How do you find out that f(x)= 25\3x^3 is a pdf at the intervals of
[2,10]? It should equal to one when doing the anitderivative.

the antiderivative of x^3 is (x^4)/4

f(x)= (25/3)* (x^3) become F(x)= (25/3)* (x^4)/4 
F(10) = (25/3) *(10^4)/4 =20833.3333333333
F(2) = (25/3) *(2^4)/4 =33.3333333333
20833.333333-33.333333= 20800 far from 1 then it's not a pdf

now if you mean 25/(3x^3)

the antiderivative of 1/3x^3 is -1/x^2

f(x)= (25/3x^3) become F(x)= -(25/x^2)
F(10) =-(25/10^2) =-0.25
F(2) = -(25/2^2) =-6.25
-6.25--0.25= -6 far from 1 then it's not a pdf

The integral from 2 to 10 of (25/3)x^(-3) is -(25/6)x^(-2) evaluated
from 2 to 10.  That's -25/600 + 25/24 = 1.