Answer
$\mu=\frac{1}{3}$
$Var(X)=\frac{1}{18}$
$\sigma=\frac{\sqrt 2}{6}$
Work Step by Step
We are given $f(x)=2(1-x); [0;1]$
The expected value
$\mu=\int^{1}_{0}xf(x)dx$
$=2\int^{1}_{0}x(1-x)dx$
$=2\int^{1}_{0}(x-x^{2})dx$
$=2(\frac{x^{2}}{2}-\frac{x^{3}}{3})|^{1}_{0}$
$=2(\frac{1}{6})=\frac{1}{3}$
The variance is
$Var(X)=\int^{1}_{0}(x-\frac{1}{3})^{2}f(x)dx$
$=2\int^{1}_{0}(x^{2}-\frac{2}{3}x+\frac{1}{9})(1-x)dx$
$=2\int^{1}_{0}(x^{2}-x^{3}-\frac{2}{3}x+\frac{2}{3}x^{2}+\frac{1}{9}-\frac{1}{9}x)dx$
$=2\int^{1}_{0}(-x^{3}+\frac{5}{3}x^{2}-\frac{7}{9}x+\frac{1}{9})dx$
$=2(\frac{-x^{4}}{4}+\frac{5x^{3}}{9}-\frac{7}{18}x^{2}+\frac{1}{9}x)|^{1}_{0}$
$=\frac{1}{18}$
The standard deviation of X is
$\sigma=\sqrt Var(X)$
$=\sqrt \frac{1}{18}=\frac{\sqrt 2}{6}$
