If `f(x+f(y))=f(x)+yAAx ,y in Ra n df(0)=1,` then prove that `int_0^2f(2-x)dx=2int_0^1f(x)dxdot`

Explanation

It is given that `f(x+f(y))=f(x)+y`.
Putting `y=0`, we get `f(x+f(0))=f(x)+0`
or `f(x+1)=f(x)`
Now, using the property
`int_(0)^(2a)f(x)dx=int_(0)^(a)f(x)dx+int_(0)^(a)f(2a-x)dx`, we get
`int_(0)^(2)f(2-x)dx=int_(0)^(1)f(2-x)dx+int_(0)^(1)f(2-(2-x))dx`
`=int_(0)^(1)f(2-(1-x))dx+int_(0)^(1)f(x)dx`
`=int_(0)^(1)f(1+x)dx+int_(0)^(1)f(x)dx`
`=2int_(0)^(1)f(x)dx`
Alternative method
It is given `f(x+f(y))=f(x)+y`.
Putting `y=0`, we get `f(x+f(0))=f(x)+0`
or `f(x+1)=f(x)`
Thus, `f(x)` is periodic with period 1.
Now, `I=int_(0)^(2)f(2-x)dx`
Putting `2-x=t`, we get
`I-int_(0)^(2)f(t)dt=2int_(0)^(1)f(t)dt`
`=2int_(0)^(1)f(x)dx`