# Integrals and continuity

1. May 20, 2012

### mike1988

Let f : R to R be a continuous function, and suppose that definite integral from m to n |∫(m to n)f(x)dx|≤(n-m)^2 for every closed bounded interval [m, n] in R. Then is it the case that f(x) = 0 for all x in R?

I tried using fundamental theorem of calculus but got stuck, since I only got that F'(x)=f(x)≤ 0.

Any help/suggestion would be appreciated.

Last edited: May 20, 2012
2. May 20, 2012

### xAxis

Why do you think f(x) would be 0? Doesn't make much sense.

3. May 20, 2012

### deluks917

If f is non-zero at a point a then there exists some b, K > 0 s.t. f(x) > K for all x in [a-b,a+b]. Can you finish the proof?

4. May 20, 2012

### mike1988

At first I thought it was wrong, so was trying to find a counterexample. But I saw this question in a book asks to show that f(x)=0 for all x in R, which means it should be true. I was using Mean Value theorem and fundamental theorem of calc but I am not really heading anywhere with those.

Thanks and any further help will be appreciated!

5. May 20, 2012

### mike1988

I guess I don't really get what this means (are you using the sing-change property of continuity?). I am really stuck!

Thanks

6. May 20, 2012

### DonAntonio

Using continuity of f is that the condition is fulfilled, so $$\,f(x)>K \,\,\forall\,x\in [a-b,a+b]\Longrightarrow \int_{a-b}^{a+b}f(x)dx>K\int_{a-b}^{a+b}dx = 2Kb$$ Can you see now how to finish with a contradiction?

DonAntonio

7. May 20, 2012

### deluks917

note also if 0<b<c then f(x) >K on [a-c,a+c].