- #1
Ara macao
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Prove that if f is continuous on [a,b] and
[tex]\int_a^b |f(x)|\,dx = 0[/tex]
then f(x) = 0 for all x in [a,b].
so I'll have to use an epsilon delta proof by contradiction here. I'll have to assume that there exists a c such that f(c) != 0 and for all x = f(c)/2, there exists a delta such that |f(x)-f(c)|< epsilon for |x-c| < delta. and then I should make |f(x)| > epsilon /2. This would contradict the original hypothesis...
But I'm getting confused here...
Thanks!
[tex]\int_a^b |f(x)|\,dx = 0[/tex]
then f(x) = 0 for all x in [a,b].
so I'll have to use an epsilon delta proof by contradiction here. I'll have to assume that there exists a c such that f(c) != 0 and for all x = f(c)/2, there exists a delta such that |f(x)-f(c)|< epsilon for |x-c| < delta. and then I should make |f(x)| > epsilon /2. This would contradict the original hypothesis...
But I'm getting confused here...
Thanks!
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