Proof by basic principles of Riemann Integration Let $f:[a,b] \rightarrow \mathbb R$ be Riemann Integrable on $[a,b]$ and $f(x)\ge0$ for all $x\in[a,b]$. Show that $$\int_a^b f(x)dx\ge0$$ using basic principles of Riemann Integration.
I'm new to this concept of Riemann Integration. Apparently, I have to show that  $U(P,f)\ge0$ for any partition in $[a,b]$, which will prove that the integral itself is greater than or equal to zero.
But I'm unable to frame the proper statements, and I'm afraid I might make an assumption without really justifying it.
So, can somebody please show me how to write a formal proof in such a case?
 A: Let $\int_a^b f\,$ be defined as a limit of Riemann sums.
Then for every $\varepsilon > 0$ there exists $\delta > 0$ such that for every tagged partition $P$ of $[a,b]$ with $\|P\| < \delta$ one has $$\left |\int_a^b f- R(P,f) \right| < \varepsilon$$ So, if $P$ is a tagged partition such that $\|P\| < \delta$, then $$\int_a^b f > R(P,f) - \varepsilon$$ Hence, if $f(x) \ge 0$ for very $x \in [a,b]$, then $$\int_a^b f > -\varepsilon$$
But $\varepsilon$ is arbitrary $\dots$
A: Guessing the notation you are using you have $$L(P,f)\le\int f\le U(P,f)$$
You can use the function which is identically $=0$ as a step function from below, and for this, quite obviously, the integral is $=0$. So you are done.
A: If $f$ is a Riemann integrable function, then for any partition $P$, the lower sum $L(P,f)$ is less than or equal to the integral, while the upper sum $U(P,f)$ is greater than or equal to the integral. If you pay attention to the quantifiers, you find that to show that the integral is larger than some number $c$, it is enough to find a single partition whose lower sum is larger than $c$. 
In this case you can take the simplest partition of all, where there is just the single subinterval $[a,b]$. Over this interval there is some infimum, which must be nonnegative; then the lower sum is this infimum times $b-a$, which will again be nonnegative.
