Integral f(3x)dx for x=0 to 3 is equivalent to integral f(x)dx for x=0 to 9. It is assumed that f(x) and x are continuous over this range. Integral f(x)=5 for x=0 to 9, so f(3x)=5 for x=0 to 3.
To illustrate the validity, let f(x)=3x^2+2x+1, so integral of f(x)=x^3+x^2+x+k and f(3x)=27x^3+9x^2+3x+k. Apply the limits: 9^3+9^2+9; 27*27+9*9+3*3. Both these are the same as 3^6+3^4+3^2. Since f(x) was chosen randomly, this implies the validity of the solution. It's clear also that f(3x) is not the same as 3f(x).