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

*print*Print*list*Cite

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

Put y = -x. Then dx = -dy.

So {Int f(x) dx from x= -a to x= a}= {Intf(y)(-dy) from y = a to y = -a.} = - {Int f(y)dy from y = a to y = -a.} = Int f(x) dx from x= -a to x= a.

Therefore {Int f(x) dx from -a to a} = { Int f(-x)dx from x= -a to a} = I.

Now consider Int Int{(x)+f(-x)}dx = {Int 1 dx from x= -a to a}

2I = (a- (-a)) = 2a.

So. I = a.

Therefore Integral f(x) dx = 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**