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Analysis 2 , HELP! with integrals

  1. Apr 6, 2008 #1
    If f is continuous on [a,b], f(x) [tex]\geq[/tex] 0 for each x in [a,b] and there is a number p in [a,b] such that f(p) > 0, then [tex]\int[/tex] f dj > 0 ( Note : the integral is from a to b)
  2. jcsd
  3. Apr 6, 2008 #2
    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.
  4. Apr 6, 2008 #3
    well I thought about putting a "box" around it and picking variable to the right and left of p. Is that it?
  5. Apr 6, 2008 #4
    That's pretty much it. Then just use the definition of continuous.
  6. Apr 6, 2008 #5
    but how does the def of continuous show that the integral is greater than zero
  7. Apr 6, 2008 #6
    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
  8. Apr 6, 2008 #7
    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].
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