Question :

expected value-->

E(g(x))= $\displaystyle \int_0^{1 }g(x^2) ~dx $ for any given g(x);

find P($\displaystyle 1/4$ <X <$\displaystyle 9/16 $)

my answer (am i right...?):

put g(x)= x,suppose there is a p.d.f h(x)= x;

then E(g(x))= $\displaystyle \int_0^{1 }g(x)h(x) ~dx $

= $\displaystyle \int_0^{1 }(x^2) ~dx $

so, P($\displaystyle 1/4$ <X <$\displaystyle 9/16 $)

=$\displaystyle \int_{1/2}^{3/4} (x^2) ~dx $