# Analysis 2 , HELP! with integrals

1. Apr 6, 2008

### Misswfish

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)

2. Apr 6, 2008

### slider142

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.

3. Apr 6, 2008

### Misswfish

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

4. Apr 6, 2008

### slider142

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

5. Apr 6, 2008

### Misswfish

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

6. Apr 6, 2008

### slider142

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.

Last edited: Apr 6, 2008
7. Apr 6, 2008

### Misswfish

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].