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This is clear to me thinking about integration like area under a curve, but how can it be proven? (assuming $f$ is continuous and using the fundamental theorem of calculus, and $\forall x\in[a,b]$)

user42
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    For any partition $P$ you must have $L(f,P) \ge 0$, hence the integral must be non negative. Neither continuity nor fundamental theorem of calculus are necessary here. Only integrability of $f$. – copper.hat Mar 17 '15 at 18:44
  • thanks. Is it possible to somehow use the fundamental theorem of calculus for this? – user42 Mar 17 '15 at 18:47
  • Define $F(x)=\int_a^x f(t) dt$. What is $F(a)$? What is $F'$? Once you know $F'$, can you see that $F$ is increasing? Given $F(a)$ and that $F$ is increasing, can you see that $F(x)=\int_a^x f(t) dt\geq 0.$ – Randy E Mar 17 '15 at 18:48
  • Got it,thank you. But what is $F(0)$? – user42 Mar 17 '15 at 18:50
  • I guess I should say "non-decreasing" rather than "increasing", in the event that $f(x)=0$ over some interval of positive length in the domain of $f.$ – Randy E Mar 17 '15 at 18:54
  • This problem is in desperate need of quantifiers. – Git Gud Mar 17 '15 at 18:55
  • I added the quantifier. Thank you everyone for the responses – user42 Mar 17 '15 at 18:57
  • @RandyE This. – Git Gud Mar 17 '15 at 18:58
  • @GitGud Thanks! I've wondered about this terminological problem for a while, but never found a good answer. So I should say "weakly increasing" rather than "increasing" or "non-decreasing." – Randy E Mar 17 '15 at 19:10

3 Answers3

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Assume that $x\ge a$ in the integral $\int_a^x f(t)dt$

Let $f(x)$ be the derivative of a function $F(x)$

Since $f(x)\ge0$, the function $F(x)$ must be weakly increasing, and since $x\ge a$ we have $F(x)\ge F(a)$

We know from the fundamental theorem of calculus that $$ \int_a^x f(t)\,\mathrm dt=F(x)-F(a) $$ Can you see it now?

Alice Ryhl
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You know that $$f\geq g\implies \int f\geq \int g,$$ therefore, $$f\geq 0\implies \int f\geq \int 0=0$$

Surb
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  • To the downvoter: what's wrong here ? – Surb Mar 17 '15 at 18:53
  • I'm not the downvoter and your answer is correct and the simplest way of proving it, however the question stated using the fundamental theorem of calculus – Alice Ryhl Mar 17 '15 at 18:55
  • It's correct in the sense that it obviously answer the problem, but - in addition to what Kristoffer said - it's also obvious that such a problem wouldn't come up after this result was proven. – Git Gud Mar 17 '15 at 19:00
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Simply use $\int_{a}^{x} (f-0)dt \geq0$. Now split the integrals and take the second one to the RHS

Sid
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