Prove $\displaystyle \int^{a}_{0} f(x)dx = \int^{a}_{0}f(a-x)dx$

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

Any help would be appreciated!

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

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 AMearboth
- Feb 6th 2008, 02:28 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):

$\displaystyle \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:

$\displaystyle \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 AMslevvio
thank you! (Handshake)