Solution
= Solution
The <normalization of a probability distribution> requires
$$
1=\int_0^b ax\,dx=\frac{ab^2}{2},
\qquad a=\frac2{b^2}.
$$
The prescribed <mean> then gives
$$
1=\mathbb E[X]=\int_0^b x\cdot ax\,dx
=\frac{ab^3}{3}=\frac{2b}{3}.
$$
As $a,b>0$, these equations have the unique solution
$$
\boxed{b=\frac32,\qquad a=\frac89.}
$$