# If $\int_a^b f(x)\,dx > 1$, then is there a $c$ such that $\int_a^c f(x)\, dx = 1$?

I have a question which is:

If $f$ is an integrable function on $[a,b]$ and $\int_{a}^{b}{f(x)\,dx}>1$, then there exists a point $c\in(a,b)$ such that $\int_{a}^{c}{f(x)\,dx}=1$.

This seems true to me intuitively. But I can't seem to prove or disprove it. Can someone help me? Thanks :)

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Consider the function $F(s) = \int_a^sf(x)dx$, where $s \in [a,b]$. Clearly, $F(a)=0$, and we know that $F(b)>1$. Since $F(s)$ is continuous, then by Bolzano's theorem there must be some $s \in (a,b)$ such that $F(s) = 1$. That is your $c$.
Yes. The mapping $x\mapsto \int_a^x f(t)\,dt$ is continuous. Apply the intermediate value theorem.
Sorry, I don't understand. How can you map $x$. You don't know what $f(x)$ is.. –  Jason Feb 11 '12 at 15:16