Let g(x) be a non negative continuous function on R such that $g(x) +g(x+\frac{1}{3})=5$.. Problem : 
Let g(x) be a non negative continuous function on R such that $g(x) +g(x+\frac{1}{3})=5$ then calculate the value of integral $\int^{1200}_0 g(x) dx$ 
My approach : 
$g(x) +g(x+\frac{1}{3})=5$.....(1)
put x = x +$\frac{1}{3}$ 
we get $g(x+\frac{1}{3})+g(x+\frac{2}{3})=5$...(2) 
Subtracting (2) from (1) we get 
$g(x) = g(x+\frac{2}{3})$ 
$\Rightarrow $ g(x) is periodic with period $\frac{2}{3}$
So we can write the integral as 
$1800 \int^{2/3}_0 g(x)dx $
now what to do further please suggest , will be of great help. thanks 

I don't know whether this solution is the fastest but I am pretty sure it is the most accurate one. 
Step 1: multiply the differential infinitesimal (dx) on $g(x) +g(x+\frac{1}{3})=5$
$(g(x) +g(x+\frac{1}{3}))dx= 5dx$
Because of the multiplication's disturbution law:
$g(x)dx +g(x+\frac{1}{3})dx= 5dx$
Step 2: integral two side with from low limit(0) to upper limmit(1/3):
$\int_0^{\frac{1}{3}}g(x)dx + g(x+\frac{1}{3})dx = \int_0^{\frac{1}{3}} 5 dx$
By the integration's indentitiy:
$\int_0^{\frac{1}{3}}g(x)dx + \int_0^{\frac{1}{3}}g(x+1/3)dx = \int_0^{\frac{1}{3}} 5 dx$
substitude x for x+1/3
$\int^{2/3}_0 g(x)dx = \int_0^{\frac{1}{3}}g(x)dx + \int_\frac{1}{3}^{\frac{2}{3}}g(x)dx = \int_0^{\frac{1}{3}} 5 dx = \frac{5}{3}$
