# What's wrong in this integration equation?

does anyone know what's wrong in this expression?

$$\int_0^a f(a - x) \; \mathrm d x = \int f(0) \; \mathrm dx - \int f(a) \;\mathrm dx = \int_a^0 f(x) \; \mathrm dx$$

I tested values and it's indeed wrong. It's supposed to have a minus sign, but I don't see which step has error.. :(

did you mean $\int_0^a f'(a-x)dx = f(0)-f(a) = -\int_0^a f'(x) dx$ or something like this? – Guest 86 Feb 9 '13 at 23:42
anything else you forgot to mention about $f$? why do you think these equalities hold? – Guest 86 Feb 9 '13 at 23:44
No, the middle part is nonsense. If you set $u=a-x$, then $du=-dx$. Also, when $x=0$, you have $u=a$; and when $x=a$, you have $x=0$. After substitution, the integral on the left becomes $-\int_a^0 f(u)\,du$; which is the negative of the integral on the right. – David Mitra Feb 9 '13 at 23:50
OK, let's take it one step at a time: \begin{align*} \int_0^a f(a - x) d x &= - \int_a^0 f(y) d y \qquad \text{by $$y = a - x$$, $$d y = - dx$$} \\ &= \int_0^a f(y) d y \qquad \text{turn around limits} \end{align*} Your derivation works if $f(a - x) = f(a) - f(x)$, and that isn't always true.