Given the function f(x) and the identity 3f(x)-f(-x)=x+2, determine the area bounded by f(x), x and y axis and the line x=1?

Given the identity 3f(x)-f(-x)=x+2, that is verified for any
real value of x, then we can replace x by -x and we'll get an identity,
also.

3f(-x) - f(x) = -x + 2

We'll
create a system formed from the identities above:

-f(-x) + 3f(x) = x
+ 2 (1)

3f(-x) - f(x) = -x + 2
(2)

We'll eliminate f(-x) by multiplying (1) by 3 and adding the
expression resulted to (2):

-3f(-x) + 9f(x) + 3f(-x) - f(x) = 3x + 6
- x + 2

We'll combine like terms:

8f(x)
= 2x + 8

f(x) = x/4 + 1

Since we know
the expression of the function f(x), we can evaluate the area under the curve f(x) and bounded by
x and y axis and the line x = 1.

We'll calculate the definite
integral of f(x), whose limits of integration are x = 0 and x =
1.

Int f(x)dx = Int (x/4 + 1)dx

Int
(x/4 + 1)dx = (1/4)*Int xdx + Int dx

Int f(x)dx = x^2/8 +
x

We'll apply Leibniz Newton:

Int
f(x)dx = F(1) - F(0)

F(1) - F(0) = 1/8 + 1 - 0/8 -
0

F(1) - F(0) = 9/8

The
area bounded by the curve f(x) = x/4 + 1, the x lines and the limits x = 0 to x =1, is A = Int
f(x)dx = 9/8 square units.