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

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

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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?

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Hhhh great thanks –  Rawhi Mar 17 '12 at 22:28
Use change of variables $u=a-x$