Solution
= Solution
{c}
Normalization of the <probability density function> and the specified <mean> give two equations:
$$
1=\int_0^b ax\,dx=\frac{ab^2}{2},\qquad 1=\mathbb E[X]=\int_0^b ax^2\,dx=\frac{ab^3}{3}.
$$
Dividing the second by the first gives $2b/3=1$. Consequently
$$
\boxed{b=\frac32,\qquad a=\frac{2}{b^2}=\frac89.}
$$
Both are positive, as required for the stated <probability density function>.