Solution
= Solution
{c}
The second <moment> is
$$
\mathbb E[X^2]=\int_0^b ax^3\,dx=\frac{ab^4}{4}=\frac{b^2}{2}=\frac98.
$$
Using the <mean> $\mathbb E[X]=1$, the <variance> is therefore
$$
\boxed{\operatorname{var}(X)=\frac98-1=\frac18.}
$$
= Solution
{c}
The second <moment> is
$$
\mathbb E[X^2]=\int_0^b ax^3\,dx=\frac{ab^4}{4}=\frac{b^2}{2}=\frac98.
$$
Using the <mean> $\mathbb E[X]=1$, the <variance> is therefore
$$
\boxed{\operatorname{var}(X)=\frac98-1=\frac18.}
$$