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

Let F(x) = integral f(x)

Then the area between f(x) and x= 0 and x=1 is:

A = F(1) - F(0)

Let us find F(x):

F(x) = intg f(x)

= intg 3x^2 - 2x + 1 dx

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

= x^3 - x^2 + x

==> F(1) = 1-1 + 1 = 1

==> F(0) = 0

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

= 1- 0= 1

**Then the area = 1 square unit.**

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

Solution:

The area A under the curve f(x) between x= a and x = b is given by the definite integral:

A = integral f(x) dx between x = a and x = b.

Let F(x) = integral f(x) dx.

Then the integral f(x) dx between x= a and x= b is F(b) = F(a).

F(x) = Integral {3x^2-2x+1}dx

F(x) = Integral (3x^2) dx - 2integral (2x)dx+ Integral dx

F(x) = 3x^3/3 -2x^2/2 +x +C , as Integral X^n dx = (1/(n+1))x^(n+1).

F(x) = x^3-x^2+x+C.

F(1) - F(0) = {1^3-1^2+1+C}-(0^3-0^2+0+C}

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

Therefore the area between the curve , x axis and x=0 and x= 1 is 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.**