Question: Let $f:[a.b]\to \Bbb R \in R[a,b]$ s.t. $f(x)>0 \ \forall x \in \Bbb R.$ Is $\int _a^b f(x)\,dx>0$ ?
What We thought: We know how to prove it for weak inequality, for strong inequality - no clue :-)
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@ user: Hi what is the notation R[a,b] standing for ? Regards – TheBridge Nov 05 '13 at 15:42
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R[a,b] means Riemann integrable in [a,b] – jreing Nov 05 '13 at 15:45
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Do you know measure theory? – Clayton Nov 05 '13 at 15:51
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only calculus I, now starting calc 2. – jreing Nov 05 '13 at 15:52
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@user1685224: In your definition of Riemann integrable, do you require the function $f(x)$ to be continuous? – Clayton Nov 05 '13 at 15:53
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Smells false. Try $a=b$. – user7530 Nov 05 '13 at 15:53
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no we don't require it to be continuous. And we've been told that the statement is true. But we cannot figure it out. And also $a\ne b$ of course. – jreing Nov 05 '13 at 15:55
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$f$ is Riemann integrable iff $f$ is continuous almost everywhere. – GA316 Nov 05 '13 at 16:05
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Review https://math.stackexchange.com/questions/351157/is-the-riemann-integral-of-a-strictly-positive-function-positive – Daniel Camarena Perez Dec 11 '18 at 01:59
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Show that the integral of a positive function is positive and Integral of a positive function – PinkyWay May 12 '21 at 22:07
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This is a surprisingly rough question! I don't think there's any easy way around using (or proving as part of the solution) that every Riemann integrable function on $(a,b)$ is continuous at at least one point. You can show this fairly easily using e.g. Darboux sums, but I would consider this beyond the scope of an introductory Calculus class... maybe another answer can fill in how one might easily see this?
In any case, let's say that $f$ is continuous at some point $p\in (a,b)$, with $f(p) = k > 0$. This means that there exists some $\delta > 0$ such that $|f(p+h) - k| < \frac{k}{2}$ for $|h| < \delta$. Let $L = \min(\delta, p-a, b-p) > 0$. Then $$\int_a^b f(x)\, dx \geq \int_{p-L}^{p+L} f(x)\, dx \geq \ldots$$ can you take it from there?
user7530
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Thanks for the answer, We did learn lower and upper Darboux sums, we thought maybe using reductio ad absurdum of I=0... By using these things do you think it's possible? – jreing Nov 05 '13 at 17:17
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If f is continuous, we can argue like this, Let $m$ be the infemum of the function on$[a,b]$. show that $m \gt 0$ and $\int _a^b f(x)\,dx \gt m(b-a) \gt 0$
GA316
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$f$ is not necessarily continuous, so the infemum could well be zero. – user7530 Nov 05 '13 at 15:57
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