If f(x)+f(x+1/2) =1

Find limit 0 to 2 integration of f(x)dx.

ans is 1..plzz explain.

f(x)+f(x+1/2) =1

f(x)=1-f(x+1/2)

₀∫²f(x)dx=₀∫²1-f(x+1/2)dx

₀∫²f(x)dx=2-₀∫²f(x+1/2)dx

take (x+1/2)=v

dx=dv

₀∫²f(v)dv=2-₀∫²f(v)dv

2(₀∫²f(v)dv)=2

₀∫²f(v)dv=1

₀∫²f(x)dx=1

But when u took x+1/2 as v , the limitss will be changed