Let f be a continuous function on R satisfying
f(x+y)=f(x)+f(y) for all x,y∈R with f(1) =2 and g be a function satisfying f(x)+g(x)= ex then the value of the integral ∫01 f(x)g(x)dx is
1e−4
14(e−2)
2/3
12(e−3)
First show that f(x)=2x,x∈R
Now ∫01 f(x)g(x)dx
=∫01 2xex−2xdx=2xex01−2∫01 exdx−43x301=2e−2(e−1)−43=23.