# Analysis 2 , HELP! with integrals

## Main Question or Discussion Point

If f is continuous on [a,b], f(x) $$\geq$$ 0 for each x in [a,b] and there is a number p in [a,b] such that f(p) > 0, then $$\int$$ f dj > 0 ( Note : the integral is from a to b)

Basically it is asking you to show that there is a neighborhood around p where f is non-zero. All you need for this is the definition of continuous.

well I thought about putting a "box" around it and picking variable to the right and left of p. Is that it?

That's pretty much it. Then just use the definition of continuous.

but how does the def of continuous show that the integral is greater than zero

but how does the def of continuous show that the integral is greater than zero
There is a non-zero volume [p-e, p+e] over which the function is everywhere non-zero (and non-negative) (show by continuity of f), so going back to partitions, the upper and lower sums will also be non-zero. The only reason for this to be true is that for some e-interval around p, the upper and lower sums (dependent on the product of infimums and supremums of f with the partition volumes) are non-zero and non-negative.
Now since you have shown it is non-zero on [p-e, p+e] for some e, and f is never negative, you know the integral over a larger interval cannot be less.

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ahhhhhh that doesnt sound right. I thought about using the definition of uniformaly continuous. I picked my eplision to be f(p)/2. I then used the theorem "Suppose S is a set of numbers and M is the least upper bound of S. If p is a number such that p< M , then there is an x in S such that p < x < or = M. Then I picked an q in (p-d, x) and a t in (x, p+d) then I have p-d<q<x<t<p+d. That is where Im stuck. I need to show that the interval of (q,t) is above zero and then show that for [a,b].