Exercise · ZW-INTEGRABILITY-OF-CONTINUOUS-FUNCTIONS-006 · ★★☆ Intermediate
Let \(f\) be continuous on \([a,b]\) and \(\varepsilon>0\). If \(|x-y|<\delta\) implies \(|f(x)-f(y)|<\tfrac{\varepsilon}{b-a}\), and \(P\) has subintervals of length less than \(\delta\), what follows for \(U(f,P)-L(f,P)\)?
Choose one option.
A It is greater than \(\varepsilon\) B It equals \(\delta\) C It is less than \(\varepsilon/(b-a)\) D It is at most \(\varepsilon\)
Show answer D. It is at most \(\varepsilon\)
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Explanation On each subinterval \(M_i-m_i\le\tfrac{\varepsilon}{b-a}\), so \(U-L=\sum(M_i-m_i)\Delta x_i\le\tfrac{\varepsilon}{b-a}(b-a)=\varepsilon\).
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