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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?

$$int^a_{0}f(x)dx=int^a_{0}f(a-x)dx$$

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
$$x_1max_{0leq xleq x_1} f(x)+(x_2-x_1)max_{x_1leq xleq x_2} f(x) +ldots + (a-x_n)max_{x_nleq xleq a} f(x)$$
The upper partial sums for the second integral in the second partition are
$$(a-x_n)max_{0leq xleq a-x_n} f(a-x)+(x_n-x_{n-1})max_{a-x_nleq xleq a-x_{n-1}} f(a-x) +ldots \+ (a-x_1)max_{a-x_1leq xleq a} f(a-x)$$
We have
$$max_{0leq xleq a-x_n} f(a-x)= max_{x_nleq xleq a} f(x)$$
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.

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
$$int_{0}^af(x);dx=int_{-a}^0f(-x);dxtag{1}$$

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

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

• From $(1)$ and $(2)$ we get

$$int_0^af(x);dx=int_{0}^af(a-x);dx$$

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.

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".