Is $1 / \sqrt{x}$ Riemann integrable on $[0,1]$? If $\int_0^1 1 / \sqrt{x}\,\mathrm dx$ Riemann integrable then using second fundamental theorem of calculus I can easily say that $\sqrt{x}$ is uniformly continuous.
Basically it has one point i.e. $0$ where it diverges. Otherwise I can use the theorem that a function with finite number of discontinuous points is Riemann integrable. Can I take zero as point of discontinuity?
 A: That function is unbounded (whatever way you choose it to extend it to $0$), and therefore it is not Riemann-integrable (by definition, every Riemann-integrable function is bounded). However, the improper integral $\int_0^1\frac1{\sqrt x}\,\mathrm dx$ converges.
The theorem that you mentioned holds for bounded functions, not in general.
A: $\frac{1}{\sqrt{x}}$ is not bounded on $(0,1)$, and therefore, it cannot be Riemann integrable.
However, $\sqrt{x}$ is continuous on $[0,1]$ and $[0,1]$ is a compact set, therefore, $\sqrt x$ is uniformly continuous on $[0,1]$. Also, the derivative of $\sqrt{x}$ is bounded on $[1, \infty)$, so $\sqrt{x}$ is uniformly continuous on $[1, \infty)$.
A: It's unbounded so no it's not Riemann integrable. It is improperly Riemann integrable though. 
A: The problem here is that the integrand is unbounded in the domain of integration so it is not Riemann integrable.
The improper integral does exist if it is perrceived as the limit $$\displaystyle\lim_{a\to0^+}\int_a^1\dfrac{1}{\sqrt{x}}=\lim_{a\to0^+}(2-2\sqrt{a})=2$$
