Given $f(x)$ is a probability density function it must hold that,
$\int_{- \infty }^{\infty}f(x)dx=1$
it is given that $0<x<1$ which implies $f(x)=0$ outside the given range of $x.$ Thus we have,
$\int_{0 }^{1}f(x)dx=1$
$\implies \int_{0 }^{1}cx^{2}(1-x)dx=1$
$\implies c\left \{ \int_{0}^{1}x^{2}dx -\int_{0}^{1}x^{3}dx\right \}=1$
$\implies c\left \{ \frac{1}{3}-\frac{1}{4}\right \}=1$
$\implies c=12
The mean of a PDF $\mu$=E(x)=$\int_{-\infty }^{\infty }xf(x)dx$
$E(X)=\int_{0 }^{1 }12*xx^{2}(1-x)dx$
$\qquad =12\left \{ \int_{0}^{1}x^{3}dx-\int_{0}^{1}x^{4}dx \right \}$
$\qquad =12*\left \{ \frac{1}{4}-\frac{1}{5} \right \}$
$\qquad =12*\frac{1}{20}$
$\qquad =0.6$
correct answer is option C.
