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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.. :(

Thank you in advance!

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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
    
@Guest86 nop, there's no derivative –  user1561559 Feb 9 '13 at 23:43
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anything else you forgot to mention about $f$? why do you think these equalities hold? –  Guest 86 Feb 9 '13 at 23:44
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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
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In the middle you have two indefinite integrals of constants, while on the ends you have definite integrals. Huh? –  Ross Millikan Feb 9 '13 at 23:56

1 Answer 1

up vote 3 down vote accepted

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.

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