A=2, 2π2ω=7-4 can be obtained from the image, and ω=π3 can be obtained from the solution.

So f(x)=2sin(π3x+φ) is substituted into (4,0),

We can get 0=2sin(4π3+φ), that is, 4π3+φ=kπ, k∈Z,

The solution is φ=kπ-4π3, and by the same token, we can get φ=-π3 by substituting it into point (52,2).

Therefore, f(x)=2sin(π3x-π3) can be obtained.

So the period of the function is 6, f( 1)+f(2)+…+f(6)=0.

So f (1)+f (2)+…+f (2013) = 335× 0+f (1)+f (2)+f (3).

=2sin0+2sinπ3+2sin2π3=23

So the answer is: 23.