1
$\begingroup$

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 :)

$\endgroup$

2 Answers 2

4
$\begingroup$

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 $c$.

$\endgroup$
1
  • $\begingroup$ Ah thank you :) I was missing the Bolzano part :) $\endgroup$
    – Jason
    Feb 11, 2012 at 15:22
3
$\begingroup$

Yes. The mapping $x\mapsto \int_a^x f(t)\,dt$ is continuous. Apply the intermediate value theorem.

$\endgroup$
2
  • $\begingroup$ Sorry, I don't understand. How can you map $x$. You don't know what $f(x)$ is.. $\endgroup$
    – Jason
    Feb 11, 2012 at 15:16
  • $\begingroup$ Yes. Edited for that change. $\endgroup$ Feb 11, 2012 at 18:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .