# How to show that $f(x)\ge 0 \implies \int_a^x fdt \ge 0$?

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 $\forall x\in[a,b]$)

• 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 $f$. Commented Mar 17, 2015 at 18:44
• thanks. Is it possible to somehow use the fundamental theorem of calculus for this? Commented Mar 17, 2015 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 $F(x)=\int_a^x f(t) dt\geq 0.$ Commented Mar 17, 2015 at 18:48
• Got it,thank you. But what is $F(0)$? Commented Mar 17, 2015 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 $f.$ Commented Mar 17, 2015 at 18:54

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?

• I do see it now, thank you! Commented Mar 17, 2015 at 19:01
• If $\int_a^x fdt=0$ for all $x\in[a,b]$, would this imply $f(x)=0$? Commented Mar 17, 2015 at 19:02
• If $f$ is continuous, yes. But not in general; take a function which is zero except at a single point.
– Neal
Commented Mar 17, 2015 at 19:06
• @user42 My proof works if $f(x)\ge0$ for all $x$ in the interval used as limits on the integral. Commented Mar 17, 2015 at 19:08
• Yes it does,thank you Commented Mar 17, 2015 at 19:08

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

• To the downvoter: what's wrong here ?
– Surb
Commented Mar 17, 2015 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 Commented Mar 17, 2015 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. Commented Mar 17, 2015 at 19:00

Simply use $\int_{a}^{x} (f-0)dt \geq0$. Now split the integrals and take the second one to the RHS

• How do you assume that $\int_{a}^{x} (f-0)dt \geq0$? Commented Mar 17, 2015 at 18:51