Prove that if $f$ is continuous and nonnegative in the interval $[a,b]$, then $$A=\int_a^b f(x) \, dx \ge 0$$

My attempted proof: Suppose otherwise i.e. $$\int_a^b f(x) \, dx < 0$$ Then by definition $\exists \delta$ such that $\forall \delta$-fine subdivisions of $[a,b]$ and any choice of $\xi_i[x_{i-1},x_i]$ then $$\left|\sum_{i=0}^n f(\xi_i)\Delta x_i-A\right|<0<\epsilon$$ This implies that $$\sum_{i=0}^n f(\xi_i)\Delta x_i-A$$ is negative.

I'm stuck at this point, any help? Also, is it possible to prove this directly? I think (I could be wrong) it's somewhat trivial since the function is nonnegative on $[a,b]$ i.e. $f(x)=0$ or $f(x)>0$ then then it should just follow that the integral would be just that.

UPDATE: John Don pointed out how I was defining the integral. I am not using Darboux definition (unfortunately) but the following:

Let$f(x)$ be a function on $[a,b]$. We say that $\int_a^b f(x) \, dx$ exists and equals A if $\forall \epsilon > 0, \exists \delta > 0$ such that $\forall$ subdivisions of [a,b] which are $\delta$-fine (i.e. $\Delta x_i < \delta$, $\forall i$) and $\forall \xi_i\in [x_{i-1},x_i]$, then $$\left|\sum_{i=0}^n f(\xi_i)\Delta x_i-A\right|<\epsilon$$

  • 2
    $\begingroup$ Any lower sum is non negative... proof done. $\endgroup$ – user251257 Nov 30 '17 at 21:20
  • $\begingroup$ Isn't it immediate from the fact that for any partition, each term in the sum is nonnegative? $\endgroup$ – MPW Nov 30 '17 at 21:20
  • $\begingroup$ Is this part of what they want to prove? A=∫ b a f(x)dx Or just A≥0? $\endgroup$ – user78090 Nov 30 '17 at 21:21
  • $\begingroup$ $| \sum_{i=0}^n f(\epsilon_i)\Delta x_i -A |<0$ really has no sense since the left side is $\ge 0$ by definition. Now if you want to demonstrate that maybe you could use the fact that $f$ has a minimum in $[a,b].$ $\endgroup$ – chak Nov 30 '17 at 21:23
  • $\begingroup$ @user78090 $A \ge 0$ because the former already follows from the assumptions $\endgroup$ – Tomás Palamás Nov 30 '17 at 21:25

Without loss of generality, assume that $[a,b]=[0,1]$. Because $f$ is Riemann integrable, by the definition you can easily deduce that $A=\lim_{n\rightarrow\infty}\displaystyle\sum_{k=1}^{n}f\left(\dfrac{k}{n}\right)\dfrac{1}{n}$. Now $f(k/n)\geq 0$ so $\displaystyle\sum_{k=1}^{n}f\left(\dfrac{k}{n}\right)\dfrac{1}{n}\geq 0$, then so is its limit.


Let $m$ be the minimum value and $M$ the maximum value of the continuous function $f$ on the interval $[a,b]$. Then $$ m(b-a)\le\int_a^b f(x)\,dx\le M(b-a) $$

More generally, for a Riemann integrable function $f$ over $[a,b]$, if $l$ is a lower bound for $f$, then $$ l(b-a)\le \int_a^b f(x)\,dx $$ Just take a Riemann sum relative to the trivial subdivision.

  • $\begingroup$ But wouldn't your last step just be sufficient? $\endgroup$ – Tomás Palamás Nov 30 '17 at 21:44
  • $\begingroup$ @ContraModernistae Yes, it would. But you don't really need Riemann sums for integrals of continuous functions and I don't know what you're precisely interested in. $\endgroup$ – egreg Nov 30 '17 at 21:49
  • $\begingroup$ I see and to show that the integral nonnegative. $\endgroup$ – Tomás Palamás Nov 30 '17 at 21:52
  • $\begingroup$ @ContraModernistae In your setting, a lower bound for $f$ is $0$. $\endgroup$ – egreg Nov 30 '17 at 21:53

Be $F$ one primitive of $f$ ( If $f$ is conntinuous, $F$ is differentiable and $F'(x)=f(x)$)

$\displaystyle A=\int_a^b f(x) \, dx= F(b)-F(a)$ but as $f(x)\ge 0$ on $[a,b]$ then $F$ is increasing so $A=F(b)-F(a)\ge0$


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