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!
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.