If `mgt0, ngt0`, the definite integral `I=int_0^1 x^(m-1)(1-x)^(n-1)dx` depends upon the values of `m` and `n` is denoted by `beta(m,n)`, called the beta function.For example, `int_0^1 x^3 (1-x)^4dx=int_0^1 x^(4-1) (1-x)^(5-1) dx=beta(4,5)` and `int_0^1 x^(3/2) (1-x)^((-1)/2)dx=int_0^1 x^(5/2-1) (1-x)^(1/2-1)dx=beta(5/2,1/2)`.Obviously, `beta(n,m)=beta(m,n)`.Now answer the question:If `int_0^2 (8-x^3)^((-1)/3)dx=kbeta(1/3,2/3)`, then `k` equals to (A) `1` (B) `1/2` (C) `1/3` (D) `1/4`