# Proof

• Feb 6th 2008, 01:14 AM
slevvio
Proof
Prove $\int^{a}_{0} f(x)dx = \int^{a}_{0}f(a-x)dx$

I managed to prove that LHS = F(a) -F(0) and RHS = F(0) - F(a).

Any help would be appreciated!
• Feb 6th 2008, 01:38 AM
earboth
Quote:

Originally Posted by slevvio
Prove $\int^{a}_{0} f(x)dx = \int^{a}_{0}f(a-x)dx$

I managed to prove that LHS = F(a) -F(0) and RHS = F(0) - F(a).

Any help would be appreciated!

Hi,

you have nothing else to do than use the definitions:
LHS:

$\int_0^a f(x)dx = F(a) - F(0)$

RHS:
$\int_0^a f(a-x) dx = F(a-a) - F(a-0) = F(0) - F(a)$
• Feb 6th 2008, 02:28 AM
mr fantastic
Quote:

Originally Posted by earboth
Hi,

you have nothing else to do than use the definitions:
LHS:

$\int_0^a f(x)dx = F(a) - F(0)$

RHS:
$\int_0^a f(a-x) dx = F(a-a) - F(a-0) = F(0) - F(a)$

Not quite, EB.

Unfortunately, your RHS = - LHS ...... a consequence of the small matter of f(a-x) = f(-[x - a]) being got from f(x) by a translation (a units from the y-axis) and a reflection (in y-axis) .... The definite integral is invariant under the translation, but not under the reflection.

Therefore (correction in red):

$\int_0^a f(a-x) dx =$ - [F(a-a) - F(a-0)] = - F(0) + F(a)

= LHS.

Personally, rather than explicitly using transformations, I'd implicitly use them by making the substitution u = a - x. Then:

du = -dx.
x = 0 => u = a.
x = a => u = 0.

Therefore:

$\int_0^a f(a-x) \, dx = \int_a^0 f(u) \, (- du) = - \int_a^0 f(u) \, du = \int_0^a f(u) \, du = F(a) - F(0) = \,$ LHS

Note: When reversing the integral limits you multiply by a negative. So the resulting double negative becomes a positive.
• Feb 6th 2008, 03:29 AM
slevvio
thank you! (Handshake)