Probability density of a function x

Question

Suppose that $z(t)={\int_{0}^{t} {t'}^2dW(t')}$ and $x=az+t$, where $dW(t)$ a Wiener process. I want to calculate the probability density for x. Since the probability density for x determines the mean of a function of x, $f(x)$ , through the relation $$<f(x)> = {\int_{0}^{t} P(x)f(x)dx}$$ and since $x=az+t$ we have $$<f(x)>={\int_{0}^{t}P(z)f(x)dz}={\int_{0}^{t}P(z)f(az+t)dz}$$. With a change in variables i get $$<f(x)>={\int_{0}^{t}P(z)f(az+t)dz}={\frac{1}{a}}{\int_{0}^{t}P(x/a-t/a)f(x)dx}$$ and thus the density for x is $$Q(x)={\frac{1}{a}P({\frac{z}{a}}-{\frac{t}{a}})}$$ Is this correct, and if so, how can i get an exact solution?