Exercise · ZW-INTEGRABILITY-OF-CONTINUOUS-FUNCTIONS-009 · ★★★ Advanced
Let \(f(x)=\sin(1/x)\) for \(x>0\) and \(f(0)=0\). Which statement is correct on \([0,1]\)?
Choose one option.
A Not integrable because it is discontinuous at \(0\) B Integrable, although the theorem for continuous functions does not apply directly C Integrable because it is continuous D Not integrable because it oscillates infinitely Show answer B. Integrable, although the theorem for continuous functions does not apply directly← Back to the question
Explanation It is bounded and discontinuous only at \(0\): enclose that point in a short interval and it is continuous elsewhere, so \(U-L\) can be made as small as desired.
Result stating that every function \(f\) continuous on a closed , bounded interval \([a,b]\) is Riemann integrable there.
\[f\text{ continuous on }[a,b]\ \Rightarrow\ \int_a^b f(x)\,dx\text{ exists}\]
More practice problems: integrability of continuous functions