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

let F(x) = intg f(x)

The area is :

A = F(1) - F(0)

Let us integrate f(x):

F(x) = intg (3x^2 - 2x + 3)

= intg 3x^2 - intg 2x + intg 3

= 3x^3/3 - 2x^2/2 + 3x +C

= x^3 - x^2 + 3x +C

==> A = F(1) - F(0)

= 1-1 + 3 +c - C

= 3

Then the area is:

**A = 3 square units.**

To find the area under f(x) = 3x^2-2x+3 between x= 0 and x=1.

The area under f(x) betweeb x= a and x = b is Int f(x)dx from x= a to x= b.

So the area under 3x^2-2x+3 from x = 0 to x=1 is:

Int(3x^2-2x+3)dx from x= 0 to x =1. F(x) from x= 0 to x = 1

F(x) =3x^3/3 -2x^2/2 +3x = x^3 - x^2 +3x

F(1) -F(0) = (1-1+3)-0 = 3.

The limits of the area are incomplete. Supposing that the x axis is the other limit, we'll calculate the area using Leibniz Newton formula:

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

We'll aclculate the definite integral for the function f(x):

Int f(x)dx = 3Int x^2dx -2Int xdx +3Int dx

Int f(x)dx = 3x^3/3 - 2x^2/2 + 3x

We'll simplify and we'll get:

Int f(x)dx = x^3 - x^2 + 3x

F(1) - F(0) = (1^3 - 0^3) - (1^2 - 0^2) + 3(1-0)

F(1) - F(0) = 1 - 1 + 3

F(1) - F(0) = 3

**So, the area located between the curve of f(x), the x axis and x = 0 to x = 1 is 3.**