# f:[0;1]->R is continuous and int^1_0 f(x)dx=1/2. Proof that there exists an c in [0,1] such that f(c)=c.

$f:\left[0;1\right]\to \mathbb{R}$ is continuous and

Proof that there exists an $c\in \left[0,1\right]$ such that $f\left(c\right)=c$.
Idea: So i would define a function $g\left(x\right)=f\left(x\right)-x$ and then find points so that $g\left(x\right)$ would be $g\left(y\right)<0$ and $g\left(z\right)>0$. And proceed with my proof and use the intermediate value theorem.
However, i am confused how to do that since i don't have the exact function but only the integral.
Can i just define an $f\left(x\right)$ for which the integral would work like $f\left(x\right)=\frac{1}{2}$?
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darkangel5991we
Your idea was just right! If $g\left(x\right):=f\left(x\right)-x$ then $g\left(x\right)$ is continuous on $\left[0,1\right]$ and:
${\int }_{0}^{1}g\left(x\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x={\int }_{0}^{1}f\left(x\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x-{\int }_{0}^{1}x\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=\frac{1}{2}-\frac{{1}^{2}}{2}=0$
You know by the mean value theorem for Riemann integrals of continuous functions that there exists a $c\in \left[0,1\right]$ such that:
$0={\int }_{0}^{1}g\left(x\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x={\int }_{0}^{1}\left(f\left(x\right)-x\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=g\left(c\right)\left(1-0\right)=f\left(c\right)-c$
That is, $f\left(c\right)=c$ for at least one $c\in \left[0,1\right]$.
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Eliza Gregory
Let using the definition above (i.e. $g\left(x\right)=f\left(x\right)-x$)
$G\left(0\right)=0,G\left(1\right)=0$
$G\left(x\right)$ is continuous and differentiable with ${G}^{\prime }\left(x\right)=g\left(x\right)$
From the mean value theorem, we know that for all $a,b\in \left[0,1\right]$ with $a there exists a $a such that $g\left(c\right)=\frac{G\left(b\right)-G\left(a\right)}{b-a}$
With, $a=0$ and $b=1$ there exists a $c\in \left[0,1\right]$ such that $g\left(c\right)=0$