Proof of f(x)=0 for All x in [a,b] via Integral Form

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The discussion centers on proving that if a continuous function f on the interval [a,b] satisfies the condition that for every continuous function g, the integral \(\int_a^b fg \, dj = 0\), then f(x) must equal 0 for all x in [a,b]. The participants reference the theorem stating that if f is continuous and non-negative on [a,b] with some point p where f(p) > 0, then \(\int_a^b f \, dj > 0\). This leads to the conclusion that the contrapositive of the theorem is essential for the proof, emphasizing the need for a clever choice of g(x) to demonstrate that f(x) = 0.

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Misswfish
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Suppose f is continuous function on [a,b] such that for each continuous function g, [tex]\int[/tex](fg)dj = 0 (Note: integral is from a to b) , then f(x) = 0 for each x in [a,b].

I know that I should use the theorem If is continuous on [a,b], f(x)[tex]\geq[/tex]0 for each x in [a,b] and theree is a number p i n [a,b] such that f(p) > 0, THen [tex]\int[/tex]f dj > 0.

I just don't understand how they tie together.
 
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Misswfish said:
I know that I should use the theorem If is continuous on [a,b], f(x)[tex]\geq[/tex]0 for each x in [a,b] and theree is a number p i n [a,b] such that f(p) > 0, THen [tex]\int[/tex]f dj > 0.

What is the contrapositive of this statement? It pretty much falls out of it.
 
Ahhh my teacher told me to pick a "clever" g(x) so that we can use this theorem and therefore f(x) = 0. I was thinking contradiction but my teacher shot that down
 

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