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?
 A: Breaking the integral is a good idea to start. Proceed as follows:
$\int_{0}^{3}f(x)dx=\int_{0}^{1}f(x)dx+\int_{1}^{3}f(x)dx=1+\int_{1}^{3}f(1+(x-1))dx$
$=1+\int_{0}^{2}f(1+u)du$ 
by the change of variables theorem and by identity $1$.
Then we have that,
$=1+\frac{1}{2}\int_{0}^{2}f(u)du=1+\frac{1}{2}\int_{0}^{1}f(u)du+\frac{1}{2}\int_{1}^{2}f(u)du=\frac{3}{2}+\frac{1}{2}\int_{1}^{2}f(1+(u-1))du$.
by identity $1$ and $3$. Now by the change of variables theorem and identities $1$ and $3$ we get
$\frac{3}{2}+\frac{1}{2}\int_{0}^{1}f(1+z)dz=\frac{3}{2}+\frac{1}{4}=\frac{7}{4}$.
A: That's a good start, and what you noticed is the right idea to finish; for any function $g$ and any values $a,b,c$, it is true that
$$\int_a^bg(x)\,dx=\int_{a-c}^{b-c}g(x+c)\,dx.$$
A: Hint:
$$\int_1^2f(t)\,dt=\int_0^1f(x+1)\,dx={1\over 2}\int_0^1f(x)\,dx={1\over 2}$$
A: The key here is to recognize that
$$\int_1^2 dx \, f(x) = \frac12 \int_0^1 dx \, f(x)$$
etc.  Because
$$\int_0^1 dx \, f(x) = 1$$
then
$$\int_0^3 dx \, f(x) = 1+\frac12+\frac14 = \frac{7}{4}$$
