Show that `int_0^a f(x) g(x) d x=2 int_0^a f(x) d x`, if `f` and `g` are defined as `f(x)=f(a-x)` and `g(x)+g(a-x)=4`

Explanation

`|x-1|` = {(x-1), x ge 1
`-(x-1), x lt 1`}
therefore `int_0^4 |x-1| dx`
=`int_0^1 (1-x) dx + int_1^4 (x-1) dx`
=`(x-x^2/2)_0^1 + (x^2/2 -x)_1^4`
`(1-1/2) +((16)/2 -4)-(1/2 -1)`
`=1/2+4+1/2 = 5.`