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This result is so obvious intuitively, when the function is partly positive and partly negative, but I cannot find a mathematical way to prove it. Assuming that the function is continuous.

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    You should, no you must require your function to be continuous, for this to be true. – 2017-01-08
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    you forgot 2 assumptions: $a\neq c$ and $f$ must be continous. in that case, assume $f(b)\neq 0$ for all $b$. Thus either $f>0$ or $f<0$ and $\int_{[a,c]}f>0$ or $\int_{[a,c]}f<0$. – 2017-01-08
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    Okay, I forgot about that. I have edited it now. But even then how to prove the result? – 2017-01-08
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    i already gave you a proof! – 2017-01-08
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    @kolobokish not true. you do not necessairily need that assumption: take any function and modify it to be zeron on $\mathbb{Q}$ – 2017-01-08
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    I'm sorry but I do not understand your proof. Also I do however understand that at some point the function has to cross zero, because the positive area will cancel out the negative area. But I want a mathematical proof that the function crosses zero at one point. Thanks. – 2017-01-08
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    it is an indirect proof. ($A\rightarrow B\Leftrightarrow \neg B\rightarrow\neg A$) – 2017-01-08
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    @Max Oh Max, what you say, just takes $f$ to be 0 at some point. Which was in the question. – 2017-01-08
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    @kolobokish you are saying the assumption of continuity is necessary, whereas it is just sufficient. – 2017-01-08
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    @kolobokish take any discontinous function $f$ and 2 reals $a,c$ such that $\int_{[a,c]} f=0$. now modify $f$ such that it is zero on $\mathbb{Q}$. Clearly $\int_{[a,c]}=0$ and $f$ has infinitely many zeros. so the *must* you wrote is wrong. the assertion also holds for some discontinous functions. It would also be enough to assume $f$ to have no jumps crossing zero. EDIT: shame on me, i also wrote "must" in my first comment. – 2017-01-08
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    I understand what you all saying, i.e., If the function changes sign it must cross zero at some point which is fine and the way I thought about the proof. But is there a step by step mathematical proof? – 2017-01-08

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It is not true. Take $ f(x)=-1 $ on the first half of the interval and $1$ on the second half.

It is however true for continuous functions.

edit: Given the discussion above, I think what is bugging you is where the continuity assumption comes into play.

First assume $f\ne 0$ on the interval for the sake of contradiction. Then I claim with the continuity assumption, we can take $f<0$ or $f>0$ on the interval. This follows directly from the intermediate value theorem. If your function changes signs, it has a root.

Then $f(x)>0$ for any $x\in [a,c]$ without any loss of generality. Then we use the monotonicity of the integral to find $$ \int_a^cf(t)dt>0 $$ a contradiction of the assumption $\int_a^cf(t)dt=0$.

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    When the function is continuous how to prove the result? – 2017-01-08
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    What @max said in the above is fine – 2017-01-08
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Suppose your function is continious. Then let $\int f(x) $ - area of then consider $f(\frac{a+c}{2})$. If it's zero then we find it. If it's more then zero then we should go left , less then zero - right. Suppose we have infinite sequence of $|[a_{i},c_{i}]| \to0$ , so there is point $b $ in intersaction of all segments. And because of continuity : $\lim_{i \to \infty}f(a_{i}) \le 0 $, $\lim_{i \to \infty}f(c_{i}) \ge 0 $, so there is $f(b) = 0$.

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Consider the function $$ F(t)=\int_{a}^{t} f(x)\,dx $$ Then $F(a)=0$ and $F(c)=0$. Apply Rolle's theorem and the fundamental theorem of calculus.