Question

If F(x) be a periodic function such that F(x) + F((1/x) + 2) = 10, find the value of ∫F(x) dx for x ∈ [0,201].

Answer 1

Given F(x) + F((1/x) + 2) = 10 ...(i) 

Replacing x by (x + (1/2)), we get 

F(x + (1/2)) + F(x + 1) = 10 ...(ii) 

Subtracting Eq. (ii) from Eq. (i), we get 

F(x) – F(x + 1) = 0 

F(x) = F(x + 1) 

Thus, the period of F(x) is 1. 

Now,

Putting x = (y + (1/2)) 

dx = dy in the 2nd integral, we have

= 201 × 10 

= 2010