Let f(x) be a non negative continuous function on R such that $f(x) +f(x+\frac{1}{3})=5$ then calculate ........ Problem : 
Let f(x) be a non negative continuous function on R such that $f(x) +f(x+\frac{1}{3})=5$
then calculate the value of the integral $\int^{1200}_0 f(x) dx$ 
My approach :
Given that :  $f(x) +f(x+\frac{1}{3})=5.....(1) $
We replace x with $x +\frac{1}{3}$ so we get the given equation as : 
$f(x+\frac{1}{3})+f(x+\frac{2}{3}).....(2)$
Now subtracting (1) from (2) we get :
$f(x+\frac{2}{3}) = f(x) $
$\Rightarrow f(x) $ is function with period $\frac{2}{3}$
Now how to move further please help on how the period of this function will impact the limit of integration. Thanks.

Hint: Note
$$\int_{1/3}^{2/3}f(x)dx= \int_0^{1/3}f(x+\frac{1}{3})dx$$
and hence
$$ \int_0^{1/3}f(x)dx+\int_{1/3}^{2/3}f(x)dx=\int_0^{1/3}(f(x)+f(x+\frac{1}{3}))dx=\frac{5}{3}.$$
Then write
$$\int_0^{1200}f(x)dx=\int_0^{1/3}f(x)dx+\int_{1/3}^{2/3}f(x)dx+\cdots$$
and you will get the answer.
