# How to prove $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$

How can I prove that $$\int_0^af(x) \; dx=\int_0^a f(a-x) \; dx$$

-

It's pedagogically better to give hints:

1. Change of variables: If $a-x = u$ then $f(a-x) = f(u)$

2. Change of variables: If $a-x = u$ then $dx = -du$

3. Change of variables: If $a-x = u$ then $x =0 \iff u =\ \color{red}{??}$ and $x=a \iff u =\ \color{red}{??}$

4. Interchange of boundary: $\displaystyle\int_a^{b} f(x) dx = - \displaystyle\int_{b}^{a} f(x) dx$

5. Formal variable renaming: $\displaystyle\int_a^{b} f(x) dx = \displaystyle\int_{a}^{b} f(z) dz$

Can you fill in the $\color{red}{??}$ and put it all together?

-
Hhhh great thanks – Rawhi Mar 17 '12 at 22:28

Use change of variables $u=a-x$

-
Apply this change to the integral on the right. You'll see that the change brought by the limits of integration and changing from dx to -du cancel out giving you the integral on the left with u instead of x. – Steven-Owen Mar 17 '12 at 22:11
@Rawhi: Seriously, I guess that this problem can appear only after change of variables was taught, and it is maybe the most basic change of variables. – Beni Bogosel Mar 17 '12 at 22:29
@Rawhi: I was surprised that you asked 'How this could help!!?', that's all. – Beni Bogosel Mar 17 '12 at 22:44

When you think about the function $f(a-x)$ as a reflection of the 'original region' about the line $x=\frac{a}{2}$, then the result becomes rather apparent...

-