrac{|\sin x|}{\sqrt x} \ dx$
Since, $|\sin x |\leq 1$
This yields:
$\dfrac{|\sin x|}{\sqrt x} \leq \dfrac{1}{\sqrt x} $
Consider the integral
$\int_{0}^{1} \dfrac{1}{\sqrt x} dx=\int_0^1 \dfrac{dx}{x^{1/2}} \\=[2x^{1/2}]_0^1 \\=2$
Thus, the integral $\int_{0}^{1} \dfrac{1}{\sqrt x} dx$ converges to $2$. Therefore, by the comparison test, the integral $\int_{0}^{1} \dfrac{|\sin x|}{\sqrt x} \ dx$ converges as well.
