let $f:[0,1]\mapsto[0,1]$ be continuous how prove this equation has one root in $[0,1]$ $$2x-\int_0^xf(t)dt=1$$
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Let $$ g(x):=2x-\int_0^xf(t)dt. $$
We have $g'(x)=2 -f(x)\geq 1$ for all $x\in(0,1)$.
So $g$ is a bijection from $[0,1]$ onto its range $[g(0),g(1)]$.
Now $g(0)=0$ and $g(1)=2 -\int_0^1f(t)dt\geq 2-1=1$.
So there exists a unique $x_0\in[0,1]$ such that $g(x_0)=1$.
PS: If you simply want the existence of such an $x_0$, and don't care about its uniqueness, see Chris's sister's answer.
If $h(x)= 2x-\int_0^xf(t)dt-1$ then we need $$h(0)\cdot h(1)\le0$$ that is obviously without doing further computations.