Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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 – John Lee Feb 9 '13 at 23:43
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
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
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.