For a laboratory assignment, if the equipment is working, the density function of the observed outcome $X$ is
$$ f(x) = \begin{cases} 2(1-x), & 0 <x<1, \\ 0, & \text{otherwise.} \end{cases} $$
Find the variance and standard deviation of $X$.
We know that the variance is related to the mean and the second moment. I am stuck on how to set up the integrals for the both of them.
The mean is given by $$E(X)=\int_0^1 (x)(2(1-x))\,dx.$$ For the calculation, note that $(x)(2(1-x))=2x-2x^2$.
For the variance, if $\mu$ is the mean, then $$\text{Var}(X)=E((X-\mu)^2)=\int_0^1 (x-\mu)^2(2(1-x))\,dx.$$
It is often the case that this is not the most convenient way to evaluate the variance. An often more useful formula is $$\text{Var}(X)=E(X^2)-(E(X))^2.$$
To find $E(X^2)$, we use $$E(X^2)=\int_0^1 (x^2)(2(1-x))\,dx.$$
The integrals are $$ \text{mean} = \mu = \int_0^1 x f(x) \, dx $$ and $$ \text{variance} = \sigma^2 = \int_0^1 (x-\mu)^2 f(x)\,dx. $$ In order to evaluate the second integral, one must find $\mu$ by evaluating the first integral.
The second integral is the variance $\sigma^2$; the standard deviation is the square root of that.
