TITLE: Why $\int_0^af(x)dx=\int_0^af(a-x)dx$?
QUESTION [6 upvotes]: Is it true that: 
\begin{equation*}
\int_0^af(x)dx=\int_0^af(a-x)dx?
\end{equation*}
Because I know that:
\begin{equation*}
\int_0^af(x)\,dx=F(a)-F(0)~\text{and}~\int_0^af(a-x)\,dx=F(a-a)-F(a-0)=F(0)-F(a)
\end{equation*}
So I got: $F(a)-F(0)=F(0)-F(a)$ and that's not necessarily true.

REPLY [5 votes]: You neglected the chain rule: if
$$
\frac d {dx} F(x) = f(x)
$$
then
$$
\frac d {dx} F(a-x) = -f(a-x), \text{ not } f(a-x).
$$

REPLY [4 votes]: You can prove it by taking $y=a-x$ and $dy=-dx$,
$$
\int _0^a f(x)dx = \int _{a}^0f(a-y)(-dy) = \int _0^a f(a-y)dy 
$$
Note:  your second statement in the question is not true.

REPLY [3 votes]: Hint: Take $y = a - x$ then $dy = -dx$ and $$- \int_a^0 f(y) dy = \int_0^a f(y) dy$$

REPLY [3 votes]: Let $u=a-x$, we have $du=-dx$, then
$$\int_{0}^{a}f(a-x)dx=\int_{u=a}^{u=0}-f(u)du=-\int_a^0f(u)du=\int_0^af(u)du$$