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If $ f $ is continuous over $ [1,3] $ and $ \int_1 ^ 3f (x) dx = 0 $, then $ f (x) = 0 $ for $ 0 \leq x \leq 1 $.

I think that is false. I'm right?

asd asd
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    That's the right answer; how would you convince someone that you're right? – Greg Martin Mar 26 '21 at 20:28
  • My reason is, that a definite integral gives you $ 0 $ in an interval, it does not mean that the function is $ 0 $, for the simple fact that you are evaluating, and there are functions such as the odd functions that certain integrals in them give $ 0 $ but the functions are not null. I'm trying to find the counterexample. – asd asd Mar 26 '21 at 20:30
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    Excellent, you're right that a counterexample is all the proof you need! – Greg Martin Mar 26 '21 at 20:32
  • Thanks for answering :). – asd asd Mar 26 '21 at 20:33
  • To find a counter-example, draw a graph of a function with equal "area" above and below the $x$ axis on the interval [1,3]. Writing an explicit formula is not hard if you consider only linear piecewise functions. – tangentbundle Mar 26 '21 at 20:42
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    Well, your integral condition is on $[1,3]$ and you want to prove something for $f$ on $[0,1]$... I would be $f$, I would party high on $[0,1]$ ! – zwim Mar 26 '21 at 21:00
  • Take $f(x)=x-2.$ Notice that you just need to remind that "an integral is the area under the curve", so for this to be $0$ you need a positive area and then a "negative" area. Then identity shifted seems to be the simplest as candidate. – Verónica Rmz. Mar 26 '21 at 21:24

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The recipe is already hinted in the comments, and there are infinitely many choices here. One of them is to take $f(x) = 2x+c$, then $\displaystyle \int_{1}^3 f(x)dx = 0\implies c = - 4\implies f(x) = 2x-4$. Observe that $f(x)$ is not identically $0$ on $[1,3]$.