Evaluate the definite integral of the function f(x) if f(x)+f(-x)=1 and x=-a to x=a.

Given the data from enunciation, we conclude that the
domain of definition of the function is the closed interval [-a ;
a].

f(x):[-a;a]->R

If f
is a continuous function over the closed interval, then the identity is
true:

Int f(x)dx (x=-a -> x=a) = Int [f(x) +
f(-x)]dx, (x=0 -> x=a)

We'll apply this sentence to
the given function:

Int f(x)dx (x=-a -> x=a) = Int
[f(x) + f(-x)]dx

But, from enunciation, [f(x) + f(-x)] =
1

Int [f(x) + f(-x)]dx = Int 1*dx = x (from x = 0 to x =
a)

We'll apply
Leibniz-Newton:

Int 1*dx = F(a) -
F(0)

Int 1*dx = a - 0

Int 1*dx
= a

Int f(x)dx (x=-a -> x=a) =
a