$\int_{\epsilon}^{2\epsilon}f(x) dx \geq C$ 
Let $f:[\epsilon, 2\epsilon] \to \mathbb{R}^{+}$ continuous. Is it possible to show that
$$
\int_{\epsilon}^{2\epsilon}f(x) dx \geq C ?
$$
where $C$ is a positive constant that does not depend on $\epsilon$.

 A: Your question is ambiguous.
One interpretation is

Can we show that for all functions $f:(0,1) \to \mathbb{R}^+$ there is a constant $C$ such
that for all $\epsilon > 0$ we have $\int_\epsilon^{2\epsilon} f(x)\textrm{ d}x > C$?

The answer to this question is "No".
If $f(x) = 1$ then $f$ is continuous and positive on $[0,1]$, and $\int_\epsilon^{2\epsilon} 1 \textrm{ d} x  = \epsilon $.  Clearly $\epsilon$ cannot be bounded from below by a positive constant $C$ independent of $\epsilon$.
However, your question could also be interpreted to be asking:

Can we find an example of a function $f: (0,1) \to \mathbb{R}^+$ and a constant $C$ such
that for all $\epsilon > 0$ we have $\int_\epsilon^{2\epsilon} f(x)\textrm{ d}x > C$?

The answer to this question was answered in the affirmative by @user619894.  Their example is
$C={1\over 2}$, $f(x)={1\over x}$, $$\begin{align*}
\int _{\epsilon}^{2\epsilon} {1\over x} dx &= \log(2\epsilon) - \log(\epsilon)\\ &= \log(\frac{2\epsilon}{\epsilon}) \\&= \log(2)\\ &=0.69... \\&> {1\over 2}
\end{align*}$$
