Definite integral of unknown function given some additional info Given 


*

*$f$ integrable on [0,3]

*$\displaystyle\int_0^1 f(x)\,\mathrm{d}x = 1$, 

*$f(x+1) = \frac{1}{2}f(x)$ for all x $\in [0, 2]$ 


How can I find $\displaystyle\int_0^3 f(x)\,\mathrm{d}x$ ? 
I tried breaking it into as follows:
$\displaystyle\int_0^3 f(x)\,\mathrm{d}x = \displaystyle\int_0^1 f(x)\,\mathrm{d}x + \displaystyle\int_1^2 f(x)\,\mathrm{d}x + \displaystyle\int_2^3 f(x)\,\mathrm{d}x$
I'm not sure how to proceed from here...it's been a long time since I took calculus, and I am probably forgetting something really basic. I feel like the solution has something to do with substitution or the fundamental theorem. I notice that 1 and 2 are 0+1, 1+2 so substituting x+1 into f(x), or something along those lines?

Hint:
$$\int_1^2f(t)\,dt=\int_0^1f(x+1)\,dx={1\over 2}\int_0^1f(x)\,dx={1\over 2}$$
