Integral Property: $\int^a_{0}f(x)dx=\int^a_{0}f(a-x)dx$ [Proof by definition of Riemann Sums] This link: Why $\int_0^af(x)dx=\int_0^af(a-x)dx$? addresses this question but I do not follow the proofs in the answers: Each proof starts off with variable $x$ but ends the right hand side with a different variable. How does that work? I can't find this discussed anywhere else on the internet.
For example, how do we get from 
$\int_0^a f(x)dx$ to $\int_a^0 f(a-x)(-dx)$?(I apology if I'm overlooking some silly thing but for some reason I am stuck):
Does anyone know the proof or where else I can find it? Also are there any special conditions on it?
$$\int^a_{0}f(x)dx=\int^a_{0}f(a-x)dx$$
 A: Good answers were already given. Let us look some pictures.


*

*$f(-x)$ is the reflection of $f(x)$ about the $y$-axis.



So, from geometric interpretation of integral, we "see" that
$$\int_{0}^af(x)\;dx=\int_{-a}^0f(-x)\;dx\tag{1}$$


*

*$f(a-x)$ is the horizontal translation $a$ units to the right of $f(-x)$ .



So,
$$\int_{-a}^0f(-x)\;dx=\int_{0}^af(a-x)\;dx\tag{2}$$


*

*From $(1)$ and $(2)$ we get


$$\int_0^af(x)\;dx=\int_{0}^af(a-x)\;dx$$
A: Let $u=a-x$, we have $du=-dx$, then
$$\int_{0}^{a}f(a-x)dx=\int_{u=a}^{u=0}-f(u)du=-\int_a^0f(u)du=\int_0^af(u)du$$
Since $\int_0^a f(x)dx=\int_0^a f(u)du$ (independent of the relationship between $x$ and $u$) we are done. This is because the variable of integration is a dummy variable
Intuitively, what's going on is we are reversing the order that we are looking at terms. Integrals are summations, and what this $u$-substitution highlights is the fact that when we look at $f(a-x)$ what we are really doing is changing the order in which we add up the terms.
A: Let $ u = a-x $.  Then $du = -dx$ and when $x = 0$, $u = a$ and when $x=a$, $u=0$.
Then write the second integral:
$$ \int_{x=0}^{x=a} f(a-x) \,dx = -\int_{x=0}^{x=a} f(u) \, du \\
= -\int_{u=a}^{u=0} f(u) \, du = \int_{u=0}^{u=a} f(u) \, du = \int_{x=0}^{x=a} f(x) \, dx
$$
The last step is just using a different symbolfor a "dummy variable".
A: The following perhaps groady argument shows that, using the definition of Riemann integrals, we have:
$$\int_0^af(x)\,dx=\int_0^af(a-x)\,dx$$
If we have a partition of $[0,a]$, say $0, x_1, x_2, \ldots, x_n, a$, then there is a corresponding partition $0, a-x_n, a-x_{n-1}, \ldots a-x_n, a$, also of $[0,a]$.  The upper partial sums for the first integral in the first partition are
$$x_1\max_{0\leq x\leq x_1} f(x)+(x_2-x_1)\max_{x_1\leq x\leq x_2} f(x) +\ldots + (a-x_n)\max_{x_n\leq x\leq a} f(x)$$
The upper partial sums for the second integral in the second partition are
$$(a-x_n)\max_{0\leq x\leq a-x_n} f(a-x)+(x_n-x_{n-1})\max_{a-x_n\leq x\leq a-x_{n-1}} f(a-x) +\ldots \\+ (a-x_1)\max_{a-x_1\leq x\leq a} f(a-x)$$
We have
$$\max_{0\leq x\leq a-x_n} f(a-x)= \max_{x_n\leq x\leq a} f(x)$$
and similar equations for the other intervals making up the partitions.  We thus see that for each upper partial sum for the first integral, there is a corresponding and equal upper partial sum for the second integral.  The same is true for lower partial sums -- just replace max with min.  Thus, because the two Riemann integrals have the same collections of lower and upper partial sums, they must have the same value.
