You need to evaluate the area of the region between the given curves, using definite integral, such that:

`int_0^1 (3x^2 - 2x +1) dx`

Using the property of linearity of integral yields:

`int_0^1 (3x^2 - 2x +1) dx = int_0^1 (3x^2) dx - int_0^1 (2x) dx + int_0^1 dx`

`int_0^1 (3x^2 - 2x +1) dx = 3*x^3/3|_0^1 - 2*x^2/2|_0^1 + x|_0^1`

`int_0^1 (3x^2 - 2x +1) dx = (x^3 - x^2 + x)|_0^1`

Using the fundamental theorem of calculus yields:

`int_0^1 (3x^2 - 2x +1) dx = (1^3 - 1^2 + 1 - 0^3 + 0^2 - 0)`

`int_0^1 (3x^2 - 2x +1) dx = 1`

**Hence, evaluating area of the region between the given curves, as an applications of definite integrals, yields `int_0^1 (3x^2 - 2x +1) dx = 1` .**

To compute the area between the given curve and lines, we'll apply Leibniz Newton formula:

Int f(x) dx = F(a) - F(b)

First, we'll compute the indefinite integral of f(x):

Int (3x^2 - 2x +1)dx

We'll apply the property of integral to be additive and we'll get:

Int (3x^2 - 2x +1)dx = Int 3x^2dx - 2Int xdx + Int dx

Int (3x^2 - 2x +1)dx = 3*x^3/3 - 2*x^2/2 + x

We'll simplify and we'll get:

Int (3x^2 - 2x +1)dx = x^3 - x^2 + x

We'll apply Leibniz Newton formula for b = 1 and a = 0.

F(b) = F(1) = 1 - 1 + 1 = 1

F(a) = F(0) = 0 - 0 + 0 = 0

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

Int (3x^2 - 2x +1)dx = 1 - 0

Int (3x^2 - 2x +1)dx = 1

The area located under the curve and between the line is A = 1 square unit.