let f:[a,b] and be continous for f(x)>=0 for all x on [a,b].
Suppose for some point s in ]a,b] f(s)>0
show that the integral of f from a to b is>0

Hi again fmunshi!
In order to answer this question, we must use the definition of
continuity. You can find a version in the following page

The epsilon-delta definition of continuity
http://www.gap-system.org/~john/analysis/Lectures/L13.html

Thus, if f is continuous at s (it is, since f is continuous in [a,b]
and s belongs to [a,b]), then, given any number e>0, there exists a
number d>0 such that, for every x belonging to [s-d,s+d], f(x) belongs
to [f(s)-e,f(s)+e]. That's one of the definitions of continuity.

Now, we have to prove that if f(s)>0, then f evaluated in a
"neighborhood" of s is also >0. To do this, we use the definition of
continuity. Suppose

f(s) = L > 0

Let's use now the definition of continuity. Given e = L/4 (which is
clearly greater than 0), there exists D>0 such that f evaluated in any
point in [s-D,s+D] belongs to [(3/4)L,(5/4)L]. Clearly, this interval
is greater then 0 (since L is). Thus, in [s-D,s+D], f is >0.

Finally, this, and the fact that f(x)>=0 for all x in [a,b] imply that
the integral is greater than 0:

 int(f) from a to b
=[int(f) from a to s-D] + [int(f) from s-D to s+D] + [int(f) from s+D
to b]

Now, using the fact that definite integrals preserve inequalities:

f>=0 implies that
int(f) from a to s-D >= int(0) from a to s-D = 0

and the same for int(f) from s+D to b.

Also, since f>0 in [s-D,s+D] then,

int(f) from s-D to s+D > int(0) from s-D to s+D = 0   (*)

Therefore,

[int(f) from a to s-D] + [int(f) from s-D to s+D] + [int(f) from s+D
to b] > 0

So int(f) from a to b > 0, which is the desired result. Notice that
the inequality (*) also uses the fact that D>0. If D=0, then we get
int(f) from s to s, which is always 0, no matter what f(s) is.


Google search strategy:
definition of continuity
://www.google.com/search?hl=en&ie=UTF-8&oe=UTF-8&q=definition+of+continuity


This should answer the question. If there's anything unclear, please
request a clarification, I'll be more than happy to further assist you
in this topic.


Best wishes!
elmarto