Is it true that if ∫_-1^1 f(x) dx=0 then f(x)=0 for all x∈[-1,1]?

No: f(x)=x has integral 0 on [-1,1] without being zero

Yes, if f is positive on the whole interval

Exercise · ZW-DEFINITE-INTEGRAL-008 · ★★★ Advanced

Is it true that if \(\int_{-1}^{1} f(x)\,dx=0\) then \(f(x)=0\) for all \(x\in[-1,1]\)?

Choose one option.

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B. No: f(x)=x has integral 0 on [-1,1] without being zero

Explanation

With f(x)=x the negative part cancels the positive one and the integral is 0, although f is not zero, because the integral is net.

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Review the concept: Definite integral

The definite integral of \(f\) on \([a,b]\) is the limit of the Riemann sums as \(n\to\infty\). It represents the signed net area between the graph of \(f\) and the \(x\)-axis, from \(x=a\) to \(x=b\).

\[\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^{n}f(x_i^{*})\,\Delta x\]

Is it true that if \(\int_{-1}^{1} f(x)\,dx=0\) then \(f(x)=0\) for all \(x\in[-1,1]\)?

Explanation

Review the concept: Definite integral

More practice problems: definite integral