Let f(x)=0 if 0≤ x<tfrac12 and f(x)=1 if tfrac12≤ x≤1. Is it integrable on [0,1], and what is its integral?

Not integrable because it is discontinuous

Exercise · ZW-INTEGRABILITY-OF-CONTINUOUS-FUNCTIONS-008 · ★★☆ Intermediate

Let \(f(x)=0\) if \(0\le x<\tfrac12\) and \(f(x)=1\) if \(\tfrac12\le x\le1\). Is it integrable on \([0,1]\), and what is its integral?

Choose one option.

Show answer

C. Integrable, equal to \(\tfrac12\)

Explanation

It has one jump and is bounded, so it is integrable even though it is not continuous. The area is that of a rectangle with base \(\tfrac12\) and height \(1\): \(\tfrac12\).

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Review the concept: Integrability of continuous functions

Result stating that every function \(f\) continuous on a closed, bounded interval \([a,b]\) is Riemann integrable there.

\[f\text{ continuous on }[a,b]\ \Rightarrow\ \int_a^b f(x)\,dx\text{ exists}\]

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