I assume that f is continuous. In that case, the general solution is f(x)=kx
Hence,
1=\int_0^1 kx\ \mathrm dx = \dfrac{k}{2}
Hence k = 2.
As such f(1)=2.
If f(x+y)=f(x)+f(y) where x,yε R and ∫(0 to 1) f(x).dx=1
find f(1)
I assume that f is continuous. In that case, the general solution is f(x)=kx
Hence,
1=\int_0^1 kx\ \mathrm dx = \dfrac{k}{2}
Hence k = 2.
As such f(1)=2.