show that if "f" is integrable then
f(x) < or equal to 0 on [a,b]
is greater than or = to zero
I am not sure what that implies.
In most basic calculus textbooks there is a theorem that states:
If $\displaystyle \left( {\forall x \in \left[ {a,b} \right]} \right)\left[ {f(x) \leqslant g(x)} \right]$ then it is true that $\displaystyle \int_a^b {f(x)dx} \leqslant \int_a^b {g(x)dx} $.
if your text has that theorem then let $\displaystyle g(x)=0$ and use the theorem.
Otherwise, you need to use upper sums.
The reason Plato asked if you knew about Riemann sums is that how you prove such a basic property of integrals depends upon how you define integrals in the first place! And "Riemann sums" is a common method of defining integrals. You mention of "upper sums" indicates that's what you have learned, although "upper sums" is only a very small part of that.
Given that f(x) is integrable on [a, b], then for any partition of [a, b] (dividing it into smaller subintervals), choosing any $\displaystyle x_i$ in the $\displaystyle i^{th}$ interval, with width $\displaystyle \Delta x_i$, the Riemann sum is $\displaystyle \sum f(x_i)\Delta x_i$ and, as we take "finer and finer" partitions, meaning that the largest $\displaystyle \Delta x_i$ goes to 0 as we take more and more intervals in the partition, and that sum goes to some fixed value, I, whic is the integral, $\displaystyle \int_a^b f(x)dx$.
Now, if $\displaystyle f(x)\le 0$ for all x in [a, b], in particular, $\displaystyle f(x_i)\le 0$ for every partition and for every $\displaystyle x_i$ in a partition. That, in turn, means that $\displaystyle f(x_i)\Delta x_i\le 0$ for every partition and every i in a partition, and so $\displaystyle \sum f(x_i)\Delta x_i\le 0$ for every partition. The integral, the limit of negative numbers can't be positive: $\displaystyle \int_a^b f(x)dx\le 0$.