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## Main Question or Discussion Point

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)

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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)

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

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

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

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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