y= 4x + 2

x = 2 and x = 3

Let f(x) = 4x+ 2

Then letr F(x) = integral f(x)

Then we know that the area between f(x) and x = 2 and x= 3 is:

A = F(3) - F(2)

Let us integrate f(x):

F (x) = intg (4x+2 ) dx

= 4x^2/2 + 2x

= 2x^2 + 2x

= 2(x^2 + 1)

Now we will find F(3) and F(2):

F(3) = 2( 3^2 + 1) = 2*10 = 20

F(2) = 2(2^2 + 1) = 2*5 = 10

==> A = F(3) - F(2)

= 20 -10 = 10

**Then the area between y , x =2 and x = 3 is 10 square units**

To find the area between y= 4x+2 and the ordinates at x=2 and x= 3.

We know that y=4x+2 at x= 2 has the y value y = 4*2+2 = 10.

Similarly for x=3, the y value of y= 4x+2 is y = 4*3+2 = 14.

Therefore the area under y= 4x+2 and x axis and the ordinates at x=2 and x= 3 is the same as the trapezium (2, 10), (2,0) , (3,0) and (3,14).

So the parallel sides (2,0) To (2,10) and (3,0) to(3,14) are 10 and 14 and the distance between the parallel sides = (3-2) = 1.

Therefore area of the trapezium = (1/2)(sum of the || sides)(distance between the parallel sides) = (1/2)(10+14)*1 = 12.

To calculate the area located under the given curve and between the lines, we'll determine the definite integral:

Int (4x + 2)dx

We'll apply the additivity of integral:

Int (4x + 2)dx = Int (4x)dx + Int (2)dx

Int (4x)dx + Int (2)dx = 4*x^2/2 + 2x

4*x^2/2 + 2x = 2x(x + 1)

F(x) = 2x(x + 1)

We'll apply Leibniz-Newton formula:

Int (4x + 2)dx = F(3) - F(2)

F(3) = 6(3 + 1)

F(3) = 24

F(2) = 12

**The area under the curve and between the lines is:**

F(3) - F(2) = 24 - 12

**F(3) - F(2) = 12 square units**