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.

Check this
$\int_0 ^{1200}f=\int_0 ^{\frac{2}{3}}f+\int_{\frac{2}{3}}^{\frac{4}{3}}f+...$
$\int _{\frac{2}{3}}^{\frac{4}{3}}f(x)=\int_0 ^{\frac{2}{3}}f(x)$
$\int _0^ {1200} f=800\times \int _0^{\frac{2}{3}}f$
$\int _0^{\frac{2}{3}}f=\int _0^{\frac{1}{3}}f+\int _{\frac{2}{3}}^{\frac{4}{3}}f$
Let $y=x-{\frac{1}{3}} \implies dy=dx$
 As $x={\frac{2}{3}} y={\frac{1}{3}};x={\frac{1}{3}}y=0$
$=\int _0^{\frac{1}{3}}f(y)dy+\int _0^{\frac{1}{3}}f(y+\frac{1}{3})dy$
$=\int _0^\frac{1}{3}5dy$
