# Prove integral is greater than $0$

$f(x)$ is Riemann integrable on $I=[a,b]$ and $f(x)>0$ for all $x \in I$, prove $\int_a^b f(x) dx >0$ .

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Ok, so I hope that you know that (with obvious meaning of notation) (all depending on your definition of the integral) $$\int_a^b f(x)\; dx \geq \sum_{i}[\inf_{[x_i, x_{i+1}]} f(x)]\Delta x$$ for all partitions of $[a,b]$, where $\inf$ is the infimum of $f$ over each subinterval.

Now the claim is that there is a $\delta >0$ and a subinterval $[x_1 , x_{i+1}]$ such that $f(x) \geq \delta$ on that interval. Suppose this wasn't true. Then $$\sum_{i}[\inf_{[x_i, x_{i+1}]} f(x)]\Delta x = 0$$ for all pertitions. And so (taking the limit) the integral would be $0$.

So there is an interval $[x_i, x_{i+1}]$ and a $\delta > 0$ such that $f(x)\geq \delta$ for all $x\in [x_i, x_{i+1}]$.

So now you have that $f(x)\geq \delta >0$ on an interval $[x_i , x_{i+1}]$. So then $$\int_{x_i}^{x_{i+1}} f(x)\; dx \geq \delta(x_{i+1} - x_i) > 0.$$ So $$\int_{a}^b f(x)\; dx = \int_a^{x_i} f(x)\; dx + \int_{x_i}^{x_{i+1}} f(x)\; dx + \int_{x_{i+1}}^b f(x)\; dx > 0.$$ (Assuming that you are fine with saying that the first and last integral definitely are greater then or equal to $0$).

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The integral is at least as big as any lower sum that you compute. Show that every lower sum is greater than zero.

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but we need inf f(x) to compute lower sum.Since inf f(x) could be 0, so could lower sum. –  user63219 Feb 24 '13 at 23:41

If $f$ is Riemann integrable on $[a,b]$, then there is a point $c\in [a,b]$ such that $f$ is continuous at $c$ (in fact, $f$ is continuous almost everywhere in $[a,b]\,$). Since $f(c)>0$, it follows that there is a non-degenerate interval $I\subset[a,b]$ and a positive number $\alpha$ such that $f(x)\ge \alpha$ for all $x\in I$. We then have, by the nonnegativity of $f$, that $\int_a^b f(x)\,dx\ge \int_{I} f(x)\,dx>0$.

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