Understanding the proof of why E[t|x] = y(x,w) (linear regression) In "Pattern recognition and Machine learning" (Bishop) the following is stated:

assume the target value $t$ is given by a deterministic function $y(x,w)$ plus white noise $\epsilon$ i.e
$t=y(x,w) + \epsilon.\tag{1}$
When $\displaystyle p(t|x,w,\beta) = N(t|y(x,w),\beta^{-1})$ then
$\displaystyle E[t|x] = \int tp(t|x)dt = y(x,w) \tag{2}$

where $N$ indicates the normal-distribution and $E$ the expected value.
I struggle to see why the integral just leaves us with $y$.
One of my thougts was just to subsitute $t$ from (1) into (2) such that
$\displaystyle E[t|x] = \int y(x,w)p(t|x)dt = y(x,w)\int p(t|x) dt = y(x,w)\cdot1 = y(x,w) \tag{3}$
but I feel like that's not correct since we would then just integrate with respect to $y$ instead of $t$ (when swapping those two) i.e the integrations would be
$\displaystyle \int y(x,w)p(y(x,w)|x)dy(x,w) \tag{4}$
thus we can't drag it outside the integration

$E[t|x]=\int{tp(t|x)dt}$
This above notation simply gives the expected value or mean of $t$ subject to the probability distribution $p(t|x)$; just the definition of "expectation" being stated.
Now $p(t|x)$ is a normal distribution with mean $y(x,w)$ and precision $\beta^{-1}$. Hence $E[t|x]=\int{tp(t|x)dt}=y(x,w)$ i.e., mean of the normal distribution $p(t|x)$.
