Let's say we have a continuous random variable $X\_{1}$ with a probability density function $p\_{X\_{1}}$. The expectation is: $$E(X\_1) = \int\_{X\_{1}} dx\_{1} \Big ( x\_{1} \cdot p\_{X\_{1}}(x\_{1}) \Big)$$.

Now say we have a random variable $Y\_{1}$, which is obtained as a transformation $g\_{1}:X\_{1} \rightarrow Y\_{1}$ where $y\_{1} = (x\_{1} - 4)^{2} $. The expectation for $Y\_{1}$ is: $$E(Y\_{1})=\int\_{X\_{1}} dx\_{1} \Big( (x\_{1}-4)^2 \cdot p\_{X\_{1}}(x\_{1}) \Big) $$. 

Why is the expectation not: $$E(Y\_{1})= \int\_{Y\_{1}} dy\_{1} \Big( y\_{1} \cdot p\_{Y\_{1}}(y\_{1}) \Big) =\int\_{X\_{1}} dx\_{1} \Big( (x\_{1}-4)^2 \cdot p\_{Y\_{1}}((x\_{1}-4)^2) \Big)$$? Does it mean $p\_{Y\_{1}}\big((x\_{1}-4)^2\big) = p\_{X\_{1}}(x\_{1}) $ and if so why is that the case?