= Solution
A <probability density function> must integrate to one. Therefore
$$
1=c\int_0^1x(1-x)\,dx=c\left(\frac12-\frac13\right),
\qquad\boxed{c=6}.
$$
The first two <moments> are
$$
\mathbb EX=6\int_0^1x^2(1-x)\,dx=6\left(\frac13-\frac14\right)=\frac12,
\qquad
\mathbb EX^2=6\int_0^1x^3(1-x)\,dx=6\left(\frac14-\frac15\right)=\frac3{10}.
$$
Consequently the <mean> and <variance> are
$$
\boxed{\mathbb EX=\frac12,\qquad \operatorname{var}(X)=\frac3{10}-\frac14=\frac1{20}.}
$$
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