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?
Best Answer
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.