Definite Integration Question 2

[@dhruvchaudhary5379]

Ok so if we look at the identity replace x by x+1/2 and subtract the two equations to get F(x)=F(x+1) . So the integral is nothing but 1500I =9000/ u . So I = 6/u where I = integral from limits 0 to 1 . Now the jntergal can be broken over limits 0to 1/2 and from 1/2 to 1 in the second interval substitute x= t+1/2 to get revised limits from limit t from 0 to 1/2 and then replace F(t+1/2) as 3-F(t) and 2 integrals cancel to get 3/2 in LHS to get value of Lambda as 4