0
$\begingroup$

This question already has an answer here:

I don't really know where to start with this one. Can you just ignore the $f(..)$ and deal exclusively with what's inside the brackets?

$\endgroup$

marked as duplicate by N3buchadnezzar, Surb, Daniel Fischer, Aditya Hase, Hakim Dec 12 '14 at 13:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ Make an appropriate substitution in the RHS integral. $\endgroup$ – David Mitra Dec 12 '14 at 13:12
  • $\begingroup$ let $y=a-x$ then usual substitution $\endgroup$ – Santosh Linkha Dec 12 '14 at 13:13
1
$\begingroup$

Hint Substitute $y = a-x$ and remember that $\int_{c}^dg(x)dx = -\int_{d}^c g(x)dx$.

$\endgroup$
1
$\begingroup$

Use the $u$-substitution $u=a-x$, which gives $\mathrm{d}u=\mathrm{d}x$, so we get $$ \displaystyle\int_{x=0}^{x=a} f(x) \, \mathrm{d}x = \displaystyle\int_{u=a}^{u=0} -f(u) \, \mathrm{d}u = \displaystyle\int_{a}^{0} -f(a-x') \, \mathrm{d}x' = \displaystyle\int_{0}^{a} f(a-x) \, \mathrm{d}x. $$

$\endgroup$
1
$\begingroup$

Let $a-x=t$ and $dx=-dt$ so we have $$\int_{0}^{a}f(a-x)dx=-\int_{a}^{0}f(t)dt=\int_{0}^{a}f(t)dt$$

$\endgroup$
0
$\begingroup$

Changing the name of the integration variable on the rhs can help. You want to prove that

$$ \int_0^a f(x)dx = \int_0^a f(a-y) dy $$

From the rhs, the substitution $a-y = x$ gives $$ \int_{a}^0 f(x) (-dx) $$ which is equal to the lhs.

$\endgroup$

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