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Let $f$ be continuous on $[a,b]$ with $f(x)>0$ for all $x\in [a,b]$. Prove that $\int_{a}^{b}f(x)dx>0$. (Note that needs strictness on inequality, could prove $\ge0$ easily but need $>0$).

I know that since $f$ is continuous on a closed interval then it is uniformly continuous, and I know that uniformly continuous functions are integrable. I previously had it where $\int_{a}^{b}f(x)dx =lim_{n\to \infty}\sum_{i=1}^{\infty} f(x_{i}) \Delta x$ and then said that since $f(x)>0$ then $\Delta x >0$ but then we do not use that definition of an integral in class and don't have something simple I could write it out equal to so then I am back to square one.

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As defined, $f(x)$ is continuous in $[a,b]$, a closed interval, so its image is also a closed interval. Therefore, $\exists \alpha \in \mathbb{R}$ such that $f(x) \gt \alpha \gt 0 \forall x\in[a, b]$, so:

$$\int_a^bf(x)dx \geq \int_a^b\alpha dx = \alpha(b-a) \gt 0$$

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    $\begingroup$ How do you know such $\alpha$ exist? We only know that $f(x)>0$ for all $x\in [a,b]$. The infimum of $f$ in this interval might still be zero. $\endgroup$ Commented Apr 18, 2019 at 16:16
  • $\begingroup$ @Mark Yeah, you need a theorem that the continuous image of a closed interval is a closed interval, and in particular, has a minimum and maximum value. $\endgroup$ Commented Apr 18, 2019 at 16:19
  • $\begingroup$ Sorry, I didn't read the question well and missed that the function is continuous. Because in general the statement which appears in the title is true for any integrable function, and then the proof is harder. Nevermind then. $\endgroup$ Commented Apr 18, 2019 at 16:21
  • $\begingroup$ I've edited the answer to remove the confusion $\endgroup$ Commented Apr 18, 2019 at 16:48
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The following method of proof is to show that $[a, b]$ has a subinterval on which $f$ is not only positive, but is bounded away from zero; i.e., that there exists a positive constant $c$ such that $f > c$ on that subinterval.

Once we've established what, we can say: $$ \int_{a}^{b} f(x) \; dx \geq \int_{\mbox{the subinterval}} f(x) \; dx \geq \int_{\mbox{the subinterval}} \, c \; dx = \mbox{(length of subinterval)} \, c > 0. $$

To carry out this program, ick a point $x_{0} \in [a, b]$. Let $\epsilon = f(x_{0})/2$.

Since $f(x_{0}) > 0$ and since $f$ is continuous at $x_{0}$, there exists a neighborhood $N = ]x_{0} - \delta, x_{0} - \delta[$ such that $$ |f(x) - f(x_{0})| < \epsilon \quad \mbox{for all $x$ in $N \cap [a, b]$}. $$ This last inequality means that $f(x)$ is within distance $\epsilon$ from $f(x_{0})$, hence, in particular, $$ f(x) > {1 \over 2} f(x_{0})\quad \mbox{for all $x$ in $N \cap [a, b]$}. $$ Therefore, $N \cap [a, b]$ is our subinterval, and ${1 \over 2} f(x_{0})$ is our $c$.

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