Arc length (L) of the function y=f(x) on the interval [a,b] is given by the formula,

`L=int_a^bsqrt(1+(dy/dx)^2)dx`  , if y=f(x) and a `<=`  x `<=`  b ,

`y=3/2x^(2/3)+4`

Now let's differentiate the function with respect to x,

`dy/dx=3/2(2/3)x^(2/3-1)`

`dy/dx=1/x^(1/3)`

Plug in the above derivative in the arc length formula,

`L=int_1^27sqrt(1+(1/x^(1/3))^2)dx`

`L=int_1^27sqrt(1+1/x^(2/3))dx`

`L=int_1^27sqrt((x^(2/3)+1)/x^(2/3))dx` 

`L=int_1^27sqrt(x^(2/3)+1)/x^(1/3)dx`

Now let's first evaluate the definite integral by using integral substitution,

Let `u=x^(2/3)+1`

`(du)/dx=2/3x^(2/3-1)`

`(du)/dx=2/(3x^(1/3))`

`intsqrt(x^(2/3)+1)/x^(1/3)dx=intsqrt(u)3/2du`

`=3/2intsqrt(u)du`

`=3/2((u)^(1/2+1)/(1/2+1))`

`=3/2(u^(3/2)/(3/2))`

`=u^(3/2)`

Substitute back `u=x^(2/3)+1`  and add a constant C to the solution,

`=(x^(2/3)+1)^(3/2)+C`

`L=[(x^(2/3)+1)^(3/2)]_1^27`

`L=[(27^(2/3)+1)^(3/2)]-[(1^(2/3)+1)^(3/2)]`

`L=[(9+1)^(3/2)]-[2^(3/2)]`

`L=[10^(3/2)]-[2^(3/2)]` 

`L=31.6227766-2.828427125`

`L=28.79434948`

Arc length of the function over the given interval is `~~28.79435`