Change of variables with a square Can someone help me understand this a bit better:
$\int (x-y)^2 dx = \int(y-x)^2dx$
as $(y-x)^2 = (x-y)^2$. Now, if I make the change  $z = x-y$ in the one on the LHS I get:
$\int z^2 dz$ 
as $dz = dx$. Now, if I make the change $z=y-x$ in the one on the RHS I get:
$\int z^2 (-dz)$ 
as $dz = -dx$. This implies that
$\int z^2 dz = - \int z^2 dz = 0$
which is clearly not true. I don't understand - can someone help me understand changing the variable of integration when you have a square or a square root please?
 A: In the first case, $z=x-y$. In the second case $z=y-x$. It doesn't follow that $\int z^2 dz = - \int z^2 dz = 0$, because these are two different $z$'s. The confusion comes from the fact that you used the same variable name for two different values. Look at it again, but with a different variable for the second case:
$$z=x-y$$
$$u=y-x$$
$$\begin{align}
\int{z^2}dz&=-\int{u^2}du \\
\frac{1}{3}z^3+C &= -\frac{1}{3}u^3+C
\end{align}$$
This follows because $z=-u$.
A: You are using $z$ on both sides of the equation as if it contains the same value, when in fact it does not.
On the LHS, $z = x-y$, and on the RHS, $z = y-x$.  These are two different z values.  To clarify, you should use $z1 = x-y$ and $z2 = y-x$, which makes it obvious that $z1 = -z2$.
A: With a definite integral you get
$$\int_a^b (x-y)^2\,\mathrm dx=\int_{a-y}^{b-y}z^2\,\mathrm dz $$
and
$$\int_a^b (y-x)^2\,\mathrm dx=\int_{y-a}^{y-b}z^2\,\mathrm d(-z) =\int_{y-b}^{y-a}z^2\,\mathrm dz$$
which both give the same value $\frac13(b-y)^3-\frac13(a-y)^3$.
As you also tagged the question as measure-theory (presumably Lebesgue integral), note that in that case you need to take the absolute value of the derivative in substitution, i.e. again you get
$$\int_A z^2\,\mathrm dz=\int_{-A}z^2\,\mathrm dz$$
