Prove

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

Any help would be appreciated!

Printable View

- February 6th 2008, 02:14 AMslevvioProof
Prove

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

Any help would be appreciated! - February 6th 2008, 02:38 AMearboth
- February 6th 2008, 03:28 AMmr fantastic
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):

- [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:

LHS

Note: When reversing the integral limits you multiply by a negative. So the resulting double negative becomes a positive. - February 6th 2008, 04:29 AMslevvio
thank you! (Handshake)