The area between the curve 3x^2 - 4x and the lines x=2 and x=1 is the definite integral for the curve between 1 and 2.

Let us determine the integral.

==> F = Int (3x^2 - 4x) dx = Int (3x^2 dx - Int 4x dx

==> F = 3x^3/3 - 4x^2/2 + C

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

Now we will determine F(1) and F(2).

==> F(1) = 1-2 +c = -1 + C

==> F(2) = 8 - 8 + c = c

==> Then the area is given by :

A = F(2) - F(1) = c -(-1+c) = 1

**Then the area between the curve and the lines is 1 square units.**

The area between the curve 3x^2 - 4x and the lines x=1 and x=2 is the definite integral of f(x) = 3x^2 - 4x from x = 1 to x = 2.

Int [ f(x) dx] = Int [ 3x^2 - 4x dx] , x = 1 to x = 2

=> (3x^3/ 3 - 4x^2 / 2) , x = 1 to x = 2

=> (x^3 - 2x^2), x = 1 to x = 2

=> 2^3 - 2*2^2 - 1^3 + 2*1^2

=> 8 - 8 - 1 + 2

=> 1

**The required area is equal to 1.**